# Writing an Expression for the Volume of a Spherical Shell

The question is as follows:

Write an expression for the volume of the spherical shell formed between two concentric spheres, the inner one of radius $r$, the outer one of radius $R$. Factor your answer so that it has the form $4 π · (trinomial) · (binomial)$. In this situation, what is the meaning of the binomial? What can be said about the value of the trinomial when the binomial has a very small value? Make a conjecture concerning the surface area of a sphere of radius $R$.

I know that the volume of the sphere with the inner radius of r would be $\frac{4}{3}\pi r^3$ and the volume of the sphere with the outer radius of R would be $\frac{4}{3}\pi R^3$. Therefore, the volume of the spherical shell formed between the concentric circles would be the difference between the outer sphere and the inner sphere--which, when mathematically written, would be $\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3$. That expression, after it's factored, would be $\frac{4}{3}\pi(R^3 - r^3)$. My question is how can I represent $(R^3 - r^3)$ the product of a trinomial and binomial. Any help will be greatly appreciated!

EDIT: I think that I may have found the answer to the question that I have presented above: $(R^2 + Rr + r^2)(R - r)$. Therefore, my complete answer to the volume of the spherical shell is $\frac{4}{3}(R^2 + Rr + r^2)(R - r)$.

I still don't know what the binomial is representing for this particular problem, nor its relation to the trinomial when the binomial value gets smaller. I think that I can use the help for understanding what the binomial and trinomial represents as a way to formulate a conjecture regarding the surface area of a sphere with radius R.

EDIT #2: The binomial, as I understand it, represents the thickness of the spherical shell formed by the concentric circles.

• Since this is the kind of thing you have to just know, I don't mind telling you flat-out that one of the terms is $R-r$. You can find the other term by long division. – dfan Jan 27 '18 at 18:12
• Factorize as $a^3-b^3=(a-b)(a^2+b^2+ab)$. – MalayTheDynamo Jan 27 '18 at 18:13
• Thank you @MalayTheDynamo and @dfan! I have found the answer right after I posted the problem. However, would anyone of you help in guiding me in the questions that I posted in the third paragraph? – geo_freak Jan 27 '18 at 18:17

Note that $$\frac{4}{3} \pi (R^2 + Rr + r^2)(R - r)$$

has three parts.

You have already figured out the $(R-r)$ part as the thickness of the spherical shell.

The trinomial part is very close to $3r^2$ in case that the thickness is very small.

Now when you multiply this $3r^2$ by $\frac{4}{3} \pi$, you come up with $$4\pi r^2$$ which is the surface area of the shell( Approximate of course)

Thus The volume of the shell is approximately, the surface area of the shell multiplied by the thickness of the shell.

• How did you get that the trinomial part will be very close to $3r^2$? – geo_freak Jan 27 '18 at 18:35
• Writing $r$ in place of $R$ as these are near to each othert – Narasimham Jan 27 '18 at 18:45
• Later you can also find that surface area is $4 \pi r^2$ as differential coefft. of $4/3 \pi r^3$ with respect to $r$ – Narasimham Jan 27 '18 at 18:47
• I see! Thank you so much for clarifying @Narasimham! – geo_freak Jan 27 '18 at 18:51
• Thank you for your attention. – Mohammad Riazi-Kermani Jan 27 '18 at 18:53

Let $V(r)$ be the volume of the sphere with radius $r$ and $A(r)$ its surface area.We have $$V(r) = V(1)\cdot r^3$$ since concentric spheres are homothetic. In particular $$V(R)-V(r) = V(1)(R-r)(R^2+Rr+r^2).$$ On the other hand $A(r) = A(1)\cdot r^2$ and $$A(R) = \lim_{r\to R^-}\frac{V(R)-V(r)}{R-r} = \lim_{r\to R^-} V(1)(R^2+Rr+r^2)=3V(1)\,R^2.$$ This proves $A(1)=3\,V(1)$ and the same argument in dimension $n$ proves $A(1)=n\,V(1)$: to find the volume or the surface area of a Euclidean ball are equivalent problems. Remark: for $n=3$, the same can be shown by approximating a sphere with the union of a large number of cones and recalling that the volume of a cone is given by one third of the product between the height and the base area.

• Thank you for the answer that you have posted. I, unfortunately, am having a hard time understanding it since I am just taking Alg II/Trig for my maths class. Is it possible to explain it in more simpler terms? – geo_freak Jan 27 '18 at 18:28
• @geo_freak: $A(R)$ is the derivative of $V(R)$ with respect to $R$ by elementary geometric considerations. In particular the ratio between $A(1)$ and $V(1)$ exactly equals the dimension of the space. – Jack D'Aurizio Jan 27 '18 at 18:33

Write an expression for the volume of the spherical shell formed between two concentric spheres, the inner one of radius $r$, the outer one of radius $R$. Factor your answer so that it has the form $4 \pi \cdot (\text{trinomial}) \cdot (\text{binomial})$

$$V(R, r) = \frac{4}{3}\pi (R^3-r^3) = \frac{4}{3}\pi(R-r)(R^2+r^2 + Rr) = 4 \pi \, t(R,r) \, b(R,r)$$ with \begin{align} t(R,r) &= \frac{1}{3}(R^2+r^2 + Rr) \\ b(R, r) &= R-r \end{align}

What can be said about the value of the trinomial when the binomial has a very small value?

In this case we have $r \approx R$ and we can estimate the volume by $$V(R,r) \approx A(r) \, b(R,r)$$ where $A(r)$ is the area of a sphere with radius $r$. So we get $$t(R,r) = \frac{1}{3}(R^2+r^2 + Rr) = \frac{V(R,r)}{4\pi b(R,r)} \approx \frac{A(r) \, b(R,r)}{4\pi b(R,r)} = \frac{A(r)}{4\pi}$$

Make a conjecture concerning the surface area of a sphere of radius R.

$$A(t) \approx 4\pi\, t(R,r) = 4\pi \, \frac{1}{3}(R^2 + r^2 + Rr) \approx 4\pi \, \frac{1}{3}(r^2 + r^2 + r^2) = 4 \pi r^2$$


# Computing determinant of this matrix

I have a very specific kind of matrix and I have to find the formula to find the determinant of these matrix.

a(i,j)=a if(i==j) and a(i,j)=0 if(floor(i/2)=floor(j/2) and i!=j) and n is odd

$$A=(a_{i,j})_{n \times n}=\left( \begin{array}{ccccc} a&0&b&b \cdots &b&b\\ 0& a&b&b \cdots& b&b\\ b& b& a&0 \cdots& b&b\\ b& b& 0&a \cdots& b&b\\ \vdots& \vdots& \vdots& \ddots&\cdots\\ b&b&b & \cdots&b&a \end{array} \right)?$$

-
Try writing down its eigenvectors. – Qiaochu Yuan May 7 '13 at 9:50
What's the rule for the placement of the $0$'s ? – Raskolnikov May 7 '13 at 9:54
a(i,j)=a if(i==j) and a(i,j)=0 if(i/2=j/2 and i!=j) – Alen May 7 '13 at 9:59
Please notice n is odd – Alen May 7 '13 at 10:01

The problem is closely related to this one posted and answered yesterday, with $p\leftarrow2$, $np\leftarrow n-1$, $c\leftarrow a/b$, with the entire matrix divided by $b$.

Since according to a comment under the question $n$ is odd, we need to deal with the last component separately. All eigenvectors in the linked question except for the one filled with $1$s sum to $0$. Thus we can append a $0$ to them to obtain eigenvectors of the present matrix. That leaves a two-dimensional subspace to be dealt with, spanned by the vector $x$ that has a $1$ in the last component and the vector $y$ that has $1$s everywhere else. Applying $A$ to these vectors yields $Ax=ax+by$ and $Ay=(n-1)bx+((n-3)b+a)y$. Thus the product of the remaining two eigenvalues is

$$\left|\matrix{a&b\\(n-1)b&(n-3)b+a}\right|=((n-3)b+a)a-(n-1)b^2\;.$$

Multiplying this by the $n-2$ eigenvalues from the linked question, with the above substitutions and the factor $b^{n-2}$ that was divided out, yields the determinant of the present matrix:

\begin{align} \det A &= \left(((n-3)b+a)a-(n-1)b^2\right)b^{n-2}\left(\frac ab\right)^{(n-1)/2}\left(\frac ab-2\right)^{(n-3)/2} \\ &= \left(((n-3)b+a)a-(n-1)b^2\right)a^{(n-1)/2}\left(a-2b\right)^{(n-3)/2}\;. \end{align}

-
It sounds pretty close but $n$ is odd... – achille hui May 7 '13 at 11:38
@achille: Aha -- thanks for pointing that out -- yet another reminder why it's a bad idea to bury clarifications of the question deep in the comments... – joriki May 7 '13 at 11:42
Can this matrix have a particular formula? – Alen May 7 '13 at 11:58
@Alen: How do you mean, can it? – joriki May 7 '13 at 12:24
Means, can I derive a formula for calculating the det A? – Alen May 7 '13 at 12:33

Since $n$ is odd, let $n = 2m+1$ and $u_n$ be the $n \times 1$ column vector with all entries one.

The matrix $A$ can be rewritten to have the form $X + b u_n u_n^{T}$ where $X$ is a block diagonal matrix consists of $m$ $2\times 2$ block $\tilde{X}$ and one $1 \times 1$ block:

$$X = \begin{pmatrix} \tilde{X} & 0 & 0 & \dots & 0\\ 0 & \tilde{X} & 0 &\dots & 0\\ 0 & 0 & \tilde{X} &\dots & 0\\ \vdots &\vdots & \vdots & \ddots & 0\\ 0 & 0 & 0 & 0 & a - b \end{pmatrix} \quad\text{ and }\quad \tilde{X} = \begin{pmatrix}a - b & -b\\-b & a-b\end{pmatrix}$$

When one perturb an invertible matrix $X$ by a rank one matrix like $b u_n u_n^T$, it is known that the determinant of the updated matrix $X + b u_n u_n^{T}$ is given by:

$$\det( X + b u_n u_n^{T} ) = \det(X)( 1 + b u_n^{T} X^{-1} u_n )\tag{*}$$

Substitute our special form of $X$ into $(*)$ and notice:

\begin{align} \det(X) &= \det(\tilde{X})^m (a - b) = ((a-b)^2 - b^2)^m (a - b)\\ u_n^{T} X^{-1} u_n &= m\;u_2^T \tilde{X}^{-1} u_2 + \frac{1}{a-b} = \frac{2m}{a-2b} + \frac{1}{a-b} \end{align}

We get: \begin{align} \det A &= ((a-b)^2 - b^2)^m (a - b)\left\{1 + b \left(\frac{2m}{a-2b} + \frac{1}{a-b} \right)\right\}\\ &= a^m (a-2b)^{m-1}\left(a^2 + 2 (m-1)ab - 2mb^2\right) \end{align}

Update

For the general case when the matrix elements is given by:

$$a_{i,j} = \begin{cases} a, & \text{ for } i = j\\ 0, & \text{ for } \lfloor\frac{i}{q}\rfloor = \lfloor\frac{j}{q}\rfloor \wedge i \neq j\\ b, & \text{ otherwise }. \end{cases}$$

where $0 \le i, j < n$ and $q$ is an integer $> 1$. Let $n = mq + r$ where $0 \le r < q$. Once again, we rewrite $A$ into the form $X + b u_n u_n^T$ where X is now a block diagonal matrix consists of $m$ $q \times q$ block $\tilde{X}_q$ and one $r \times r$ block $\tilde{X}_r$.

