Fishermen

• Feb 28th 2010, 08:52 PM
harish21
Fishermen
Three fisherman come back from a day of fishing and put their fish together and go to bed. The 1st one wakes up goes to the pile of fish, throws one away and takes a 3rd and leaves. The 2nd one does the same. The last one does the same as well. What is the minimum number of fish that this can work with?
• Feb 28th 2010, 09:56 PM
Soroban
Hello, harish21!

Quote:

Three fisherman come back from a day of fishing and put their fish together and go to bed.
The 1st one wakes up, goes to the pile of fish, throws one away and takes a 3rd and leaves.
The 2nd one does the same. The last one does the same as well.
What is the minimum number of fish that this can work with?

Let $N$ = total number of fish.

The 1st discards one fish and take a third of the remainder.
Then $N$ must be of the form: . $3A + 1$
. . $N \:=\:3A + 1$ .[1]

He discards one fish . . . $3A$ fish left.
He take a third of the fish ( $A$ fish) and leaves.
There are $2A$ fish left.

The 2nd discards one fish and takes a third of the remainder.
Then $2A$ must be of the form $3B + 1$
. . $2A \:=\:3B+1$ .[2]

He discards one fish . . . $3B$ fish left.
He take third of the fish ( $B$ fish) and leaves.
There are $2B$ fish left.

The 3rd discards one fish and takes a third of the remainder.
Then $2B$ must be of the form $3C + 1$
. . $2B \:=\:3C + 1 \quad\Rightarrow\quad B \:=\:\frac{3C+1}{2}$ .[3]

Substitute [3] into [2]: . $2A \;=\;3\left(\frac{3C+1}{2}\right) + 1 \quad\Rightarrow\quad A \:=\:\frac{9C+5}{4}$ .[4]

Substitute [4] into [1]: . $N \;=\;3\left(\frac{9C+15}{4}\right) + 1 \;=\;\frac{27C + 19}{4}$

We have: . $N \;=\;\frac{24C + 16 + 3C + 3}{4} \;=\;\frac{24C+16}{4} + \frac{3C+3}{4}$

Hence: . $N \;=\;6C+4 + \frac{3(C+1)}{4}$

Since $N$ is an integer, $C+1$ must be divisible by 4.
The first time this happens is when $C = 3.$

Therefore: . $N \;=\;6(3) + 4 + \frac{3(3+1)}{4} \quad\Rightarrow\quad \boxed{N \:=\:25}$

• Mar 1st 2010, 12:26 AM
harish21
Quote:

Originally Posted by Soroban
Hello, harish21!

Let $N$ = total number of fish.

The 1st discards one fish and take a third of the remainder.
Then $N$ must be of the form: . $3A + 1$
. . $N \:=\:3A + 1$ .[1]

He discards one fish . . . $3A$ fish left.
He take a third of the fish ( $A$ fish) and leaves.
There are $2A$ fish left.

The 2nd discards one fish and takes a third of the remainder.
Then $2A$ must be of the form $3B + 1$
. . $2A \:=\:3B+1$ .[2]

He discards one fish . . . $3B$ fish left.
He take third of the fish ( $B$ fish) and leaves.
There are $2B$ fish left.

The 3rd discards one fish and takes a third of the remainder.
Then $2B$ must be of the form $3C + 1$
. . $2B \:=\:3C + 1 \quad\Rightarrow\quad B \:=\:\frac{3C+1}{2}$ .[3]

Substitute [3] into [2]: . $2A \;=\;3\left(\frac{3C+1}{2}\right) + 1 \quad\Rightarrow\quad A \:=\:\frac{9C+5}{4}$ .[4]

Substitute [4] into [1]: . $N \;=\;3\left(\frac{9C+15}{4}\right) + 1 \;=\;\frac{27C + 19}{4}$

We have: . $N \;=\;\frac{24C + 16 + 3C + 3}{4} \;=\;\frac{24C+16}{4} + \frac{3C+3}{4}$

Hence: . $N \;=\;6C+4 + \frac{3(C+1)}{4}$

Since $N$ is an integer, $C+1$ must be divisible by 4.
The first time this happens is when $C = 3.$

Therefore: . $N \;=\;6(3) + 4 + \frac{3(3+1)}{4} \quad\Rightarrow\quad \boxed{N \:=\:25}$

Awesome Soroban! I was given this problem in one of my classes and found it very interesting. I thought it would be nice to share. 25 is indeed the logical answer. However -2 is also a correct answer to this problem. I forgot the name of the scientist who gave an answer of -2 to this problem!!
• Mar 1st 2010, 03:08 PM
icemanfan
If the total number of fish is 25, then:

The first person throws one away (total down to 24) and takes a third (total down to 16).

The second person throws one away (total down to 15) and takes a third (total down to 10).

The third person throws one away (total down to 9) and takes a third (total down to 6).

If every time someone takes a third of the fish, the number remaining must be a whole number, then 25 is the minimum number. Without that restriction, the minimum number of fish is 4.75. In both cases I assume you're not allowed to have a negative number of fish.
• Mar 1st 2010, 07:26 PM
Wilmer


# Convergence and divergence of $\frac{a_n}{n}$ as $n\to\infty$

Suppose that a sequence $\{a_n\}_{n\geqslant 1}$ satisfies:

$$a_{m+n}\leqslant a_m+a_n$$

for all integers $m,n\geqslant 1$. Show that $\frac{a_n}{n}$ either converges or diverges to $-\infty$ as $n\to \infty$.

Resolution: For an arbitrarily fixed positive integer $k$ we put $n=qk+r$ with $0\leqslant r<k$. Applying the given inequality for $q$ times we get $a_n=a_{qk+r}\leqslant qa_k+a_r$; so,

$\frac{a_n}{n}\leqslant\frac{a_k}{k}+\frac{a_r}{n}$ Taking the limit as $n\to\infty$, we get

$\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant\frac{a_k}{k}$.

The sequence $\frac{a_n}{n}$ is therefore bounded above. Since $k$ is arbitrary, we conclude that:

$\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant \inf_{k\geqslant 1}\frac{a_k}{k}\leqslant\lim \inf_{k\to\infty}\frac{a_k}{k}$,

which concludes the proof.

Question:

1) First the author says that $k$ is a fixed positive integer. However in the last expression $\lim \inf_{k\to\infty}\frac{a_k}{k}$ the author makes $k$ vary. How is this supposed to prove that $\frac{a_n}{n}$ converges since the behaviour of $\lim \inf_{k\to\infty}\frac{a_k}{k}$ is not known? Why is $k$ allowed to vary?

2)On the question it is mentioned the divergence to $-\infty$. How was that point addressed in the answer or resolution?

• Is this from Hata's problem book ? Sep 5 '18 at 13:06
• Yes it is. What do you think of the problems in that book? I am trying to solve them but I find them hard. Sep 5 '18 at 13:07
• Sep 5 '18 at 13:12
• @RobertZ It may be the same problem, but the questions seem to differ in my opinion. Sep 5 '18 at 13:13
• Answer: that is not taking the limit. Actually you could use other alphabets, that does not affect the $\liminf a_\nu / \nu$. By definition, $$\liminf \frac {a_n} n = \lim_n \inf_{k \geqslant n} \frac {a_k}k,$$ and you could prove that for any sequence $x_n$, $\inf_{k \geqslant n} x_k$ is increasing, hence $\inf_{k \geqslant 1} x_k \leqslant \inf_{k \geqslant n} x_k \to \liminf x_k [n \to \infty].$
– xbh
Sep 5 '18 at 13:21

The author first considers a fixed positive $k$ and proves $\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant\frac{a_k}{k}$. Consequently, $\sup_n \frac{a_n}n < \infty$, hence $\left(\frac{a_n}{n}\right)_n$ is bounded above.