It is not hard to verify the entries on $\tilde{X}_s$ is given by:

$$(\tilde{X}_s)_{i,j} = \begin{cases} a-b, &\text{ for } i = j\\-b, &\text{ otherwise. }\end{cases} \quad\implies\quad \begin{cases}\det(\tilde{X}_s) &= a^{s-1}(a - sb)\\ u_s^{T} \tilde{X}_s^{-1} u_s &= \frac{s}{a-sb}\end{cases}$$ Repeating the same procedure as above, we obtain the formula for $\det(A)$ in the general case:

$$\det A = (a^{q-1}(a - qb))^m (a^{r-1}(a-rb)) \left\{1 + b \left(\frac{mq}{a-qb} + \frac{r}{a-rb}\right)\right\}$$

-
hue:r can be -1 here? – Alen May 7 '13 at 16:44
@Alen that is a typo, should be $0 \le r < q$. In principle, one should treat $r = 0$ as a special case. However, it turns out the final expression also work for the $r = 0$ case. – achille hui May 7 '13 at 16:53
I was actually thinking of (pn-1)*(pn-1) dimension matrix – Alen May 7 '13 at 17:12
@Alen, your $pn-1$ case is equivalent to setting the parameters $(m,q,r)$ to $(n-1,p,p-1)$ in my answer. – achille hui May 7 '13 at 17:24


# How to compute this determinant as quickly as possible (without using any software or calculator)?

I sat an Algebra test yesterday, which consisted of 30 questions and a total time of 45 minutes (an average of 1 min 30 secs per question). One question of the test was this:

Given the matrix: $$A=\begin{bmatrix} -2 & 4 & 2 & 1 \\ 4 & 2 & 1 & -2 \\ 2 & 1 & -2 & 4 \\ 1 & -2 & 4 & 2 \end{bmatrix}$$ Which of the following options is correct?

(A) $A^{-1}=\dfrac{A}{25}$

(B) $A^{-1}=\dfrac{A}{5}$

(C) $A^{-1}=\dfrac{A}{15}$

(D) It has no inverse.

I do know how to compute the inverse of a matrix. For example, using the Gauss-Jordan elimination method. Or for example, using this formula: $$A^{-1}=\dfrac{\text{Adj}(A^T)}{\text{det}(A)}$$

I calculated the determinant and it is $625$. However, this won't help me pick the correct option (of course I can eliminate option D, which is false).

How in the world am I supposed to solve this problem in around 90-100 seconds, without using a calculator? Is there any shortcut or trick or something I missed? 90 seconds was the average time per question in the test. Given how little time I was given to solve the problem, this leads me to think that the structure of A could simplify the answer.

• Strategy: take the RHS's and multiply by $A$ to see if you get $I$. – Adrian Keister Jun 19 '18 at 14:57
• The short answer is $(-2)^2 + 4^2 + 2^2 + 1^2 = 25$. (For reasons given below.) – Mateen Ulhaq Jun 20 '18 at 0:40
• You're assuming that because you have, on average, 90 seconds to solve a problem that each problem must be solved in 90 seconds. Some will undoubtedly take less than 90 seconds. Some will undoubtedly take more. Do your best. – Bob Jarvis - Reinstate Monica Jun 20 '18 at 16:15

If $A^{-1}=\dfrac{A}{25}$ then $A^2=25I$. Similar for other options. Now you need to calculate only one diagonal element of $A^2$ to find the correct option.

• If $A$ has no inverse, there will be no (nonzero) solution $\lambda$ to $A^2 = \lambda I$, so computing $A^2$ gives this information, too. – Travis Willse Jun 19 '18 at 15:01
• Indeed it will but it would require to compute more than just one diagonal element of $A^2$ (the answer states that "you need to calculate only one diagonal element of $A^2$ to find the correct option."). Sorry for being so petty :). – yakobyd Jun 19 '18 at 15:11
• @alephzero: To be specific, if you calculate one diagonal element and it's 25, that doesn't tell you whether $A^2=25I$ or $A^2=\begin{bmatrix}25 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}$ ... – psmears Jun 19 '18 at 21:23
• For checking if $A^{-1}$ exists, this matrix is very amenable to the trick of computing the determinant modulo 2 – jld Jun 19 '18 at 22:47
• @JoseLopezGarcia for an integer-valued matrix, the determinant is non-zero if the determinant with every entry taken modulo 2 is also non-zero. So in your case $$A \mod 2=\begin{bmatrix}0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}$$and this matrix is clearly full rank and so has non-zero determinant (it will certainly be $\pm 1$ but the actual value isn't important so you can stop as soon as you recognize it's not singular). – jld Jun 20 '18 at 13:54

We can also solve this problem quickly without using the multiple choice options as hints to cut short any computations. So, suppose that we're given the matrix $A$ and asked to compute its inverse, without being told that the inverse is proportional to $A$ itself. Then observe that the columns of the matrix are orthogonal, so the matrix is an orthogonal matrix up to a normalization factor. The norm of the columns is 5. The inverse of $A/5$ is therefore equal to its transpose, and the transpose is easily seen to be equal to the matrix itself. So, the inverse of $A$ is $A/25$.

The matrix has the form $A=\begin{bmatrix} C & D \\ D & C \end {bmatrix}$

For this kind of block matrices,

$det (A) = det( C.C -D.C^{-1}.D.C) =$

$= det (C.C + D.D ) = det \left ( \begin{bmatrix} 20 & 0 \\ 0 & 20 \end {bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end {bmatrix} \right ) = 25^2$

since $C.D = -D.C$

$A^2=\begin{bmatrix} C & D \\ D & C \end {bmatrix} \begin{bmatrix} C & D \\ D & C \end {bmatrix} = \begin{bmatrix} C^2 + D^2 & 0 \\ 0 & C^2 + D^2 \end {bmatrix} = \begin{bmatrix} 25 & 0 & 0 & 0 \\ 0 & 25 & 0 & 0 \\ 0 & 0 & 25 & 0 \\ 0 & 0 & 0 & 25 \end{bmatrix}$

You're there with the computation of the determinant. The $\det A^2 = 5^8$, so the answer must be (A) in order for $\det \left(A\cdot A^{-1}\right) = 1$

But PJK's answer provides a more efficient approach.

NB you don't need to compute $A^{-1}$ in order to solve this problem, you just need to determine which option is correct---presumably the question also tests your recognition of this fact.

If $A$ is invertible, then multiplying both sides of each of the first three options, $A^{-1} = \frac{A}{\lambda}$, gives an equivalent equation $$A^2 = \lambda I .$$ If $A$ is not invertible, then there is no (nonzero) $\lambda$ for which the equation holds. So, to determine the answer, it's enough to compute $A^2$, which is computationally much cheaper than applying G.-J. elimination.

EDIT: I realized after writing that the essence of my answer is already in the answer provided by Count Iblis. Perhaps one can consider this answer as containing supplementary material to his.

We will make use of the observation that in the given options, multiplying $A$ on both sides gives the result $\frac{1}{c}I$ for some constant $c$.

First, notice that $A$ is a circulant real symmetric matrix. Because it is a circulant matrix, all its rows have the same norm of 5.

We can therefore write $A = 5 B$ where B is also real symmetric and each row (and column) of B has unit norm. This gives us the following

$A^2 = 25B^2 = 25 B^TB$.

At this point, this should provide enough hint to inspect the matrix $A$ more carefully and you will notice that it is actually orthogonal. Therefore $B$ is orthonormal so $B^TB = I$ and we have $A^2 = 25I$ as required.

If the orthogonality of $A$ still eludes you, we can use the spectral theorem and write

$A^2 = 25B^TB = 25 U\Lambda^2U^T = 25(\Lambda^{\frac{1}{2}}U^T)^T(\Lambda^{\frac{1}{2}}U^T) \implies B = \Lambda^{\frac{1}{2}}U^T$.

Since $U^T$ is orthonormal and each row and column of $B$ has unit norm, this means $\Lambda^{\frac{1}{2}} = I$ and $B^2 = B^TB = I$ as required.

Also, notice that $\Lambda = I$ so all the eigenvalues of $A$ are non-zero (they are all 1) so $A$ is invertible and this excludes the last option.

## Determining the determinant

Now that we know $B$ is orthonormal, we can obtain the determinant of $A$ easily using the property $\text{det}(cM) = c^n \text{det}(M)$ for $M \in \mathbb{R}^{n \times n}$ and det($Q$) = 1 if $Q$ is a square orthonormal matrix

$A = 5B \implies \text{det}(A) = 5^4 \text{det}(B) = 625$

• nice answer: the matrix in fact is symmetric and orthogonal (+1) – G Cab Jun 20 '18 at 15:03

You write

I calculated the determinant and it is 625

which is enough to conclusively answer.

For option $\mathbf A$ we have $\frac{1}{det(A)} = det(A^{-1}) {\space\overset{\mathbf A}{=}\space} det(\frac{A}{25}) = \frac{det(A)}{25^4}$. Similarly for the other answers.

Here is a very simple although not as elegant answer:

Whatever the inverse would be, multiplication must be $1$ for diagonal elements, in particular for the first element. We can call it $c_{11}$.

This shields $\frac{25}{x}$ where $x$ is the one between options that makes this element 1. So it's a or d. In any case, you need to calculate all elements for choosing between this two options.


# Math Help - graphing function

1. ## graphing function

i am having trouble graphing functions i missed the lesson an i'm lost for example:

$y=\frac{x}{x^2-4}$

i know it has V.A at x=-2,x=2 and H.A. at y=0 because the degree of the denominator > degree of numerator

but where do i go from here?(figuring out odd and even asymptotes, x and y intercepts) etc....

2. Originally Posted by euclid2
i am having trouble graphing functions i missed the lesson an i'm lost for example:

$y=\frac{x}{x^2-4}$

i know it has V.A at x=-2,x=2 and H.A. at y=0 because the degree of the denominator > degree of numerator

but where do i go from here?(figuring out odd and even asymptotes, x and y intercepts) etc....
You can get the first derivative to find out on which intervals it is decresing/increasing. You can use the info to know on which side of each vertical asymptote does the function go to (positive or negative infinity).