Since $\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant\frac{a_k}{k}$ is valid for every $k$, by definition of the infinimum, $\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant\inf_k\frac{a_k}{k}$. By definition of $\liminf$, one also has $\inf_k\frac{a_k}{k}\leq \liminf_k \frac{a_k}{k}$.

Thus $\liminf_k \frac{a_k}{k} = \limsup_k \frac{a_k}{k} = \inf_k \frac{a_k}{k}$, thus $\frac{a_n}{n}$ converges to a limit in $\mathbb R\cup \{-\infty, \infty\}$.
Since $\frac{a_n}{n}$ is bounded above, $\inf_k \frac{a_k}{k}<\infty$, but it could very well be equal to $-\infty$. Thus the limit belongs to $\mathbb R\cup \{-\infty\}$

Question 1)

He proves that $\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant\frac{a_k}{k}$ is true for any and all natural numbers $k$. In proving that inequality, $k$ is fixed (so it's only proven for "one $k$ at a time"), but once it's proven, you're allowed to take it for granted for any value of $k$ you'd like, and let $k$ vary.

Question 2)

Note that the $\limsup$ on the left side of $\lim \sup_{n\to\infty}\frac{a_n}{n}\leqslant\lim \inf_{k\to\infty}\frac{a_k}{k}$ and the $\liminf$ on the right side are actually of the same sequence. So the $\limsup$ of the sequence is less than or equal to the $\liminf$ of that sequence. This implies that they must be equal, and either they're finite (in which case the sequence has a limit), or they're not (in which case the sequence diverges to $-\infty$, since it's bounded above).

Answer: that is not taking the limit $k \to \infty$. Actually you could use other alphabets, that does not affect the quantity $\liminf a_\nu / \nu$. By definition, $$\liminf \frac {a_n} n = \lim_n \inf_{k \geqslant n} \frac {a_k}k,$$ and you could prove that for any sequence $x_n$, $\inf_{k \geqslant n} x_k$ is increasing in $n$, hence for all $n \in \mathbb N^*$, $\inf_{k \geqslant 1} x_k \leqslant \inf_{k \geqslant n} x_k$ Since $\inf_{k \geqslant n}x_k \to \liminf x_k [n \to \infty]$, we have $\inf_{k \geqslant 1}x_k \leqslant \liminf x_n$.

Additionally, $\liminf x_n$ always exists or equals $\infty$, so you can write down $\liminf x_n$ whenever you want.


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# What percent of the mixture is Solution X?

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What percent of the mixture is Solution X?  [#permalink]

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08 Dec 2012, 11:49
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Question Stats:

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Solution X is 20% chemical A and 80% chemical B by volume; Solution Y is 30% chemical A and 70% chemical B. If a mixture of X and Y is 22% chemical A, what percent of the mixture is Solution X?

A) 20%
B) 44%
C) 50%
D) 80%
E) 90%

Can someone please also direct me to a page in gmatclub that has similar mixture problems? Thanks!
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Joined: 22 Dec 2011
Posts: 261
Re: What percent of the mixture is Solution X?  [#permalink]

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08 Dec 2012, 12:03
1
2
JJ2014 wrote:
Solution X is 20% chemical A and 80% chemical B by volume; Solution Y is 30% chemical A and 70% chemical B. If a mixture of X and Y is 22% chemical A, what percent of the mixture is Solution X?

A) 20%
B) 44%
C) 50%
D) 80%
E) 90%

Can someone please also direct me to a page in gmatclub that has similar mixture problems? Thanks!

you can solve this by allegation method.
Given 22% of A is the final mixture; Sol X has 20% chemical A and Sol Y has 30 % of A. (you don't need Chemical B to solve when you r solving the problem using this method)

$$X / Y = (30-22 ) / (22-20)$$
$$X / Y = 8/2 = 4/1$$

So X and Y are in the ratio 4 : 1
Question asks percent of the mixture is Solution X.

So $$X / Total = 4/5 *100 = 80%$$

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Re: What percent of the mixture is Solution X?  [#permalink]

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08 Dec 2012, 12:16
3
also you can solve this one with the concept of weighted average or balance point

you care about only of chemical A in both X and Y

$$20$$------- $$22$$ ------------------------$$30$$

Now the difference between 20 and 22 = 2; between 22 and 30 = 8

So $$8y = 2 x$$

$$\frac{8}{2}$$ $$=$$$$\frac{x}{y}$$
Our ratio is 2 and 8 and using the concept of unknown multiplier the sum is 10. Our X is 8 on a total of 10 so 80 %

much more difficult to say that to explain. In that way you can solve a question difficult like this in 30 seconds.

Also algebraic approach works fine, but this is a bit faster
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Re: What percent of the mixture is Solution X?  [#permalink]

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09 Dec 2012, 01:29
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Re: What percent of the mixture is Solution X?  [#permalink]

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09 Dec 2012, 06:31
2
JJ2014 wrote:
Solution X is 20% chemical A and 80% chemical B by volume; Solution Y is 30% chemical A and 70% chemical B. If a mixture of X and Y is 22% chemical A, what percent of the mixture is Solution X?

A) 20%
B) 44%
C) 50%
D) 80%
E) 90%

Can someone please also direct me to a page in gmatclub that has similar mixture problems? Thanks!

22% of chemical A in X+Y grams of solution comes from 20% chemical A in solution X and 30% chemical A in solution Y, thus:

0.22(X + Y) = 0.2X + 0.3Y --> X = 4Y --> X/(X+Y)=4/5=0.8.

Check out question banks for similar problems:
DS mixture problems: search.php?search_id=tag&tag_id=43
PS mixture problems: search.php?search_id=tag&tag_id=114

Hope it helps.
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Re: What percent of the mixture is Solution X?  [#permalink]

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09 Mar 2013, 00:14
1
JJ2014 wrote:
Solution X is 20% chemical A and 80% chemical B by volume; Solution Y is 30% chemical A and 70% chemical B. If a mixture of X and Y is 22% chemical A, what percent of the mixture is Solution X?

A) 20%
B) 44%
C) 50%
D) 80%
E) 90%

Can someone please also direct me to a page in gmatclub that has similar mixture problems? Thanks!

Focus on Chemical A:

X--------------------------------------Y
20%---------------------------------30%

--------------------22%-----------------

30 - 22= 8--------------------22-20 = 2

X:Y = 8:2 = 4:1.

Therefore, %age of X = 4/5 * 100 = 80%
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Re: What percent of the mixture is Solution X?  [#permalink]

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02 Feb 2015, 13:11
Hi All,

These types of mixture questions can typically be solved with a variety of approaches (the average formula, ratios, TESTing VALUES, TESTing THE ANSWERS, etc.). Here's the standard Weighted Average Formula approach:

Solution X = 20% chemical A
Solution Y = 30% chemical A

Mixture of the two solutions = 22% chemical A

X = number of ounces of Solution X
Y = number of ounces of Solution Y

(.2X + .3Y) / (X + Y) = .22

.2X + .3Y = .22X + .22Y
.08Y = .02X

We can multiply both sides by 100 to get rid of the decimals...

8Y = 2X

The question asks for the percentage of the new mixture that is Solution X....

8/2 = X/Y
4/1 = X/Y

80% X and 20% Y

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Re: What percent of the mixture is Solution X?  [#permalink]

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27 Oct 2016, 16:22
1
0.20x+(1-x)0.30 = 0.22
0.20x+0.30-0.30x=0.22
-0.10x=-0.08
x=4/5 = 80%
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27 Oct 2016, 16:40
If the solution is 22% of A we know off the bat that the majority of the solution has to be X. We can eliminate A,B,C.

Solution X is 20% A and Y is 30

So X is 2 units from the combo and Y is 8 units away.

Ration of X:Y for solution A is 8:2
8/10=80%
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# Mathematica for linear algebra course?

I'm taking a linear algebra / matrix theory course and we are free to use any software we want, and will be "expected to use MATLAB or an equivalent" for homework. The professor and textbook (Applied Linear Algebra) prefer MATLAB, some students seem to like R, but I'm already familiar with Mathematica and have a student license.

Are there any possible limitations of Mathematica to watch out for, if used as a learning device (i.e. not for work or intensive computation) for a linear algebra course? e.g. Are there some operations which the other software packages can do which Mathematica cannot out-of-the-box?

• Could you mention the textbook you're using? Sep 8, 2012 at 7:36
• There is a book focused on teaching linear algebra with Mathematica, which you can get used from Amazon. It is pretty dated, and I don't know how good it is, but if you decide to go with Mathematica, it may be of some help. Sep 8, 2012 at 9:00
• If prices of licence is the issue you might want to try python numpy library as well. Or Octave for Matlab clone. Sep 8, 2012 at 12:34
• @enedene I find freemat or scilab to have a more reasonable frontend than octave, an they're equally free (although maybe their licenses are more restrictive, I don't know). overall I'd go with matlab or python if not mathematica though. in any case as he has a student license I don't think this is important.
– acl
Sep 8, 2012 at 13:05
• @Nasser Agree 100%. This was and still is my most difficult transition coming from Excel and MATLAB was the "Lists of lists" concepts. Also, MATLAB has some cool notation for solving linear systems such as Ax=b can be solved by A\b.
– kale
Sep 8, 2012 at 13:17

Mathematica offers a pretty complete set of functionality for linear algebra, and it has improved in recent versions.

For example, since version 5, Mathematica has offered the generalised Schur decomposition (also known as the QZ decomposition). This certainly wasn't available in earlier versions. It handles sparse matrices and many other wrinkles. And if it isn't available, this example shows that it can take a lot less code than in many other languages, so you are less likely to get yourself tangled up in loop constructs and the like.

Rather than focus on limitations, I thought it worth mentioning that Mathematica offers something neither Matlab nor R can, that makes it ideal as a pedagogical tool: integrated symbolics. If you are learning linear algebra and you want to understand certain things, you can in many cases pass a symbolic matrix to the relevant function, and use the resulting output to gain a deeper understanding.

Here's a simple example:

aa = Array[a, {2, 2}]

(* {{a[1, 1], a[1, 2]}, {a[2, 1], a[2, 2]}}*)

Eigenvalues[aa]

(*{1/2 (a[1, 1] + a[2, 2] - Sqrt[
a[1, 1]^2 + 4 a[1, 2] a[2, 1] - 2 a[1, 1] a[2, 2] + a[2, 2]^2]),
1/2 (a[1, 1] + a[2, 2] + Sqrt[
a[1, 1]^2 + 4 a[1, 2] a[2, 1] - 2 a[1, 1] a[2, 2] + a[2, 2]^2])}*)

Of course, sometimes the symbolic output is pretty hard to trace through (consider SingularValueDecomposition[aa] for the aa defined above!), but in can be useful in some cases.

Some pages on the internet contain claims that Mathematica is slower than Matlab, or that Matlab is somehow "better" for numerical work (and implicitly, Mathematica is only good for a bit of symbolic solving or something). As this answer shows, this claim is no longer true. It was true until about version 5, but no longer.

• Congrats on your 10k! Sep 8, 2012 at 8:55

Having taught linear algebra using both Mathematica and Matlab, I concur with what others have said that the Mathematica's features for linear algebra include all one might need for a course in undergraduate linear algebra. Since symbolic computation is also fully integrated into Mathematica, it might be better in some ways. For example, we can solve symbolic systems of small dimension fairly easily.

LinearSolve[{{a, b}, {c, d}}, {e, f}]

Note how the determinant appears in the denominator illustrating its importance in determining when a system is solvable. This can be done just as easily for a 3x3 or 4x4 system.

Also, there is a common mis-conception that Mathematica does not or cannot distinguish between row and column vectors, as illustrated by Nasser's comment. I think this is not correct. It's just that if you want to represent a row or column vector, you should use the full matrix representation of said vector. Consider, for example, the following two dot product computations (whose orders cannot be reversed):

{{a, b}, {c, d}}.{{x}, {y}} // MatrixForm

{{x, y}}.{{a, b}, {c, d}} // MatrixForm

In addition, though, Mathematica provides a consistent notion of dot product between dimensions based on tensor products. Here's the product of a rank 3 tensor with a rank 2 tensor.

Table[a[i, j, k], {i, 1, 2}, {j, 1, 2}, {k, 1, 2}].
Table[b[i, j], {i, 1, 2}, {j, 1, 2}]

By contrast, Matlab is limited to 2D floating point matrices. When you type x=2 in Matlab, you've defined a 2D matrix; the equivalent definition in Mathematica would be x={{2}}.

• @NasserM.Abbasi I guess I disagree with the documentation. The example shows that, if you represent a vector using a rank 1 tensor, then you can't distinguish between left and right multiplication. I'm saying that you can distinguish between left and right multiplication, if you represent a vector as a rank 2 tensor. Sep 9, 2012 at 11:10
• @NasserM.Abbasi Of course, you can define higher rank objects in Matlab. You can even use cell arrays to define fairly arbitrary objects. I guess I mean to say that the default in Matlab is a 2D float and support beyond that is lacking. For example, the 3D and 4D matrices that you define (which we'd call rank 3 and 4 tensors in the Mathematica context) cannot be multiplied using the * operator in Matlab. The corresponding tensors can be multiplied in Mathematica using Dot. Sep 9, 2012 at 11:15
• For better or worse, vectors, matrices, etc., in Mathematica are just lists. For pedagogical purposes this is not necessarily optimal. Before I taught linear algebra with Mathematica, I did so with APL and later J. In each of those languages, a vector is a 1-dimensional creature, whereas a row vector or a column vector is a 2-dimensional creature, namely, a 1-row or 1-column matrix. Sep 9, 2012 at 15:04

In years past I've taught a standard-content sophomore-level (in U.S.) linear algebra course where students used Mathematica. I know, then, that if you're interested, or if it's a requirement, you can build everything up from the simplest functions for manipulating matrices or you can directly use powerful built-in Mathematica functions (or a combination of both).

E.g., you might begin by defining little utility functions to scale a row, to swap two rows, and to add one row to another; from that, you can build a function to do Gauss-Jordan row reduction. Or you can directly use RowReduce.