You can use the second derivative to get the concavity. that might help.

To get the x intercept(s), set y = 0 and solve for x.

To get the y intercept(s), set x = 0 and solve for y.

Don't forget that as the function goes to positive or negative infinity, it must converge to y = 0 (horizontal asymptote, as you've mentioned)..

Cheers!

3. Hello, euclid2!

I use a very primitive approach.
But it has always worked well for me.

Graph: . $f(x) \:=\:\frac{x}{x^2-4}$

i know it has V.A. at $x = \pm2$ and H.A. at $y = 0$
. . because (deg. of denom'r) > (deg of num'r) . . . . Right!

Sketch the asymptotes . . .
Code:
                      |
:     |     :
:     |     :
:     |     :
:     |     :
- - - - - - + - - o - - + - - - - - - -
-2     |     2
:     |     :
:     |     :
:     |     :
|

The only intercept is at (0,0).

The $x$-axis is partitioned into four intervals.
Evaluate a point in each interval.

. . $\begin{array}{ccccc}f(\text{-}3) &=& -\frac{3}{5} & \Rightarrow & (-3,\:-\frac{3}{5}) \\ \\[-3mm]
f(\text{-}1) &=& +\frac{1}{5} & \Rightarrow & (-1,\:\frac{1}{5})\\ \\[-3mm]
f(1) &=& -\frac{1}{3}& \Rightarrow & (1,\:-\frac{1}{3}) \\ \\[-3mm]
f(3) &=& +\frac{3}{5} & \Rightarrow & (3,\:\frac{3}{5})\end{array}$

Plot these points.

Using your knowledge of asymptotes, sketch the graph.

Code:
                :     |     :
:*    |     :*
:     |     :
:     |     :
: *   |     : *
:     |     :  o
:  o  |     :    *
:   * |     :           *
- - - - - - - : - - o - - : - - - - - - -
*          -2     | *   2
*    :     |  o  :
o  :     |     :
* :     |   * :
:     |     :
*:     |     :
:     |    *:
:     |

4. Originally Posted by Soroban
Hello, euclid2!

I use a very primitive approach.
But it has always worked well for me.

Sketch the asymptotes . . .
Code:
                      |
:     |     :
:     |     :
:     |     :
:     |     :
- - - - - - + - - o - - + - - - - - - -
-2     |     2
:     |     :
:     |     :
:     |     :
|

The only intercept is at (0,0).

The $x$-axis is partitioned into four intervals.
Evaluate a point in each interval.

. . $\begin{array}{ccccc}f(\text{-}3) &=& -\frac{3}{5} & \Rightarrow & (-3,\:-\frac{3}{5}) \\ \\[-3mm]
f(\text{-}1) &=& +\frac{1}{5} & \Rightarrow & (-1,\:\frac{1}{5})\\ \\[-3mm]
f(1) &=& -\frac{1}{3}& \Rightarrow & (1,\:-\frac{1}{3}) \\ \\[-3mm]
f(3) &=& +\frac{3}{5} & \Rightarrow & (3,\:\frac{3}{5})\end{array}$

Plot these points.

Using your knowledge of asymptotes, sketch the graph.

Code:
                :     |     :
:*    |     :*
:     |     :
:     |     :
: *   |     : *
:     |     :  o
:  o  |     :    *
:   * |     :           *
- - - - - - - : - - o - - : - - - - - - -
*          -2     | *   2
*    :     |  o  :
o  :     |     :
* :     |   * :
:     |     :
*:     |     :
:     |    *:
:     |
this is great. does it matter which numerical values i plug in to obtain other points on the graph, or in other words why did you use f(-1,1,-3,3) besides the fact that these are the values that the function approaches around the asymptotes?


# Alternate Solution to The Meeting Place Cannot Be Changed

Hi,

I would like to share an alternate solution to the Codeforces problem The Meeting Place Cannot Be Changed. My solution is in Java, but it can be translated to other languages. I enjoyed solving this problem.

This problem is in fact very similar to the LeetCode problem Peak Index in a Mountain Array.

The official solution does binary search on the minimum time needed for the friends to meet, but my solution does binary search on the optimal place to meet.

How do we do binary search if the positions are non-integers? We simply multiply each coordinate by 10^6, then find the optimal time for integer meeting points, then divide by 10^6. This will ensure that the difference between the meeting place returned and the actual meeting place is \leq 10^{-6}, and the difference between the minimum times is \leq 10^{-6}.

My solution relies on the two following key observations:

1. The minimum times to meet every spot on the line decreases, then increases as you sweep the line.
2. No three points contain the same minimum times.

The minimum time for each meeting place is the maximum times it takes for each friend to travel to that meeting place. We can compute the minimum time for an arbitrary meeting place in O(n) time.

Now, we can do binary search on the optimal meeting place while keeping track of the minimum time. For each meeting place \text{mid} we check, we find the minimum times for it and adjacent meeting places, and compute the total running minimum of all times. We break out of the search if \text{minTime(mid - 1)} > \text{minTime(mid)} < \text{minTime(mid + 1)}; otherwise, we use the information to narrow the bound of possible optimal meeting places.

Once we break out of the loop, we have the minimum meeting time and hence the answer.

My code below:

import java.io.*;
import java.util.*;

public class TheMeetingPlaceCannotBeChanged {
public static void main(String[] args) throws IOException {

// positions multiplied by 10^6 so we can work with integers
long[] positions = new long[numFriends];
for (int i = 0; i < numFriends; i++) {
positions[i] = Long.parseLong(stp.nextToken()) * 1000000L;
}
int[] speeds = new int[numFriends];
for (int i = 0; i < numFriends; i++) {
speeds[i] = Integer.parseInt(sts.nextToken());
}
// Find min, max position
long minPosition = Long.MAX_VALUE;
long maxPosition = Long.MIN_VALUE;
for (long p : positions) {
minPosition = Math.min(minPosition, p);
maxPosition = Math.max(maxPosition, p);
}
// Binary search to find minimum time
long low = minPosition;
long high = maxPosition;
double minTime = Double.MAX_VALUE;
while (low <= high) {
long mid = (low + high) / 2;
double midTime = time(positions, speeds, mid);
double lowerTime = time(positions, speeds, mid - 1);
double higherTime = time(positions, speeds, mid + 1);
minTime = Math.min(Math.min(minTime, midTime), Math.min(lowerTime, higherTime));
if (midTime < lowerTime && midTime < higherTime) break;
else if (lowerTime > midTime || midTime > higherTime) {
low = mid + 1;
} else {
high = mid - 1;
}
}
System.out.printf("%.6f\n", minTime / 1000000);
}
// Time is maximum time of all friends to get to point
static double time(long[] positions, int[] speeds, long position) {
double maxTime = Double.MIN_VALUE;
for (int i = 0; i < speeds.length; i++) {
maxTime = Math.max(maxTime, (double) Math.abs(positions[i] - position) / speeds[i]);
}
return maxTime;
}
}

Thanks,
rayfish

Good solution as well, this is also known as ternary search, which works a little faster by cutting into 3 equal size chunks. https://en.wikipedia.org/wiki/Ternary_search

You should add this to USACO Guide as one of the user solutions .


# Is the number of primes congruent to 1 mod 6 equal to the number of primes congruent to 5 mod 6?

(I know that there's an infinity of primes congruent to 5 mod 6, but I don't know if there is an infinity of primes congruent to 1 mod 6.)

But what I'd really like to know is whether or not the number of primes ≡1 mod 6 less than a given n can be said to be asymptotically equal to the number of primes ≡5 mod 6 less than that n. So the "total" number of primes ≡1 mod 6 would be equal to the number of primes ≡5 mod 6.

(I hope this is understandable. I'd also like to know how to phrase this inquiry in a more standard fashion, if someone would tell me how to fix it.)

-
You might ask like this: let $\pi_1(n)$ be the number of primes of the form $6k+1$ that are less than $n$, and $\pi_5(n)$ similarly. Then you can ask whether $\pi_1(n)/\pi_5(n)$ approaches a limit as $n$ increases without bound, and if so, whether that limit is 1. – MJD Aug 22 '12 at 1:13
That's precisely what I mean; thank you. – Annick Aug 22 '12 at 1:14
(You didn't ask for this, but for an elementary proof that there are infinitely many primes congruent to $1 \bmod 6$, consider the possible primes dividing a number of the form $k^2 - k + 1$.) – Qiaochu Yuan Aug 22 '12 at 1:28
Got it, thanks. Great to know. – Annick Aug 22 '12 at 2:35

Yes, this is true. A more general statement follows from a suitably strong form of Dirichlet's theorem on arithmetic progressions, namely that asymptotically the proportion of primes which are congruent to $a \bmod n$ (for $\gcd(a, n) = 1$) is $\frac{1}{\varphi(n)}$.

In the particular case that $n = 6$ it is possible to give a more elementary proof of a weaker result. Define the Dirichlet L-function

$$L(s, \chi_6) = \sum_{n=1}^{\infty} \frac{\chi_6(n)}{n^s}$$

where $\chi_6$ is the unique nontrivial Dirichlet character $\bmod 6$. This takes the form $\chi_6(n) = 1$ if $n \equiv 1 \bmod 6$, $\chi_6(n) = -1$ if $n \equiv 5 \bmod 6$, and $\chi_6(n) = 0$ otherwise. The Euler product of this L-function is

$$L(s, \chi_6) = \prod_p \left( \frac{1}{1 - \chi_6(p) p^{-s}} \right) = \prod_{p \equiv 1 \bmod 6} \left( \frac{1}{1 - p^{-s}} \right) \prod_{p \equiv 5 \bmod 6} \left( \frac{1}{1 + p^{-s}} \right).$$

It is possible to explicitly evaluate $L(1, \chi_6)$ and in particular to show that it is not zero; in fact,

$$L(1, \chi_6) = \int_0^1 \frac{1 - x^5}{1 - x^6} \, dx$$

and this can be evaluated using partial fractions (but note that the integrand is always positive so this number is definitely positive). So we conclude that

$$-\log L(s, \chi_6) = \sum_{p \equiv 1 \bmod 6} \log (1 - p^{-s}) + \sum_{p \equiv 5 \bmod 6} \log (1 + p^{-s})$$

approaches a nonzero constant as $s \to 1$ (if summed in the appropriate order) even though the first and second terms separately approach $\mp \infty$. So the contributions coming from primes in each residue class cancel out asymptotically. This is not quite as strong as the desired statement, though; if you fill in all the details in what I've said you'll show that the Dirichlet density of the primes congruent to $\pm 1 \bmod 6$ are the same but this should still be true for the natural density and this requires a further argument (I am not sure how much further, though).