Again, you might define a function to test for invertibility and, when invertible, find the inverse of a matrix by adjoining an identity matrix, row reducing, etc. Or you can directly use Inverse.

Even for something more complicated like Gram-Schmidt orthonormalization, you can build up a function for this yourself, beginning by defining a function that projects a vector onto the span of a given set of vectors. Or use Orthogonalize.

You could even define your own determinant function (in one of several ways), then use that to find eigenvalues, etc.

If your course involves advanced operations such as LU or QR decompositions, again you can define functions for these yourself or else use the built-in ones.

Of course in general you would expect built-in functions to handle round-off issues more robustly than anything you would program yourself. But this would be the same with Matlab.

So yes, Mathematica would be an entirely appropriate tool in place of Matlab.

• is there any chance you can add a link to the course notes?
– user21
Sep 9, 2012 at 1:24
• @ruebenko: I considered doing so and still could. But the notebooks are 10 years old now, and some of the files are .m packages encoded for the Windows/PC platform. (Students would define a function, then run an encoded package that tested it with various sets of data against my "correct" but encoded version of the function.) Sep 9, 2012 at 15:00


## Integral of polynomial times rational function of trig function over multiple periods

Problem: Compute

$$\int_0^{12\pi} dx \frac{x}{6+\cos 8x}$$

It looks like the OP is trying to compute the antiderivative and use the fundamental theorem of calculus. With multiple periods, that approach is paved with all sorts of difficulty that is really an artifice related to the functional form of the antiderivative. In truth, there should be no such difficulty. A better approach involves the residue theorem, which I will outline below. Note that the presence of the linear term in the numerator provides a slight complication, but one that has been treated before in this site several times.

$$\int_{0}^{12 \pi} dx \frac{x}{6+\cos{8 x}} = \frac1{64} \int_0^{96 \pi} du \frac{u}{6+\cos{u}}$$

$$\int_{n 2 \pi}^{(n+1) 2 \pi} du \frac{u}{6+\cos{u}} = \int_0^{2 \pi} dv \frac{v+2 \pi n}{6+\cos{v}} = \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} + 2 \pi n \int_0^{2 \pi} \frac{dv}{6+\cos{v}}$$

Thus, summing over $n$:

$$\int_{0}^{12 \pi} dx \frac{x}{6+\cos{8 x}} = \frac{48}{64} \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} + \frac{48 (47) \pi}{64} \int_0^{2 \pi} \frac{dv}{6+\cos{v}}$$

We may now compute each of these integrals. The second integral is straightforward by the residue theorem, i.e., let $z=e^{i v}$, then

$$\int_0^{2 \pi} \frac{dv}{6+\cos{v}} = -i 2 \oint_{|z|=1} \frac{dz}{z^2+12 z+1}$$

The only pole of the integrand inside the unit circle is at $z=-6+\sqrt{35}$. The integral is then $i 2 \pi$ times the residue of the integrand at this pole, or $2 \pi/\sqrt{35}$.

To compute the first integral, we consider the complex integral

$$\oint_C dz \frac{\log{z}}{z^2+12 z+1}$$

where $C$ is the unit circle with a detour up and back about the positive real axis. (A circular piece about the origin vanishes.) This integral is equal to

$$-\frac12 \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} + \int_1^0 dx \frac{\log{x}+i 2 \pi}{x^2+12 x+1} + \int_0^1 dx \frac{\log{x}}{x^2+12 x+1}$$

or, simplifying,

$$-\frac12 \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} – i 2 \pi \int_0^1 \frac{dx}{x^2+12 x+1}$$

The contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=-6+\sqrt{35}$. Note that the negative sign is taken to be $e^{i \pi}$. Thus we have

$$-\frac12 \int_0^{2 \pi} dv \frac{v}{6+\cos{v}} = i 2 \pi \int_0^1 \frac{dx}{x^2+12 x+1} + i 2 \pi \frac{-\log{\left ( 6+\sqrt{35} \right )}+i \pi}{2 \sqrt{35}}$$

Now,

$$\frac1{x^2+12 x+1} = \frac1{2 \sqrt{35}} \left (\frac1{x+6-\sqrt{35}} – \frac1{x+6+\sqrt{35}} \right )$$

so that

$$\int_0^1 \frac{dx}{x^2+12 x+1} = \frac1{2 \sqrt{35}} \log{\left (\frac{7+\sqrt{35}}{7-\sqrt{35}} \right )} = \frac1{2 \sqrt{35}} \log{\left (6+\sqrt{35}\right )}$$

Note the cancellation with the real part of the residue. Thus,

$$\int_0^{2 \pi} dv \frac{v}{6+\cos{v}} = \frac{2 \pi^2}{\sqrt{35}}$$

Putting these results altogether, we have

>$$\int_0^{12 \pi} dx \frac{x}{6+\cos{8 x}} = \frac{72 \pi^2}{\sqrt{35}}$$

#### One Comment

• Hello Ron,

I think the second bit is somewhat more convoluted than necessary since one could just reflect the integral across the bounds to cancel out the linear term on the numerator.

Apart from that, this simplification is great! 🙂


# Count the possible ways to seat people at a round table

I have a very simple combinatorics problem, and I am (almost) sure I did this right. However, the smartest guy in my class thinks different and this made me doubt, so I will post our views to this problem and I would like to know who is right.

In how many different ways can 10 people be seated around a round table, under the condition that there are 5 man and 5 women, and no two males or two females can be seated adjacently.

My view

We number the chairs in an arbitrary way, so the chair numbering is fixed. At chair $1$, there is seated either a man or a woman, which given $2$ possibilities. When we observe the gender of the person in the first chair, the division of the table over the two genders in given. Then there are $5!$ ways to seat the men over the $5$ male chairs and $5!$ ways to seat the women over the $5$ female chairs. So in total this gives $2.(5!)^2$ possibilities.

My classmates view

First all males are seated, which is possible in $5!$ ways. Because no males can sit adjacently and the people are seated in a circle, we can rotate the table and in this way we count all possibilities 5 times too much, so $5!/5=4!$ ways to seat the man. After this, the table division is fixed, but we still have to seat the women. This gives another $5!$ ways, so in total $5!4!$ ways.

What I think is wrong with this view

I don't think the argument about rotation of the table makes any sense. The argument can make sense though, if we use exactly my classmate's argument, but acknowledge that the first person to be seated can be any of the $10$ people, so giving an extra $10$ possibilities. Then proceed as my classmate, giving the same answer as me. However, my classmate does not agree. So who is right?

Also, if we use our arguments to the same problem but a table of $2$ people, one male, one female, there are obviously $2$ possibilities. Then my method gives $2.(1!)^2=2$ possibilities, but my classmate's method gives only $1!0!=1$ possibility.

EDIT

I think the main problem is in different interpretations of the problem, where my interpretation is that it matters who sit at 'chair 1' while my classmates thinks that only the ordering of the people matters. However, given only the above problem, what would be the right interpretation, or is this ambiguous?