For more details see any book on analytic number theory, e.g. Apostol.

-
Thanks so much for this highly detailed answer and explanatory links. – Annick Aug 22 '12 at 1:27

Look up "prime races" at http://www.dms.umontreal.ca/~andrew/PDF/PrimeRace.pdf.

Their conclusion: "It does seem that “typically” qn + a has fewer primes than qn + b if a is a square modulo q while b is not."

So, since 1 is a quadratic residue mod 6, while 5 is not, there will typically be more primes of the form 6n+5 than 6n+1 (though their ratio does tend to 1).

-


# Does the sequence $(a_n)_{n\in\mathbb{N}}$ have a convergent subsequence?

Does the sequence $(a_n)_{n\in\mathbb{N}}$ have a convergent subsequence?

$$a_n= \begin{cases} \sin(n), & \text{if n is odd} \\ n, & \text{if n is even} \end{cases}$$

I understand that $\sin(n)$ is a bounded sequence so I can use Bolzano Weierstrass theorem to state that the sequence has a convergent subsequence. Not sure where to go from here assuming I am along the right lines?

• "so I can use Bolzano Weierstrass theorem to state that the sequence has a convergent subsequence." Yes. "Not sure where to go from here". You're not sure where to go from having reached an irrefutable conclusion? You don't have to go anywhere. You were ask whether it has a convergent subsequence and you answered that it did. So ... where else is there to go? – fleablood May 30 '18 at 18:39
• @fleablood I think the question asker was missing a subtlety for this question: $\sin(n)$ has a convergent subsequence, but it is not a subsequence of $(a_n)$ - the answer is, of course, to show that $\sin(n)$ for odd $n$ has a convergent subsequence, but the work presented in the question is not complete as it stands. – Milo Brandt May 30 '18 at 18:57
• It's not that subtle is it? $a_n$ where $n$ is odd s $a_n= \sin n$ is a subsequence. That subsequence is bounded so there is a subsequence of the subsequence that is bounded. A subsequence of a subsequence is a subsequence. You have to avoid being lazy and stupid ($a_n$ itself is not bounded and $a_n; odd$ is not $\{\sin n\}$ so you have to claim that $a_n; odd$ has the subsequence; not $\{\sin n\}$) but that's not hard to do. – fleablood May 31 '18 at 0:34
• "I understand that sin(n) is a bounded sequence so I can use Bolzano Weierstrass theorem to state that the sequence has a convergent subsequence." Okay, there are two observations you must make. $\{a_n\}$ is not bounded (but the subsequence $\{a_{2k+1}\} is$. And stating $\{\sin n\}$ has a convergent subsequence isn't enough as $\{\sin n\} \not \subset \{a_n\}$. But as $\{a_{2k+1} \}\subset \{\sin n\}$ is bounded there is a convergent $\{b_j\}\subset \{a_{2k+1}\} \subset \{\sin n\}$. That's fairly obvious and easy, but enough to make a question worth 5 marks. – fleablood May 31 '18 at 0:41

Let's construct the subsequence which converges to $0$.

Consider convergents of $\pi$ continued fraction: $\left\{ \dfrac{n_j}{d_j},\; j\in\mathbb{N} \right\}$. And focus on its numerators: $$n_1 = 3,\\ n_2 = 22, \\ n_3 = 333, \\ n_4 = 355, \\ n_5 = 103993, \\ n_6 = 104348, \\ \vdots$$

Then $$\lim_{j\to \infty} \sin n_j =0.$$

Indeed, easy to estimate: $$|\sin n_j| = \left| \sin\left( n_j - \pi d_j \right) \right| \approx \left| n_j - \pi d_j \right|=d_j\left| \dfrac{n_j}{d_j} - \pi \right| < \dfrac{1}{d_{j+1}}.$$

And it remains to show that the sequence $\{n_j, \; j\in\mathbb{N}\}$ contains infinite number of odd $n_j$.

If continued fraction (of $\pi$) denote as $[a_0;a_1, a_2, \ldots, a_j, \ldots]$, then $$n_{j+1} = a_{j+1}n_j + n_{j-1}.$$

There are $8$ possibilities:

\begin{array}{|l|l|l|l|} \hline (n_{j-1}, n_j) & a_{j+1} & n_{j+1} & \rightarrow (n_j, n_{j+1})\\ \hline (odd, odd) & odd & even & \rightarrow (odd, even) \\ (odd, odd) & even & odd & \rightarrow (odd, odd) \\ \hline (odd, even) & odd & odd & \rightarrow (even, odd) \\ (odd, even) & even & odd & \rightarrow (even, odd) \\ \hline (even, odd) & odd & odd & \rightarrow (odd, odd) \\ (even, odd) & even & even & \rightarrow (odd, even) \\ \hline (even, even) & odd & even & \rightarrow (even, even) \\ (even, even) & even & even & \rightarrow (even, even) \\ \hline \end{array}

which show that if at least one of $n_j$ is odd, then there is infinite number of odd $n_j$.

A subsequence of a subsequence is a subsequence. The sequence of values of $a_n$ for which $n$ is odd is a subsequence, and you've already said that that has a convergent subsequence. That convergent subsequence is a subsequence of a subsequence of the original sequence. Therefore it is a subsequence of the original sequence.

You wrote (twice) “function” where you should have written “sequence”. Other than that, you are doing fine.

• Edited thanks. The question is worth 5 marks, I don't think this answer justifies this. Where do I go from here? Thanks. – Ben Jones May 30 '18 at 18:36
• You know that the sequence $(a_{2n-1})_{n\in\mathbb N}$ is bounded. Therefore, it has a convergent subsequence, which will be a subsequence of the original sequence, of course. – José Carlos Santos May 30 '18 at 18:37
• "Where do I go from here?" Well, you could go to breakfast. Or Tahiti, I hear Tahiti is nice this time of year. ... you've answered the question. You don't need to go anywhere. – fleablood May 30 '18 at 18:42
• I'm off to Tahiti then. – Ben Jones May 31 '18 at 18:36


1. ## Dimensions of Kindle (Electronic pad with a screen)

I created this problem and wanted to check my answers with the MathForum.

x = length
y = width

Scenario:
There are two Kindles.
K(1) has the dimensions of 3.6 in by 4.8 in. with a surface area (SA) of $17.28in^2$ and a diagonal that measures 6 in
K(2) has the dimensions of 5.4 in by 7.9 in. with a surface area of $42.66in^2$ and a diagonal that measures 9.56 in

a. How much bigger is K(2)'s SA compared to K(1)'s SA?

b. Write an equation that represents the maximum and minimum value of
x and y when the SA of K(3) is 42.66 in.
my answer: Either $x=42.66*y$ or $y=42.66*x$

Note: For 'c' I want to know if it's possible to have a minimum and maximum value of the length and width when the diagonal is 6 in. My guess is that you can only have 1 combination of 'x' and 'y' that gives you a SA of 42.66 and a diagonal of 6 in.

c. Write an equation/system of equations that represents the maximum and minimum value of x and y if the diagonal of K(3) is 6 inches.

my answer: $x^2 + y^2 = (6)^2$ --> y = -X + 6 (If I simplified that right) and $x=42.66*y$ or $y=42.66*x$; where the two lines intersect is the value of x and y when the diagonal is 6.

2. Originally Posted by Masterthief1324
I created this problem and wanted to check my answers with the MathForum.

x = length
y = width

Scenario:
There are two Kindles.
K(1) has the dimensions of 3.6 in by 4.8 in. with a surface area (SA) of $17.28in^2$ and a diagonal that measures 6 in
K(2) has the dimensions of 5.4 in by 7.9 in. with a surface area of $42.66in^2$ and a diagonal that measures 9.56 in

a. How much bigger is K(2)'s SA compared to K(1)'s SA?
Yes, that is correct.

b. Write an equation that represents the maximum and minimum value of
x and y when the SA of K(3) is 42.66 in.
my answer: Either $x=42.66*y$ or $y=42.66*x$
What? You said there were two kinds of "kindle", K(1) and K(2). Where did K(3) come from? Without any information about K(3) it is impossible to answer this. Did you leave something out?

Note: For 'c' I want to know if it's possible to have a minimum and maximum value of the length and width when the diagonal is 6 in. My guess is that you can only have 1 combination of 'x' and 'y' that gives you a SA of 42.66 and a diagonal of 6 in.

c. Write an equation/system of equations that represents the maximum and minimum value of x and y if the diagonal of K(3) is 6 inches.

my answer: $x^2 + y^2 = (6)^2$ --> y = -X + 6 (If I simplified that right)
No, you didn't. " $x^2+ y^2= z^2$" does NOT reduce to "x+ y= z". For example, $4^2+ 3^2= 16+ 9= 25= 5^2$ but $4+ 3\ne 5$.

and $x=42.66*y$ or $y=42.66*x$; where the two lines intersect is the value of x and y when the diagonal is 6.
If $x^2+ y^2= 36$ and xy= 42.66, then y= 42.66/x so $x^2+ \frac{1819.8756}{x^2}= 36$. Multiplying both sides by $x^2$, $x^4+ 1819.8756= 36x^2$ or $x^4- 36x^2+ 1819.8756= 0$. If you let $u= x^2$, this becomes the quadratic equation $u^2- 36u+ 1819.8756= 0$. Solve that for u, then take the square root to find x.

But are you sure you have interpreted the question correctly? The problem said "find maximum and minimum" values, not specific values.

3. Sorry, I did make that last problem confusing.
I hope this clarifies it.

K(3) is a hypothetical Kindle device with a surface area of $42.66in^2$ and a diagonal of 6 in.

c1. Can you have more than one dimensions that satisfy the SA & the diagonal length of 6in? Why or why not?
my answer: Not if the diagonal is 6 inch -- there is only 1 value of x and y (length and width)

c2. Can you represent c1. with and equation?
my answer: Since the diagonal is 6, if I was to use the Pythagorean theorem, there is only 1 value of x and y that would fit the specifications.

(specifications)
You know indefinitely that the K(3) has a surface area of 42.66 and a diagonal of 6 in:

my equation:
x^2 + y^2 = 6 (is it 6 or 6^2?) --> y=√(z^2 - x^2)
x*y = 42.66

If I was to graph the two equations, will where they intersect be the dimensions that fit the specifications?

4. b. Write an equation that represents the maximum and minimum value of
x and y when the SA of K(3) is 42.66 in.
If K(3) has a surface area of 42.66, then you know that

$xy=42.66$

since the diagonal is 6 inches, you know that

$x^2 + y^2 = 6^2$ (draw a picture to understand why it is 6^2 and not 6)

You have two equations, you can try to solve them both for y and graph them to approximate a solution

$y=42.66/x$
$y=sqrt(36-x^2)$

Once you look at this graph you will realize that there is no solution to your problem

K(3) is physically impossible

---------------------------------------
To understand this, think of the kindle with diagonal 6 that has the largest surface area, this kindle would be square. It would have x = y = sqrt (18), so its largest surface area would be xy=18

I created this problem and wanted to check my answers with the MathForum.
To fix your problem, you must either increase the size of the diagonal or reduce the surface area of K(3) to make the two curves intersect


# Number of ways in which boys and girls sit alternately if six boys and six girls sit randomly?

Six boys and six girls sit in a row randomly. What is the total number of ways in which the boys and the girls sit alternately?