• Excluding the one and only genious Václav Mordvinov of course. – Václav Mordvinov Jan 10 '18 at 11:10
• Are they seated at/around the table, or "on" the table, as you stated? Sitting 10 people on the table may be tricky. (I am only joking) – mbomb007 Jan 10 '18 at 23:09
• Not a native speaker. – Václav Mordvinov Jan 10 '18 at 23:10

Your solution would be okay if there is a "special chair" (you name it "chair $1$"). Then there is no essential difference with placing the $10$ persons in a row in which case there is also a special chair (for instance the utmost left, or the utmost right).

If there is no special chair then you can start by placing one man. After that there are $4!$ different arrangements for the other men and $5!$ different arrangements for the women. This leads to a total of $4!5!$ possibilities.

It is not for nothing that a round table is used, indicating that there is no special chair. So I would say that your classmate is right.

• Okay thanks, I think you are right about that my classmate's interpretation is the most natural, otherwise they wouldn't have used a circle in the question. It is still a bit ambiguous but you're right! – Václav Mordvinov Jan 10 '18 at 9:51
• You are welcome. According to your interpretation there are $2$ ways to place one man and one woman at a round table. That is correct if the two chairs are distinguishable, but incorrect if they are not (which is definitely a more likely interpretation). – drhab Jan 10 '18 at 9:56
• The round table could be so that there is no one on the "end" who is exempted from the condition "no two males or two females can be seated adjacently". – Acccumulation Jan 10 '18 at 16:48
• @Acccumulation: If you have five men and five women in a row, with no two adjacent men and no two adjacent women, then it's already guaranteed that the people at the two ends will be one man and one woman. So if that's the only reason they specified a round table, then they needn't have specified it. – ruakh Jan 11 '18 at 0:24
• @Acccumulation I fully agree with ruakh – drhab Jan 11 '18 at 6:58

Your disagreement comes from different interpretations of the question.

You're assuming the chairs around the table are indexed ${1,...,10}$ and you're looking at the number of ways to assign the numbers ${1,...,10}$ to your 5 men and 5 women.

Your classmate is looking at how many ways you can arrange 5 men and women around the table relative to each other instead of in absolute terms.

The difference between the two answers is a factor of 10 ($10\times4!5! = 2\times 5 \times 4!5! = 2\times(5!)^2$), which corresponds to the extra degree of freedom (e.g. where is chair #1).

• Good answer! Thanks – rae306 Jan 10 '18 at 19:46

The thing is, that when you are a circle, each combination $x_1, x_2 \dots x_{10}$ is counted $10$ times, namely

$x_1, x_2 \dots x_9 x_{10}$,

$x_2, x_3 \dots x_{10},x_1$,

$x_3, x_4 \dots x_1, x_2$,

$\dots$

$x_{10}, x_1, \dots x_8, x_9$

since rotating the table preserves the combination of people sitting

In the case of $2$ people you get the combinations:

man, woman

woman, man

but they are essentially the same.

Your edit is spot on, depends on whether the chairs differ from one another, but it'd assume that this is not the case

• Yes, but why wouldn't it matter who sits at chair 1? See also my edit. Is this really obvious from the problem? I understand completely what you are saying but to me the question seems ambiguous. – Václav Mordvinov Jan 10 '18 at 9:22
• @VáclavMordvinov We all passed through that. Yes, it's ambiguous, but with time you will start to see what they want with this kind of questions. Not that it will ever change, but you will get used to it, and you will see that it's not “ambiguous” once you know how they speak. But yes, you are right, that way to phrase problems is not ideal and ambiguous to a random reader. We all passed that phase at which we had a problem with probability problems badly written. Good luck. – Manuel Jan 10 '18 at 14:21
• @Manuel, you're right, especially since it is a problem and not a real-life application. I just have to read better next time :) – Václav Mordvinov Jan 10 '18 at 18:29
• @VáclavMordvinov "Read better", yes, but do not forget that you are right now thinking it's badly written. I meant that sadly it's badly written, but it's not the worst, you just have to get used to it. – Manuel Jan 10 '18 at 18:31
• Yes, generally I add what I assume for such a question and if necessary I include multiple answers for multiple assumption sets. It's not that much more effort and yields the rigjt answer for sure. – Václav Mordvinov Jan 10 '18 at 18:37

In combinatorics, the definition of "different" is often a crucial issue. There is nothing in the problem saying that the chairs are interchangeable, and you should not add assumptions to the problem. If I were only one person to be seated, I would have ten choices where to sit. Your interpretation is the correct one. If your classmate's interpretation were intended, then the problem is poorly stated.

And actually, if we to follow your classmate's reasoning, then why stop there? Your classmate is noting that rotation leaves the arrangement "essentially" the same, but what about reflections? If all that matters is who is sitting next to whom, and absolute location is unimportant, then wouldn't the answer be 5!4!/2?

• Good answer, I will send this to him too :) – Václav Mordvinov Jan 10 '18 at 18:34
• If you were the only person to be seated, you could actually tell that there is only one resulting configuration, if the table is ROUND with no orientation mark – G Cab Jan 11 '18 at 2:08
• @G Cab Yes, if the table is a non-rotating, uncharged black hole in otherwise empty space, then all locations are indistinguishable. If, on the other hand, this takes place in the real world ... – Acccumulation Jan 11 '18 at 3:15


# External bisectors of the angles of ABC triangle form a triangle $A_1B_1C_1$ and so on

If the external bisectors of the angles of the triangle ABC form a triangle $A_1B_1C_1$,if the external bisectors of the angles of the triangle $A_1B_1C_1$ form a triangle $A_2B_2C_2$,and so on,show that the angle $A_n$ of the $n$th derived triangle is $\frac{\pi}{3}+(-\frac{1}{2})^n(A-\frac{\pi}{3})$,and that the triangles tend to become equilateral.

My attempt:The angle $A_1$ of triangle $A_1B_1C_1$ is $\frac{\pi}{2}-\frac{A}{2}$,angle $A_2$ of triangle $A_2B_2C_2$ is $\frac{A}{4}-\frac{3\pi}{4}$ but my answer is nowhere look like resembling final answer.Is my approach correct.If not,what is the correct way to solve this question?Can someone guide me?

You are on the right track, you just need to continue. You have a correct formula for $A_1$, so format it differently.

\begin{align} A_1 &= \frac{\pi}2-\frac A2 \quad\text{(your formula)}\\ &= \frac{\pi}3+\frac{\pi}6+\left(-\frac 12\right)A \\ &= \frac{\pi}3+\left(-\frac 12\right)\left(A-\frac{\pi}3\right) \\ &= \frac{\pi}3+\left(-\frac 12\right)^1\left(A-\frac{\pi}3\right) \\ \end{align}

Your formula for $A_1$ from $A$ also gives $A_{n+1}$ from $A_n$, so combining your formula with my derivation above and using induction we get

\begin{align} A_{n+1} &= \frac{\pi}3+\left(-\frac 12\right)\left(A_n-\frac{\pi}3\right) \\ &= \frac{\pi}3+\left(-\frac 12\right)\left[\frac{\pi}3+ \left(-\frac 12\right)^n\left(A-\frac{\pi}3\right)-\frac{\pi}3\right] \\ &= \frac{\pi}3+\left(-\frac 12\right)\left[ \left(-\frac 12\right)^n\left(A-\frac{\pi}3\right)\right] \\ &= \frac{\pi}3+\left(-\frac 12\right)^{n+1}\left(A-\frac{\pi}3\right) \\ \end{align}

which finishes the proof by induction.