My attempt: Consider these six seats _ _ _ _ _ _

The number of ways to arrange 6 boys in 6 places is 6! Now 7 gaps are created between these 6 seats. So, we can select any 6 of these 7 gaps and make girls sit there. There are 7C6 * 6! ways to do that (since the girls can shuffle amongst themselves).

Hence the total number of ways to make boys and girls sit alternately should be 6! 7C1 * 6! but the answer is 2*6!*6!.

What am I missing here?

• You can't choose any $6$...the unfilled gap must be either first or last.
– lulu
Mar 18, 2017 at 11:34
• Note "alternately". Only two of the seven ways to pick six of the gaps result in an alternating pattern, the rest have two boys next to each other. Mar 18, 2017 at 11:34
• Oh right! Thank you so much. Mar 18, 2017 at 11:35

You are implicitly assuming that there are 13 seats for $6+6=12$ people. It is likely that the original question is making the implicit assumption that there are only 12 seats. In this latter case, when a boy sits first, there $6! \cdot 6!$ ways to sit the others. Multiply this by 2 to cover the case where the first seater is a girl.

You are over-counting.

You are also counting the cases where extreme 2 gaps are both occupied and out of 5 gaps in middle 4 are occupied.

For example, let's first arrange the boys: B_B_B_B_B_B = 6! ways.

We are left with 7 gaps so we select 6 among them and permute the girls so (7C6 6! ways.)

Hence according to you answer is 6! . 7C6 . 6! = 7.6!.6!

But it also counts the cases like GBGBGBBGBGBG two boys are sitting together and alternate pattern is disturbed so we eliminate these unfavorable cases by bijection. So, Unfavorable ways= 6! . 2C2 . 5C4. 6! = 5.6!.6! So hence total ways of favourable permutations = 7.6!.6! - 5.6!.6! =2.6!.6! (The required answer)


# How do you calculate $2^{2^{2^{2^{2}}}}$?

From information I have gathered online, this should be equivalent to $2^{16}$ but when I punch the numbers into this large number calculator, the number comes out to be over a thousand digits. Is the calculator wrong or is my method wrong?

• The number is equal to $2^r$ where $r=2^{16}$. – Aravind Mar 16 '17 at 17:17
• this is quiet not clear use parentesis – Dr. Sonnhard Graubner Mar 16 '17 at 17:17
• If parenthesis are not used, it is assumed that exponents are evaluated from top-down as opposed to bottom-up. $(a^b)^c=a^{bc}\neq a^{(b~^c)}$. Exponentiation is not associative. The answer of $2^{16}$ is for if it were evaluated bottom-up as (((2^2)^2)^2)^2 instead of top-down which is 2^(2^(2^(2^2))) which is much larger – JMoravitz Mar 16 '17 at 17:22
• Possible duplicate of What is the order when doing $x^{y^z}$ and why? – Simply Beautiful Art Mar 16 '17 at 21:11
• My TI calculator has an inline option of showing this as 2^2^2^2^2, which evaluates as 65536 (left to right evaluation). But when I switch to math print mode it shows the tower of powers, which it tries, but fails, to evaluate top down...overflow. I'm not saying that a TI calculator is the final arbiter of math truth, but it is what I get. – paw88789 Mar 17 '17 at 2:22

## 6 Answers

$$2^{2^{2^{2^2}}}=2^{2^{2^4}}=2^{2^{16}}=2^{65536}\tag1$$

The number of digits:

$$\mathcal{A}=1+\lfloor\log_{10}\left(2^{65536}\right)\rfloor=19729\tag2$$

• This is all true enough, but it doesn't really answer the question as actually asked. – hBy2Py Mar 16 '17 at 18:59
• It currently does answer the question in the title. Equation 1 shows how to evaluate the expression. It also implies, though doesn't explicitly states that the answer the asker's method (or at least the answer he got using the method) was wrong. – Fluidized Pigeon Reactor Mar 16 '17 at 21:30
• The title is a reference for searching and tracking. The question is the text beneath that. NAA. – Nij Mar 17 '17 at 8:43

What you have is a power tower or "tetration" (defined as iterated exponentiation). From the latter link, you would most benefit from this brief excerpt on the difference between iterated powers and iterated exponentials.

The comment by JMoravitz really gets to the heart of the matter, namely that exponential towers must be evaluated from top to bottom (or right to left). There actually is a notation for your particular question: ${}^52=2^{2^{2^{2^{2}}}}$. You really need to look at ${}^42$ before you get something meaningful because, unfortunately, $${}^32=2^{2^{2}}=2^4=16=4^2=(2^2)^2;$$ however, $${}^42=2^{2^{2^{2}}}=2^{2^{4}}=2^{16}\neq2^8=(4^2)^2=((2^2)^2)^2.$$ Hence, your method is wrong, but everything in those links should provide more than enough for you to become comfortable with tetration.

• It seems that the Knuth's notation $2\uparrow\uparrow n$ has gained popularity over Rucker's one $^n2$ nowadays. – zwim Mar 16 '17 at 17:47
• @Daniel W. Farlow look like you beat me to posting. Got to love it when that happens. – Sentinel135 Mar 16 '17 at 17:53
• @zwim I actually prefer Rucker's notation, but I do see the appeal of Knuth's very unambiguous notation. – Daniel W. Farlow Mar 16 '17 at 17:55

By convention, the meaning of things written $\displaystyle a^{b^{c^d}}$ without brackets is $\displaystyle a^{\left(b^{\left(c^d\right)}\right)}$ and not $\left(\left(a^b\right)^c\right)^d$.

This is because $\left(\left(a^b\right)^c\right)^d$ equals $a^{b\cdot c\cdot d}$ anyway, so it makes pragmatic sense to reserve the raw power-tower notation $\displaystyle a^{b^{c^d}}$ for the case that doesn't have an alternative notation without parentheses.

As others have explained, $\displaystyle 2^{2^{2^{2^2}}}$ interpreted with this convention is $2^{65536}$, a horribly huge number, whereas $(((2^2)^2)^2)^2$ is $2^{16}=65536$, as you compute.

• "Horribly huge number" rubs me the wrong way in this context. Huge numbers are horrible when they count something that you don't want. The number's not horrible when it's your bank balance! (Actually it might be. If you put together that many pennies it'd probably collapse into a black hole the size of the Milky Way.) – Matt Samuel Mar 16 '17 at 22:44
• @MattSamuel: Milky Way? This number exceeds the number of Planck volumes in the observable universe ... to the hundredth power! – Henning Makholm Mar 17 '17 at 1:24
• Do you really expect me to do coordinate transformations in my head while strapped to a centrifuge??? – Matt Samuel Mar 17 '17 at 1:25
• 2^65536 really isn't a horribly huge number when you consider that its binary representation fits in a mere 8 KB of memory. I mean, we are already using RSA moduli around 2^4096 already, so this number is only around 10× longer. – Nayuki Mar 17 '17 at 3:28
• Actually all finite numbers are pretty small because all but a finite number of the rest are bigger. Come to that all transfinite numbers are pretty small as well. – Martin Rattigan Apr 2 '17 at 21:55

I would calculate it using Maxima (which evaluates repeat exponentiation correctly, right-to-left), since there is no point wasting brain cells on something that a machine can do:

$maxima Maxima branch_5_39_base_2_gc9edaee http://maxima.sourceforge.net using Lisp GNU Common Lisp (GCL) GCL 2.6.12 Distributed under the GNU Public License. See the file COPYING. Dedicated to the memory of William Schelter. The function bug_report() provides bug reporting information. (%i1) 2^2^2^2^2; (%o1) 200352993040684646497907235156025575044782547556975141926501697371089405\ 955631145308950613088093334810103823434290726318182294938211881266886950636476\ 154702916504187191635158796634721944293092798208430910485599057015931895963952\ 486337236720300291696959215610876494888925409080591145703767520850020667156370\ 236612635974714480711177481588091413574272096719015183628256061809145885269982\ 614142503012339110827360384376787644904320596037912449090570756031403507616256\ 247603186379312648470374378295497561377098160461441330869211810248595915238019\ 533103029216280016056867010565164675056803874152946384224484529253736144253361\ 437372908830379460127472495841486491593064725201515569392262818069165079638106\ 413227530726714399815850881129262890113423778270556742108007006528396332215507\ 783121428855167555407334510721311242739956298271976915005488390522380435704584\ 819795639315785351001899200002414196370681355984046403947219401606951769015611\ 972698233789001764151719005113346630689814021938348143542638730653955296969138\ 802415816185956110064036211979610185953480278716720012260464249238511139340046\ 435162386756707874525946467090388654774348321789701276445552940909202195958575\ 162297333357615955239488529757995402847194352991354376370598692891375715374000\ 198639433246489005254310662966916524341917469138963247656028941519977547770313\ 806478134230959619096065459130089018888758808473362595606544488850144733570605\ 881709016210849971452956834406197969056546981363116205357936979140323632849623\ 304642106613620022017578785185740916205048971178182040018728293994344618622432\ 800983732376493181478984811945271300744022076568091037620399920349202390662626\ 449190916798546151577883906039772075927937885224129430101745808686226336928472\ 585140303961555856433038545068865221311481363840838477826379045960718687672850\ 976347127198889068047824323039471865052566097815072986114143030581692792497140\ 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810504334176954306753804111392855379252924134733948105053202570872818630729115\ 891133594201476187266429156403637192760230628384065042544174233546454998705531\ 872688792642410214736369862546374715974435494344389973005174252511087735788639\ 094681209667342815258591992485764048805507132981429935991146323991911395992675\ 257635900744657281019180584180734222773472139772321823177171691640010882611254\ 909336118678057572239101818616854910850088527227437421208652485237245624869766\ 224538481929867112945294551549703058591930719849710541418163696897613112674402\ 700964866754593456705993699546450055892162804797636568613331656390739570327203\ 438917541526750091501119885687270884819553167693168127289214303137681801644547\ 736751835349785792427646335416243360112596025210950161226411034608346564823559\ 793427405686884922445874549377675212032470380303549115754483129527589193989368\ 087632768543876955769488142284431199859570072752139317683783177033913042306095\ 899913731468456901042209516196707050642025673387344611565527617599272715187766\ 001023894476053978951694570880272873622512107622409181006670088347473760515628\ 553394356584375627124124445765166306408593950794755092046393224520253546363444\ 479175566172596218719927918657549085785295001284022903506151493731010700944615\ 101161371242376142672254173205595920278212932572594714641722497732131638184532\ 655527960427054187149623658525245864893325414506264233788565146467060429856478\ 196846159366328895429978072254226479040061601975197500746054515006029180663827\ 149701611098795133663377137843441619405312144529185518013657555866761501937302\ 969193207612000925506508158327550849934076879725236998702356793102680413674571\ 895664143185267905471716996299036301554564509004480278905570196832831363071899\ 769915316667920895876857229060091547291963638167359667395997571032601557192023\ 734858052112811745861006515259888384311451189488055212914577569914657753004138\ 4717124577965048175856395072895337539755822087777506072339445587895905719156736 (%i2) bfloat(%); (%o2) 2.003529930406846b19728 (%i3)  Of course if I just wanted to estimate the magnitude of the number without resorting to the use of arbitrary precision computer software, I'd note that the exponent is$2^{2^{2^2}}=2^{(2^{(2^2)})}=2^{(2^4)}=2^{16}=65536$; multiplying it by$\log_{10} 2\sim 0.30103$gives$19728.302$, so the result is approximately$10^{0.302}\times 10^{19728}\sim 2\times 10^{19728}$. • Wow. Someone actually posted the digits. – Matt Samuel Mar 16 '17 at 23:36 • There is a mistake in the computed result. The 2 in the middle should be a 3. Probably a typo. – augustin Mar 17 '17 at 3:43 • "2 in the middle"... can you be a bit more specific? – Viktor Toth Mar 17 '17 at 3:48 • @augustin: The digit in the middle is an 8. And yes, I checked it; with the help of the computer, of course. Given that the total number of digits (19729) is odd, the digit in the middle is well defined (it is the digit which is preceded and followed by the same number of digits). – celtschk Mar 17 '17 at 7:59 • @celtschk now you sir are fun at parties! – Pierre Arlaud Mar 17 '17 at 8:44 This looks an awfully close to what is known as a tetration (a.k.a. power tower). This is$^{(k)}a=a^{^{(k-1)}a}$where$^1a=a$. For numbers greater than one, these usually get really big really fast, and faster than exponents do. So in your case, you have$^52=2^{2^{16}}$. Now if you want to see an interesting one look at$\lim_{k\to \infty}\;^{(k)}(\sqrt{2})$. • Wouldn't that work for any$\sqrt[n]{n}$? – Random832 Mar 16 '17 at 20:04 • Wouldn't what work for any$\sqrt[n]{n}\$? I never said the answer. And yes, but it depends on what you think the answer is. ;D – Sentinel135 Mar 16 '17 at 20:13
• Looks like that limit goes to infinity really quickly. – Joshua Mar 16 '17 at 20:36
• @Joshua are you sure? can you try and prove it? – Sentinel135 Mar 16 '17 at 20:56
• Darn. I hate accumulated roundoff. – Joshua Mar 16 '17 at 21:24

Your equation can be simplified using Knuth's up arrow notation. \begin{equation*} 2^{2^{2^{2^2}}} = 2 \uparrow\uparrow 5 \end{equation*}

(because we can calculate tetration with Knuth's up arrow notation)

By definition of Knuth's up arrow notation, You can get this result.

\begin{equation*} 2\uparrow\uparrow5 = 2^{(2^{(2^{(2^2)})})} \end{equation*}

And according to web2.0calc,

\begin{equation*} 2^{(2^{(2^{2})})} = 65536 \end{equation*}

Finally, the answer would be:

\begin{equation*} 2^{65536} \end{equation*}

(correct me if I'm wrong, this was my first answer on Mathematics SE)


# block matrix multiplication

If $A,B$ are $2 \times 2$ matrices of real or complex numbers, then

$$AB = \left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]\cdot \left[ \begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array} \right] = \left[ \begin{array}{cc} a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\ a_{21}b_{11}+a_{22}b_{21} & a_{22}b_{12}+a_{22}b_{22} \end{array} \right]$$

What if the entries $a_{ij}, b_{ij}$ are themselves $2 \times 2$ matrices? Does matrix multiplication hold in some sort of "block" form ?

$$AB = \left[ \begin{array}{c|c} A_{11} & A_{12} \\\hline A_{21} & A_{22} \end{array} \right]\cdot \left[ \begin{array}{c|c} B_{11} & B_{12} \\\hline B_{21} & B_{22} \end{array} \right] = \left[ \begin{array}{c|c} A_{11}B_{11}+A_{12}B_{21} & A_{11}B_{12}+A_{12}B_{22} \\\hline A_{21}B_{11}+A_{22}B_{21} & A_{22}B_{12}+A_{22}B_{22} \end{array} \right]$$ This identity would be very useful in my research.

• Yes it does if the "blocking" is "conforming". May 9, 2014 at 14:12
• Yes. (The blocking is "confirming" in the situation you have given.) Discussed in detail in §6.12 of cip.ifi.lmu.de/~grinberg/primes2015/sols.pdf (specifically Exercise 38 and Remark 6.73; search for "block-matrix notation" if these numbers change). Dec 4, 2015 at 10:16

It depends on how you partition it, not all partitions work. For example, if you partition these two matrices

$$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}, \begin{bmatrix} a' & b' & c' \\ d' & e' & f' \\ g' & h' & i' \end{bmatrix}$$

in this way

$$\left[\begin{array}{c|cc}a&b&c\\ d&e&f\\ \hline g&h&i \end{array}\right], \left[\begin{array}{c|cc}a'&b'&c'\\ d'&e'&f'\\ \hline g'&h'&i' \end{array}\right]$$

and then multiply them, it won't work. But this would

$$\left[\begin{array}{c|cc}a&b&c\\ \hline d&e&f\\ g&h&i \end{array}\right] ,\left[\begin{array}{c|cc}a'&b'&c'\\ \hline d'&e'&f'\\ g'&h'&i' \end{array}\right]$$

What's the difference? Well, in the first case, all submatrix products are not defined, like $\begin{bmatrix} a \\ d \\ \end{bmatrix}$ cannot be multiplied with $\begin{bmatrix} a' \\ d' \\ \end{bmatrix}$

So, what is the general rule? (Taken entirely from the Wiki page on Block matrix)

Given, an $(m \times p)$ matrix $\mathbf{A}$ with $q$ row partitions and $s$ column partitions $$\begin{bmatrix} \mathbf{A}_{11} & \mathbf{A}_{12} & \cdots &\mathbf{A}_{1s}\\ \mathbf{A}_{21} & \mathbf{A}_{22} & \cdots &\mathbf{A}_{2s}\\ \vdots & \vdots & \ddots &\vdots \\ \mathbf{A}_{q1} & \mathbf{A}_{q2} & \cdots &\mathbf{A}_{qs}\end{bmatrix}$$

and a $(p \times n)$ matrix $\mathbf{B}$ with $s$ row partitions and $r$ column parttions

$$\begin{bmatrix} \mathbf{B}_{11} & \mathbf{B}_{12} & \cdots &\mathbf{B}_{1r}\\ \mathbf{B}_{21} & \mathbf{B}_{22} & \cdots &\mathbf{B}_{2r}\\ \vdots & \vdots & \ddots &\vdots \\ \mathbf{B}_{s1} & \mathbf{B}_{s2} & \cdots &\mathbf{B}_{sr}\end{bmatrix}$$

that are compatible with the partitions of $\mathbf{A}$, the matrix product

$\mathbf{C}=\mathbf{A}\mathbf{B}$

can be formed blockwise, yielding $\mathbf{C}$ as an $(m\times n)$ matrix with $q$ row partitions and $r$ column partitions.

• Use \pmatrix{a&b&c\\d&e&f\\g&h&i} May 9, 2014 at 14:37
• @Berci I know that. But getting those horizontal and vertical lines is the difficult part. May 9, 2014 at 14:38
• @PandaBear You can use $\color{blue}{\text{colors}}$ :) May 9, 2014 at 14:45
• @PavelJiranek Your comment is funny looking. May 9, 2014 at 14:46
• I think this is wrong. You can't partition both of them same way. If you partition after x rows in first matrix , you've to partition after x columns (not rows ) in the second matrix. Otherwise while multiplying you'll have to multiply mn block with another mn block which is not possible. (you need np block) Try it with your example. Oct 1, 2015 at 18:08

It is always suspect with a very late answer to a popular question, but I came here looking for what a compatible block partitioning is and did not find it:

For $$\mathbf{AB}$$ to work by blocks the important part is that the partition along the columns of $$\mathbf A$$ must match the partition along the rows of $$\mathbf{B}$$. This is analogous to how, when doing $$\mathbf{AB}$$ without blocks—which is of course just a partitioning into $$1\times 1$$ blocks—the number of columns in $$\mathbf A$$ must match the number of rows in $$\mathbf B$$.

Example: Let $$\mathbf{M}_{mn}$$ denote any matrix of $$m$$ rows and $$n$$ columns irrespective of contents. We know that $$\mathbf{M}_{mn}\mathbf{M}_{nq}$$ works and yields a matrix $$\mathbf{M}_{mq}$$. Split $$\mathbf A$$ by columns into a block of size $$a$$ and a block of size $$b$$, and do the same with $$\mathbf B$$ by rows. Then split $$\mathbf A$$ however you wish along its rows, same for $$\mathbf B$$ along its columns. Now we have $$A = \begin{bmatrix} \mathbf{M}_{ra} & \mathbf{M}_{rb} \\ \mathbf{M}_{sa} & \mathbf{M}_{sb} \end{bmatrix}, B = \begin{bmatrix} \mathbf{M}_{at} & \mathbf{M}_{au} \\ \mathbf{M}_{bt} & \mathbf{M}_{bu} \end{bmatrix},$$

and $$AB = \begin{bmatrix} \mathbf{M}_{ra}\mathbf{M}_{at} + \mathbf{M}_{rb}\mathbf{M}_{bt} & \mathbf{M}_{ra}\mathbf{M}_{au} + \mathbf{M}_{rb}\mathbf{M}_{bu} \\ \mathbf{M}_{sa}\mathbf{M}_{at} + \mathbf{M}_{sb}\mathbf{M}_{bt} & \mathbf{M}_{sa}\mathbf{M}_{au} + \mathbf{M}_{sb}\mathbf{M}_{bu} \end{bmatrix} = \begin{bmatrix} \mathbf{M}_{rt} & \mathbf{M}_{ru} \\ \mathbf{M}_{st} & \mathbf{M}_{su} \end{bmatrix}.$$

All multiplications conform, all sums work out, and the resulting matrix is the size you'd expect. There is nothing special about splitting in two so long as you match any column split of $$\mathbf A$$ with a row split in $$\mathbf B$$ (try removing a block row from $$\mathbf A$$ or further splitting a block column of $$\mathbf B$$).