So the given formula is correct. As we let $n\to\infty$ we see that $\left(-\frac 12\right)^n\to 0$ and thus $A_n\to\frac{\pi}3$. We can do the same to $B$ and $C$ to get $B_n\to\frac{\pi}3$ and $C_n\to\frac{\pi}3$, so all three angles approach $60°$, and the (increasingly large) triangles $\triangle A_nB_nC_n$ approach equilateral.

Here is a quick proof of your formula $A_1 = \frac{\pi}2-\frac A2$, though I used Greek letters for angles in this diagram.

This should be self-explanatory, and the final value for $A_1= \frac{\pi}2-\frac A2$ comes directly from $A_1=\frac{\beta+\gamma}2$ and $\alpha+\beta+\gamma=\pi$.

Your approach was a good way to start working on this problem. By working the first two or three steps, you can see what calculations are involved. Your idea of comparing your results to the desired results was also a good one. It gives you a chance to check yourself in case you go down a wrong path.

But while it's true that the values you calculated, namely, $A_1 = \frac{\pi}{2} - \frac A2$ and $A_2 = \frac A4 - \frac{3\pi}{4}$, do not look much like $\frac{\pi}{3}+\left(-\frac12\right)^n \left(A - \frac{\pi}{3}\right)$, that alone should not disturb you; general formulas very often look quite different from the results you get by direct calculations. That's part of what made it so interesting and useful when people discovered such formulas. It often takes some art to "see" a pattern in a sequence of values from some procedure such as $A_1$, $A_2$, $A_3$, and so forth.

For example, the sum of the first $n$ positive integers is $\frac12(n^2 + n)$. But the sum of the first $4$ positive integers is $1+2+3+4 = 10$, and $10$ does not look like $\frac12(n^2 + n)$. It doesn't even look like $\frac12(4^2 + 4)$, but it turns out that if you do the multiplications and additions of that formula, you can see that $\frac12(4^2 + 4) = \frac12(16 + 4) = \frac12(20) = 10$.

That is, in order to compare a particular result of yours (the values you found for $A_1$ and $A_2$) against the general formula, you can plug the appropriate value of $n$ into the formula, do whatever operations the formula says to do, and see what comes out. For example, your own calculation found that $A_1 = \frac{\pi}{2} - \frac A2$. The general formula for $A_n$ should produce $A_1$ when $n = 1$. So let's set $n = 1$ and see what happens:

\begin{align} A_n & = \frac\pi3 + \left(-\frac12\right)^n \left(A - \frac\pi3\right) \\ A_1 & = \frac\pi3 + \left(-\frac12\right) \left(A - \frac\pi3\right) & \text{(because we set $n = 1$)} \\ & = \frac\pi3 - \frac12 A + \frac\pi6 \\ & = \frac{\pi}{2} - \frac A2 \end{align}

So your calculation of $A_1$ looks OK. For $A_2$, we set $n = 2$:

\begin{align} A_n & = \frac\pi3 + \left(-\frac12\right)^n \left(A - \frac\pi3\right) \\ A_2 & = \frac\pi3 + \left(-\frac12\right)^2 \left(A - \frac\pi3\right) & \text{(because we set $n = 2$)} \\ & = \frac\pi3 + \frac14 \left(A - \frac\pi3\right) \\ & = \frac\pi3 + \frac14 A - \frac{\pi}{12} \\ & = \frac\pi4 + \frac A4 \end{align}

This is a different from the result $\frac A4 - \frac{3\pi}{4}$ that you calculated for $A_2$. So maybe there was a problem with your method, or maybe you just made an arithmetic error.

But once you figure out what is going on with $A_2$, you still need to prove the general formula for all $n$, and for that you may need something like the other answer (which used induction). But if you can see how to get $A_1$ from $A$ and how to get $A_2$ from $A_1$, that can help when you have to get $A_{n+1}$ from $A_n$, as the inductive proof requires.

• This is a good response, and the method you outline here is close to the one I used to get my answer. +1! – Rory Daulton Aug 3 '15 at 14:43


# Reasoning in an integration substitution

Evaluate: $\displaystyle\int_{0}^{1}\frac{\ln(x+1)}{x^2+1}\,\mathrm{d}x$

So I did this a completely different way than what the answer key states. I used integration by parts and some symmetry tricks and got the correct answer. However the answer key says:

Make the substitution $x=\frac{1-u}{1+u}$

The same solution was reached in about half the steps but still using symmetry, my questions are How would I know to do that? Is this a certain type of substitution? Is there something else that maybe I could use this for?

• the change of variable reminds me a bit the Weierstrass substitution $\cos(x)=\frac{1-t^2}{1+t^2}$ – Masacroso May 18 '17 at 20:31
• That is true. The weierstrauss sub is actually what I used in part of my long solution – Teh Rod May 18 '17 at 20:33
• The transformation is a Möbius Transform, not a Weierstrass substitution. – Mark Viola May 18 '17 at 20:39
• possible duplicate math.stackexchange.com/questions/155941/… – Darío A. Gutiérrez May 18 '17 at 21:09

The substitution is called a Möbius Transformation, which has the more general form

$$w=\frac{az+b}{cz+d}$$

where $ad\ne bc$. The transformation maps straight lines into circles and circles into straight lines. They have a variety of uses in applied mathematics and physics along with complex analysis.

As a simple example, the Beta function $B(x,y)$ can be represented by the integral

$$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt$$

Enforcing the Mobius Transformation $t\to \frac{t}{1+t}$ so that $dt\to \frac{1}{(1+t)^2}\,dt$. Then, we see that

$$B(x,y)=\int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}}\,dt$$

which is an alternative representation for the Beta function.

As another example, in This Question, the OP requested evaluation of the integral $I$ expressed as

$$I=\int_0^\infty\frac{\log(e^x-1)}{e^x+1}\,dx$$

In the accepted answer posted by User @FDP, the Mobius transform was used to facilitate an efficient way forward.

As a third and final example, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$1+x\le e^x\le \frac{1}{1-x}$$

for $x<1$. Then, it is trivial to see that $\log(x)\le x-1$.

Applying the Möbius Tranformation $x\to \frac{-x}{1+x}$ we find that the logarithm function is bounded below by $\log(x)\ge \frac{x-1}{x}$.

Here's an alternative method that doesn't use differentiation under the integral sign. Substitute $x = \tan \theta$, to give $$I = \int_0^{\pi/4} \log(1+\tan\theta) d\theta \\ = \int_0^{\pi/4} \log(\cos\theta+\sin\theta)-\log\cos(\theta)d\theta \\ = \int_0^{\pi/4} \frac12\log2 + \log(\cos(\theta- \pi/4))-\log\cos(\theta) d\theta \\ = \frac{\pi}{8}\log2 + \int_0^{\pi/4} \log(\cos(\theta- \pi/4)) d\theta - \int_0^{\pi/4}\log\cos(\theta) d\theta$$ We have the known property that $\int_0^a f(x) dx = \int_0^a f(a-x)dx$, and $\cos$ is an even function, so these two final integrals are equal, and $I = \frac{\pi}{8}\log2$.

Here's how I would do it. Let $$f(\alpha)=\int_0^1 \frac{\log(1+\alpha x)}{1+x^2}\ dx$$ where $f(1)$ is the integral we seek to evaluate. By differentiation under the integral sign we have $$f'(\alpha)=\int_0^1\frac{\partial}{\partial\alpha}\frac{\log(1+\alpha x)}{1+x^2}\ dx\\ =\int_0^1\frac{x}{(1+x^2)(1+\alpha x)}\ dx.$$

Now we apply partial fraction decomposition: suppose that $$\frac{x}{(1+x^2)(1+\alpha x)}=\frac{Ax+B}{1+x^2}+\frac{C}{1+\alpha x}$$ for some constants $A,B,C$. By combining the fractions and collecting the terms we obtain the linear system of equations $$\begin{cases}A+\alpha B=1\\ \alpha A+C=0\\B+C=0\end{cases}$$ which we find to be equivalent to $$\begin{cases}A=\frac{1}{1+\alpha^2}\\B=\frac{\alpha}{1+\alpha^2}\\C=-\frac{\alpha}{1+\alpha^2}\end{cases}$$ From this we conclude that $$f'(\alpha)=\int_0^1\frac{Ax+B}{1+x^2}+\frac{C}{1+\alpha x}\ dx\\ =\left[\frac12 A\log(1+x^2)+B\arctan{x}+\frac{C}{\alpha}\log(1+\alpha x)\right]_{x=0}^1\\ =\frac{\log2}{2}\frac{1}{1+\alpha^2}+\frac{\pi}{4} \frac{\alpha}{1+\alpha^2}-\color{red}{\frac{\log(1+\alpha)}{1+\alpha^2}}$$ It is easy to see by looking at the definition that $f(0)=0$. From this it follows that $f(\alpha_0)=\int_0^{\alpha_0} f'(\alpha)\ d\alpha$, and in particular $$f(1)=\int_0^1 \frac{\log2}{\alpha}\frac{1}{1+\alpha^2}+\frac{\pi}{4}\frac{\alpha}{1+\alpha^2}-\color{red}{\frac{\log(1+\alpha)}{1+\alpha^2}}\ d\alpha$$ which simplifies to $$\color{red}{\int_0^1 \frac{\log(1+x)}{1+x^2}\ dx}= \frac{\pi \log 2}{16}-\color{red}{\int_0^1 \frac{\log(1+\alpha)}{1+\alpha^2}\ d\alpha}$$ hence the result $$\int_0^1 \frac{\log(1+x)}{1+x^2}\ dx=\frac{\pi \log 2}{8}.$$

• That is a smarter way of doing it are you sure you meant $\frac{\log 2}{x}$? – Teh Rod May 18 '17 at 20:45
• @TehRod that was a typo, now amended :) – user1892304 May 18 '17 at 20:51

Set $$x=\tan t, \qquad dt=\dfrac1{1+x^2}dx,$$ Then \begin{align} \int_0^1 \frac{\ln(1+x)}{1+x^2}\, dx&= \int_0^{\pi/4} \ln(\cos t +\sin t)\, dt- \int_0^{\pi/4} \ln(\cos t)\, dt\\\\ &=\int_0^{\pi/4} \ln\left(\sqrt{2}\cos \left(\frac \pi4- t\right)\right)\, dt- \int_0^{\pi/4} \ln(\cos t)\, dt\\\\ &=\int_0^{\pi/4} \ln\left(\sqrt{2}\right)\, dt+\int_0^{\pi/4} \ln\left(\cos \left(\frac \pi4- t\right)\right)\, dt- \int_0^{\pi/4} \ln(\cos t)\, dt\\\\ &=\frac{\pi}8 \:\ln 2+\int_0^{\pi/4} \ln(\cos u)\, du-\int_0^{\pi/4} \ln(\cos t)\, dt\quad \left(u=\frac \pi4- t\right)\\\\ &=\frac{\pi}8 \:\ln 2. \end{align}


Help needed with definite integral for $\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \frac{1}{n+k}$

My foreign school book is not clearest at this point, what is really needed with this? What does it mean that "assign definite integral for $\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \frac{1}{n+k}$"? I am not sure, whether I should assign it like this $\lim_{n\rightarrow \infty} \int_{1}^{n} \frac{1}{n+x} dx$ and noticing that it is continuous then trying to find borders so that $F(a) - F(b)$?

(the book messes up all kind of Leibniz stuff at this point, a bit messy -- and just stating $\int f(x)\,dx= F(a)-F(b)$, but not even paying attention to different borders. As far as I know, it is important to specify whether borders depend on the integration factor)

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...good basic video related to this here by patrickJMT. – hhh Feb 4 '12 at 14:11

I suspect the following: $$\sum_{k=1}^n {1\over n+k}=\sum_{k=1}^n{1\over n} {1\over 1+{k\over n}}$$ Now interpret the right hand sum as a Riemann sum for the function $f(x)={1\over 1+x}$ over $[a,b]=[0,1]$ (for a fixed $n$, the partition of $[0,1]$ is $\{{1\over n}, {2\over n},\ldots, {n\over n} \}$ and the ${1\over n}$ is the common width of the subintervals).

Taking the limit as $n\rightarrow \infty$ gives the corresponding integral: $$\lim_{n\rightarrow\infty}\sum_{k=1}^n{1\over n} {1\over 1+{k\over n}} = \int_0^1 {1\over 1+x}\,dx.$$

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+1 for the illustration with picture! – user21436 Feb 4 '12 at 15:02
+1 thanks! I finally understood the formula after your updated image. – hhh Feb 4 '12 at 18:23
How about if you have a situation so that the height is in different form such as $\frac{1}{n+\frac{k}{n}}$ or $\frac{1}{n+\frac{k}{n^{2}}}$? What about can I have a first term such as $\frac{1}{n^{2}}$ or $\frac{1}{n^{3}}$? – hhh Feb 4 '12 at 21:49
@hhh Generally, for an integral over $[a,b]$, you use something of the form $f(a+[(b-a)i/n])\cdot{1\over n}$. The $1/n$ is the width of the rectangles for the partition points $a+[(b-a)i/n]$ (other widths could be used, like $1/(3n)$). The height is obtained by evaluating $f$ at the partition points. I don't think $1\over n+(i/n)$ would get you one, since there is no way to write that as $f(a+[(b-a)i/n]$ for some $f$. Other expressions would work: e.g if you have $(i/n)^3+{1\over (i/n)^2+1}$, then your $f$ would be $f(x)=x^3+{1\over x^2+1}$ – David Mitra Feb 4 '12 at 22:20
@hhh I don't think that will work. Basically, you chop $[a,b]$ into $n$ pieces so that the maximum length of the pieces is small. The widths can't all be $(1/n)^2$. If you add the lengths of the pieces, you have to get $b-a$. – David Mitra Feb 4 '12 at 22:57

The problem converts into an integral this way:

\begin{align*}\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{n+k}&=\lim_{n \to \infty} \dfrac{1}{n} \cdot\sum_{k=1}^n \dfrac{n}{n+k}\\&=\lim_{n \to \infty} \dfrac{1}{n} \cdot \sum _{k=1}^n\dfrac{n}{n+k} \\&= \lim_{n\to \infty} \dfrac{1}{n} \cdot \sum_{k=1}^n \dfrac{1}{1+\frac{k}{n}}\end{align*}

Now interpret the final limit as Riemann sum of the function $f(x)=\dfrac{1}{1+x}$

So, the limit in question equals, $\int_0^1{(\dfrac{1}{1+x}) \mathrm dx}= \ln 2$ $\blacksquare$

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...thank you, can you suggest some reading about this topic? (My book is not the best one to practise on this, it is introduces Rieman sum after many pages the questions dealing with the rieman sums and the organizaton of material is not most pedagocical-- and I cannot make head-and-tail about what it is really trying to explain, have to look for some English material on this.) – hhh Feb 4 '12 at 14:05
@hhh Did you try reading the article from Wikipedia, it is a bit Shabby, however. Look at David's Answer below. His picture should give you an idea of what is happening. And, Wolfram math world has an appplet which you could try and fiddle with until you are acquainted. – user21436 Feb 4 '12 at 15:02

Hint: You can write this as $$\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}.$$ How close is this to the Riemann sum of the integral $$\int_0^1 \frac{1}{1+x}dx?$$

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# Find the value of point A and B

1. Nov 26, 2015

### shayaan_musta

Hello experts.