The nonworking example from the accepted answer is nonworking because the columns of $$\mathbf A$$ are split into $$(1, 2)$$ while the rows of $$\mathbf B$$ are split into $$(2, 1)$$.

• So basically the formula is valid whenever it makes sense? Apr 3, 2021 at 16:10
• @Filippo pretty much, though it did not make sense to me at the time which is why I had to go looking for it Apr 5, 2021 at 11:33

The recipe for multiplication of (scalar) matrices $$(AB)_{i,j}=\sum_k A_{i,k}B_{k,j}\tag1$$ is saying: to obtain the $$(i,j)$$ element of $$AB$$, form the dot product as you walk along row $$i$$ of $$A$$ while simultaneously walking down column $$j$$ of $$B$$. In other words,

The element at row $$i$$, column $$j$$ of $$AB$$ is the product of row $$i$$ of $$A$$ with column $$j$$ of $$B$$.

Using the notation $$A_{i,\ast}$$ to denote row $$i$$ of $$A$$ and $$B_{\ast,j}$$ to denote column $$j$$ of $$B$$, this can be restated symbolically as $$(AB)_{i,j}=A_{i,\ast}B_{\ast, j}\ .\tag2$$

This all works provided you can match up any row of $$A$$ with any column of $$B$$, which means the number of columns of $$A$$ must equal the number of rows of $$B$$. (This is what it means for $$A$$ and $$B$$ to be conformable.)

You can generalize (1) to block matrices, i.e., the formula still works when every $$A_{i,k}$$ and $$B_{k,j}$$ is itself a matrix [and the subscripts enumerate rows and columns of blocks], provided every product on the RHS of (1) makes sense. This means the number of columns of $$A_{i,k}$$ must equal the number of rows of $$B_{k,j}$$.

Example: Partition matrix $$A$$ into $$A:=\pmatrix{A_1 &A_2}$$ and $$B$$ into $$B:=\pmatrix{B_1\\ B_2}$$, where #columns($$A_1$$) = #rows($$B_1$$), and #columns($$A_2$$) = #rows($$B_2$$). Then $$AB=\pmatrix{A_1 & A_2}\pmatrix{B_1\\ B_2}=A_1B_1 + A_2 B_2\ .\tag3$$ To see this, consider the element at row $$i$$, column $$j$$ of $$AB$$. This element is computed as the product of row $$i$$ of $$A$$ with column $$j$$ of $$B$$. But row $$i$$ of $$A$$ is just row $$i$$ of $$A_1$$ followed by row $$i$$ of $$A_2$$. Similarly, column $$j$$ of $$B$$ is column $$j$$ of $$B_1$$ atop column $$j$$ of $$B_2$$. Assuming conformability, the product of row $$i$$ of $$A$$ with column $$j$$ of $$B$$ is the sum of two pieces: one piece is the product of row $$i$$ of $$A_1$$ with column $$j$$ of $$B_1$$, and the other is the product of row $$i$$ of $$A_2$$ with column $$j$$ of $$B_2$$. In symbols we've argued that $$(AB)_{i,j}=A_{i,\ast}B_{\ast,j}=(A_1)_{i,\ast}(B_1)_{\ast,j}+(A_2)_{i,\ast}(B_2)_{\ast,j}=(A_1B_1)_{i,j} + (A_2B_2)_{i,j},\tag4$$ and we're done.

Example: Consider conformable partitioned matrices $$C:=\pmatrix{C_1\\ C_2}$$ and $$D:=\pmatrix{D_1 & D_2}$$. Then $$CD=\pmatrix{C_1\\ C_2}\pmatrix{D_1 & D_2} =\pmatrix{C_1D_1 & C_1D_2\\C_2D_1&C_2 D_2}.\tag5$$ Again, this can be seen by considering what happens when you multiply row $$i$$ of $$C$$ with column $$j$$ of $$D$$.


# Math Help - Challenging Integral

1. ## Challenging Integral

I was tutoring someone in Calculus today, and they had an extra credit assignment to do. These integrals appeared on past PUTNAM exams, and they can get help on these. However, I've been trying to work this one out and I (and all the other tutors I work with) couldn't figure it out.

So for the sake of me giving a hint or two to this person, I would like some guidance on how to tackle this beast:

$\int_2^4\frac{\sqrt{\ln\!\left(9-x\right)}\,dx}{\sqrt{\ln\!\left(9-x\right)}+\sqrt{\ln\!\left(x+3\right)}}$

All I know is that this equals 1 but I don't know how to get it. XD

Any input or suggestions would be appreciated!!

2. Originally Posted by Chris L T521
I was tutoring someone in Calculus today, and they had an extra credit assignment to do. These integrals appeared on past PUTNAM exams, and they can get help on these. However, I've been trying to work this one out and I (and all the other tutors I work with) couldn't figure it out.

So for the sake of me giving a hint or two to this person, I would like some guidance on how to tackle this beast:

$\int_2^4\frac{\sqrt{\ln\!\left(9-x\right)}\,dx}{\sqrt{\ln\!\left(9-x\right)}+\sqrt{\ln\!\left(x+3\right)}}$

All I know is that this equals 1 but I don't know how to get it. XD

Any input or suggestions would be appreciated!!
Consider

$
\int_2^4 \frac{\sqrt{\ln 9-x} + \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = \int_2^4 1 = 2
$

Expanding gives

$
\int_2^4 \frac{\sqrt{\ln 9-x} + \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$
$+ \int_2^4 \frac{ \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$

Let $x = 6-u$ in the second integral

$\int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$ $- \int_4^2 \frac{ \sqrt{\ln 9-u} }{\sqrt{\ln 3+u} + \sqrt{\ln 9-u}}\,dx$

$= 2 \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 2$

$\int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 1$

3. Originally Posted by danny arrigo
Consider

$
\int_2^4 \frac{\sqrt{\ln 9-x} + \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = \int_2^4 1 = 2
$

Expanding gives

$
\int_2^4 \frac{\sqrt{\ln 9-x} + \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$
$+ \int_2^4 \frac{ \sqrt{\ln x+3} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$

Let $x = 6-u$ in the second integral

$\int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx$ $- \int_4^2 \frac{ \sqrt{\ln 9-u} }{\sqrt{\ln 3+u} + \sqrt{\ln 9-u}}\,dx$

$= 2 \int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 2$

$\int_2^4 \frac{\sqrt{\ln 9-x} }{\sqrt{\ln 9-x} + \sqrt{\ln x+3}}\,dx = 1$

I would have never thought of that. Thank you very much!

4. It is very simple actually.Look at this standard property in definite integral
$
\int_a^bf(x)dx=\int_a^bf(a+b-x)dx
$

and the rest is obvious

Interesting corollary to above

$\int_a^b\frac{f(x)}{f(x)+f(a+b-x)}dx=\frac{b-a}{2}$

5. Originally Posted by pankaj
It is very simple actually.Look at this standard property in definite integral
$
\int_a^bf(x)dx=\int_a^bf(a+b-x)dx
$

and the rest is obvious

Interesting corollary to above

$\int_a^b\frac{f(x)}{f(x)+f(a+b-x)}dx=\frac{b-a}{2}$
The first part of this is in Stewart's Calculus book (problems plus).


# Probability of drawing exactly $1$ ace upon drawing $2$ cards from a deck

I got this question and answered it incorrectly. I haven't yet seen the correct answer. The possible answers were:

1. $\frac{4}{52}$
2. $\frac{16}{221}$ (my answer)
3. $\frac{2}{52}$
4. $\frac{32}{221}$

My reasoning is the following:

Event A: Card is not an ace.

Event B: Card is an ace.

$$P(A)\times P(B|A)=\frac{4}{52}\times \frac{48}{51}=\frac{16}{221}$$

This is assuming the cards are drawn sequentially. Drawing exactly one ace from a single draw is $4/52$ but to ensure that only a single ace was drawn one should consider the probability of not getting a second ace.

How am I wrong?

• What if the first card is not an ace but the second is? – Ethan Bolker Dec 25 '17 at 16:21
• If you are Lemmy, then in probability 1 you will draw the ace of spades. – Asaf Karagila Dec 26 '17 at 8:18
• You can also do it in another way (as a check). The probability of two aces is $\frac{4}{52}\cdot\frac{3}{51}$. The probability of zero aces (that is two non-aces), is $\frac{48}{52}\cdot\frac{47}{51}$. So the answer should be $$1-\frac{4}{52}\cdot\frac{3}{51}-\frac{48}{52}\cdot\frac{47}{51}$$ – Jeppe Stig Nielsen Dec 26 '17 at 13:44

You forgot to assume $P(B)\times P(A|B)=\frac{48}{52}\times\frac4{51}=\frac{16}{221}$.

So $P(A)\times P(B|A)+P(B)\times P(A|B)= \frac{16}{221}+ \frac{16}{221}= \frac{32}{221}$.

The correct answer is $4$ , i.e. $\frac{32}{221}$

• Why? Isn't it the same thing, just reversed? – Rivasa Dec 25 '17 at 21:02
• @Annabelle, there are two possible ways you can get it, so you account for both. Thinking about it intuitively, because there aren’t many aces, you would expect your chances to be better by taking two cards than by taking one. Otherwise, if you kept taking more and more, your chances wouldn’t change, which would be strange when you got to 50 cards and are guaranteed to get two aces... – Jeff Dec 25 '17 at 21:12
• @Annabelle: Consider that you have a deck of 3 cards, 2 aces. You draw two cards. What's the probability of drawing exactly one ace? Well, the event "drawing one" is equivalent to "an ace remains in the deck after drawing two cards". What is the probability of that? Clearly 2/3 as there are two aces in a deck of three if you imagine randomly picking a card to remain in the deck after drawing. But computing it as the OP did here you would end with P=1/3 because of mistakenly not considering that both drawing an ace first and drawing an ace second must be taken into account. – Jon Dec 25 '17 at 21:58
• I think I get it now... @Jon – Rivasa Dec 26 '17 at 2:46

We have: $$P_{\text{ only one ace }} = P_{\text{ ace }1} P_{\text{ non-ace }2} + P_{\text{ non-ace }1} P_{\text{ ace }2} = \frac{4}{52}\times \frac{48}{51} + \frac{48}{52}\times \frac{4}{51} = \frac{32}{221}$$

• Why the downvote? – user371838 Dec 26 '17 at 13:40

Answer 4 seems correct to me because:

$$P=\frac {\binom4 1\binom {48} 1}{\binom {52} 2}=\frac {32}{221}$$

You can figure this out without any knowledge of conditional probabilities:

There are 52 choices for the first pick and 51 choice for the second pick. So, in total, there are $52 \times 51 = 2652$ possible "hands" that consist of two cards.

Now, the question is, how many of these 2652 contain a single ace. Let's do some counting of the possible one-ace hands:

There are $4 \times 48 = 192$ hands that have an ace as their first pick and a non-ace as the second pick (because there are 4 aces and 48 non-aces). Similarly, there are $48 \times 4 = 192$ hands that have a non-ace as their first pick and an ace as the second pick. So the total number of hands that have a single ace is $192+192=384$.

So, the probability of a one-ace hand is $384/2652$. Dividing top and bottom by $12$ gives us $384/2652 = 32/221$.

• Does the downvoter care to explain his or her reasoning? – bubba Dec 27 '17 at 2:51


# Number of different permutations

Consider some text $T$

How many different permutations of this text can we achieve ?

The easiest case is when every letter appears only once in the text, so the answer is $|T|!$

But when we have for example $T=aab$ the number of different permutations is 3, because $aab$ $aba$ $baa$

How can we solve this problem without actually generating all possible permutations of this text ?

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It is given by the multinomial theorem:

$$\frac{N!}{m_1! m_2! m_3! \ldots m_k!} = \text{choose}(n, (m_1,m_2, \ldots, m_k))$$

Here $N$ is the number of elements in $T$, $|T|=N$. If there are $k$ unique elements in $T$, $m_i$ is the number of times that element shows up. From your example, $N=3$, $m_1 = 2$, $m_2 = 1$, so we get

$$\frac{3!}{2! \cdot 1!} = \frac{6}{2} = 3$$

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Thank you, i didn't know it's that easy :) –  Chris Feb 1 '12 at 16:45

An example might help. Consider the text $probability$. If we thought of all letters as distinct: $p\ r\ o\ b_1\ a\ b_2\ i_1\ l\ i_2\ t\ y$, then there would be $11!$ permutations.

But this overcounts, as not all letters are distinct. In particular, amongst the $11!$ factorial arrangements above, we have both \eqalign{ &r\ p\ o\ b_1\ a\ b_2\ \color{maroon}{i_1}\ l\ \color{maroon}{i_2}\ t\ y\cr &r\ p\ o\ b_1\ a\ b_2\ \color{maroon}{i_2}\ l\ \color{maroon}{i_i}\ t\ y\cr} For any particular arrangement amongst the $11!$ arrangements of the letters, we can find one other (by permuting the $i$'s) that is the same when we regard the $i$'s as indistinguishable.

So we've over counted by a factor of 2. Similarly, because there are two $b$'s, we overcounted by another factor of 2.

So, the correct number of arrangements is $11!\over 2\cdot2$.

A similar argument can be made to establish the formula that Hooked gives in his answer. In particular if one letter was repeated thrice, then if we found the number of arrangements where we thought of all letters as distinct, we would have over counted by a factor of $3!$; since there are $3!$ ways to permute the three repititions of the letter.

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The general formula for a permutation of a text with length $n$ with $n_1$ indistinguishable objects of one type, $n_2$ of the next type is:

$$\frac{n!}{n_1!n_2!\dots n_k!}$$

For your text: aab n=3, $n_1$(the number of times a appears)=2 and $n2=1$(Since b appears once). This formula simply divides out all of the objects that $n!$ overcounts.

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# Math Help - Simplifying fractions

1. ## Simplifying fractions

This is the problem

Simplify the following equation:
(5x/(2x^2+7x+3))-((x-3)/(2x^2-3x-2))+((2x+1)/(x^2+x-6))

The way i did it was factoring the denominators then adding them, therefore getting common terms where the denominator would be (2x+1)(x+3)(x-2). After that, I don't know what to do next, I'm stumped XD

Would appreciate the help..

2. Hello, mcroldan08!

Simplify the following expression:

. . $\frac{5x}{2x^2+7x+3} - \frac{x-3}{2x^2-3x-2} + \frac{2x+1}{x^2+x-6}$
It sounds like your intentions were correct.
But since you didn't show us what you did, we can't tell if you made any errors.

We have: . $\frac{5x}{(x+3)(2x+1)} - \frac{x-3}{(x-2)(2x+1)} + \frac{2x+1}{(x-2)(x+3)}$

You are correct . . . the LCD is: . $(x+3)(2x+1)(x-2)$

And we must convert the fractions to the LCD.

$\frac{5x}{(x+3)(2x+1)}\cdot{\color{blue}\frac{x-2}{x-2}} - \frac{x-3}{(x-2)(2x+1)}\cdot{\color{blue}\frac{x+3}{x+3}} + \frac{2x+1}{(x-2)(x+3)}\cdot{\color{blue}\frac{2x+1}{2x+1}}$

. . $= \;\frac{5x(x-2)}{(x+3)(2x +1)(x-2)} - \frac{(x-3)(x+3)}{(x-2)(2x+1)(x+3)} + \frac{(2x+1)(2x+1)}{(x-2)(x+3)(2x+1)}$

. . $= \;\frac{5x(x-2) - (x-3)(x+3) + (2x+1)(2x+1)}{(x+3)(2x+1)(x-2)}$

. . $= \;\frac{5x^2 - 10x - x^2 + 9 + 4x^2 + 4x + 1}{(x+3)(2x+1)(x-2)}$

. . $= \;\frac{8x^2 - 6x + 10}{(x+3)(2x+1)(x-2)}$

. . $= \;\frac{2(4x^2 - 3x + 5)}{(x+3)(2x+1)(x-2)}$

3. That is actually what I got... however, I was wondering if this can be simplified further. This is actually a multiple choice question and the choices are:

a.) 4/(x+3)
b.) 4/(x-3)
c.) 2/(x-3)
d.) 2/(x+3)

Could it be that the none of the choices are correct?


# Math Help - Rebound Superball Problem

1. ## Rebound Superball Problem

The rebound ratio of a speckled green superball is 75%. It is dropped from a height of 16 feet. Consider the instant when the ball strikes the ground for the fiftieth time.
a. How far downward has the ball traveled at this instant?
b. How far ( upward and downward has the ball traveled at this instant?
c. How far would the ball travel if you just let it bounce...?

I have a feeling for the first two it would be something along the lines of "16(0.75)^0+16(0.75)^1...+16(0.75)^50", and for the second 2(" "), but there must be something that would allow me to do this calculation quicker than just adding 50 numbers together. Would it have something to do with ! ?

Thanks.

2. Originally Posted by bkap
The rebound ratio of a speckled green superball is 75%. It is dropped from a height of 16 feet. Consider the instant when the ball strikes the ground for the fiftieth time.
a. How far downward has the ball traveled at this instant?
b. How far ( upward and downward has the ball traveled at this instant?
c. How far would the ball travel if you just let it bounce...?

I have a feeling for the first two it would be something along the lines of "16(0.75)^0+16(0.75)^1...+16(0.75)^50", and for the second 2(" "), but there must be something that would allow me to do this calculation quicker than just adding 50 numbers together. Would it have something to do with ! ?

Thanks.
Yes your approach to the first problem is correct, if you factorize things out a bit, it looks something like:
$16*((3/4)^0+(3/4)^1+...+(3/4)^{49})$
it ends with the 49th power because 0-49 gives a total of 50 drops
now this is just a standard progression, whose sum is:
$a(r^{n+1}-1)/(r-1)$
where a is the first term, in this case 1, r is the successive ratio, in this case 3/4, and n is how far the last term goes, in this case 49.

The second part is trivial once you got the first part, however you have to keep in mind that the first upward bounce is the same distance as the second downward bounce.

Finally for the third part, you simply let n tend to infinity and calculate. If you haven't done limit, just make n=1000 on the calculator, this will give you a pretty good approximation

3. Originally Posted by bkap
The rebound ratio of a speckled green superball is 75%. It is dropped from a height of 16 feet. Consider the instant when the ball strikes the ground for the fiftieth time.
a. How far downward has the ball traveled at this instant?
b. How far ( upward and downward has the ball traveled at this instant?
c. How far would the ball travel if you just let it bounce...?

I have a feeling for the first two it would be something along the lines of "16(0.75)^0+16(0.75)^1...+16(0.75)^50", and for the second 2(" "), but there must be something that would allow me to do this calculation quicker than just adding 50 numbers together. Would it have something to do with ! ?

Thanks.
Nah, Factorials only really occur in probability - you'll be better off with a sequence. Let s be the distance it bounces back up

$s_0: 16 = 16 \times 0.75^0$
$s_1: 16 \times 0.75^1 = 0.75s_0$
$s_2: (16 \times 0.75) \times 0.75 = 0.75s_1$
...
$s_n = 16 \times 0.75^n = 0.75s_{n-1}$

From this we can see it's a geometric sequence so use the sum of a geometric sequence to solve how far it's bounced up. To find down and up multiply by 2

4. Hello, bkap!

Everyone has given you excellent advice . . .

The rebound ratio of a speckled green superball is 75%.
It is dropped from a height of 16 feet.
Consider the instant when the ball strikes the ground for the 50th time.

a. How far downward has the ball traveled at this instant?

Total downward distance is given by:

. . $D \;=\;16 + 16(.075) + 16(.075)^2 + 16(0.75)^3 + \hdots + 16(0.75)^{49}$

. . . . . $=\;16\underbrace{\bigg[1 + (0.75) + (0.75)^2 + (0.75)^3 + \hdots + (0.75)^{49}\bigg]}_{\text{geometric series}}$

. . The sum of the geometric series is: . $\frac{1-(0.75)^{50}}{1-0.75} \:=\:3.999997735$

Therefore: . $D \;=\;16(3.999997736) \;=\;63.99996376$ feet.

b. How far ( upward and downward) has the ball traveled at this instant?
We must be very careful . . .

The balls falls 16 feet.

Then bounces up $16(0,75)$ feet and falls $16(0.75)$ feet.

Then bounces up $16(0.75)^2$ feet and falls $16(0.75)^2$ feet . . . and so on.

The total distance is:

. . $T \;=\;16 + 2\!\cdot\!16(0.75) + 2\!\cdot\!16(0.75)^2 + 2\!\cdot\!16(0.75)^3 + \hdots + 2\!\cdot\!16(0.75)^{49}$

. . . . $=\; 16 + 32(0.75)\bigg[1 + (0.75) + (0.75)^2 + \hdots + (0.75)^{49}\bigg]$

. . . . $=\; 16 + 24(3.999997735) \;=\;111.9999275$ feet.

c. How far would the ball travel if you just let it bounce?
If we let it bounce (forever), the total distance is:

. . $T \;=\;16 + 24\underbrace{\bigg[1 + (0.75) + (0.75)^2 + (0.75)^3 + \hdots\bigg]}_{\text{infinite geomtric series}}$

. . The sum of the infinite series is: . $\frac{1}{1-0.75} \:=\:4$

Therefore: . $T \;=\;16 + 24(4) \;=\;112$ feet.