I am attaching the image below. Kindly tell me how to calculate the values of A and B in the given figure?
Now I must find A and B. How to get it?
PS: P=-2.5+4.33i. But A is not at -2.5. Means PA is not making angle 90 with real axis.
Thank you

#### Attached Files:

• ###### Capture.PNG
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2. Nov 26, 2015

### Staff: Mentor

One information is missing, either about A or B or another angle. With P and the 55° at P you could still find infinitely many solutions. Or is it supposed to express A as function of B?

3. Nov 27, 2015

### shayaan_musta

No further information is given. Here is the image of the whole page below.
Here is AO=2.38. But how?

#### Attached Files:

• ###### Capture.PNG
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4. Nov 27, 2015

### Samy_A

But, as @fresh_42 rightly supposed, you have one more piece of information, namely $\frac {\overline {PA}}{\overline{PB}}=\frac {4.77}{8}$.
Then, as the text suggests, you can calculate A and B using trigonometry.

Last edited: Nov 27, 2015
5. Nov 27, 2015

### shayaan_musta

But how? This is the actual question of mine. For A and B, I posted this thread. If you are saying to solve it myself then what is the cause of posting this thread. Kindly enlighten the way for me to find A and B.

6. Nov 27, 2015

### SteamKing

Staff Emeritus
You've been posting information in dribs and drabs, and you are expecting a full solution to what is obviously a much larger problem than just figuring out this triangle.

How about some full disclosure of the complete problem or section from this text you are referencing?

7. Nov 27, 2015

### Samy_A

As both @fresh_42 and @SteamKing noted, you first didn't provide enough information to solve the triangle.
From your second attachment we learned that $\frac {\overline {PA}}{\overline{PB}}=\frac {4.77}{8}$

With this information the triangle can be solved.

Now there are probably many ways to do it.
I would begin in the following way:
If $\alpha$ is the angle in A, and $\beta$ the angle in B, $\frac {\sin(\beta)}{\sin(\alpha)}$ can be determined (using $\frac {\overline {PA}}{\overline{PB}}=\frac {4.77}{8}$ and the law of sines).
But we also know that $\alpha+\beta=125°$, so that $\sin(\beta)=\sin(125°-\alpha)=\sin(125°)\cos(\alpha)-cos(125°)\sin(\alpha)$.
These two together will allow to compute $\cot(\alpha)$. And then the other components of the triangle can be computed. It is tedious but it works.

This is just one way to do it. Hopefully someone else will come along with a less tedious method.

As this thread should probably have been posted in the homework section, we are not allowed to give the full solutions.

Last edited: Nov 27, 2015
8. Nov 27, 2015

### shayaan_musta

OK. This is what I did,

sin(125°-α)=sin(125°)cos(α)-cos(125°)sin(α)
since, sin(β)=sin(125°-α)
so,
sin(β)=sin(125°)cos(α)-cos(125°)sin(α)
dividing both sides by sin(α)
so,
sin(β)/sin(α)=sin(125°)cos(α)/sin(α)-cos(125°)sin(α)/sin(α)
sin(β)/sin(α)=sin(125°)cos(α)/sin(α)-cos(125°)
sin(β)/sin(α)=0.8192*cos(α)/sin(α)-(-0.5736)
4.77/8=0.8192*cot(α)+0.5736
0.596=0.8192*cot(α)+0.5736
0.0227=0.8192*cot(α)
cot(α)=0.0227/0.8192
cot(α)=0.028
α=88° and α+β=125° => β=37°
and
AB=sin(P)*PA/sin(β)=6.5

But now what? I still don't have point A coordinates :(

9. Nov 27, 2015

### Samy_A

Draw the line from P perpendicular to the horizontal axis. It will cross the horizontal axis in a point Q, just to the left of A.
PAQ is a right triangle and you know the coordinates of P, the coordinates of Q and the angle $\alpha$. That should lead you to the coordinates of A.

10. Nov 27, 2015

### shayaan_musta

Yes, I tried this already before posting post#8 but problem is P has iota in imaginary axis i.e. P(-2.5+4.33i).

11. Nov 27, 2015

### SteamKing

Staff Emeritus
I don't know what more you want.
So far:
1. You have determined all three angles in this triangle
2. You have calculated the length of one side of this triangle.
3. You know the coordinates of one of the vertices of this triangle.

It's time to let loose on the trigonometry and find the rest of the missing information.

Hint: Use the Law of Sines to find the lengths of the other two sides.

12. Nov 27, 2015

### shayaan_musta

OK

13. Nov 28, 2015

### Samy_A

You lost me there.

In triangle APQ:
$\cot(\alpha)=\frac {|AQ|}{|PQ|}$. The only unknown here is $|AQ|$.
Thus:
$|AQ|=\cot(\alpha)*|PQ|=0.028*4.33=0.12$
So the coordinates of A are: $(-2.5+0.12,0)=(-2.38,0)$

For B (same as above, but now in the right triangle BPQ):
$\beta=\pi-55*\frac{\pi}{180}-1.543=0.638$ (in radians).
$|BQ|=\cot(\beta)*|PQ|=\cot(0.638)*4.33=1.349*4.33=5.84$
So the coordinates of B are: $(-2.5-5.84,0)=(-8.34,0)$

Last edited: Nov 28, 2015
14. Nov 28, 2015

### shayaan_musta

Best Regards,
Pinky


# Clarifying the notion of basic/free variables in a system of equations

I have a fundamental confusion with regards to the notion of free/basic variables.

Consider the following linear system $$\begin{pmatrix} 1 & 2 & -1 \\ 2 & -4 & 0 \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 4\\ 5 \end{pmatrix}$$ This system has augmented matrix $$\begin{pmatrix} 1 & 2 & -1 & 4 \\ 2 & -4 & 0 & 5 \end{pmatrix}$$ which row reduces to $$\begin{pmatrix} 1 & 2 & -1 & 4 \\ 0 & -8 & 2 & -3 \end{pmatrix}$$ The last matrix is in echelon form. Columns 1 and 2 are pivot columns, so $$x_1,x_2$$ are basic variables. The third column is not a pivot column, so $$x_3$$ is a free variable. Finally, the last column is not a pivot column, so the system is consistent.

The above analysis tells me that $$x_3$$ is a free variable. I interpret this as saying "$$x_1,x_2$$ can be expressed as a function of $$x_3$$, while $$x_3$$ can be set equal to any number". For example, I can set $$x_3=200$$. In turn, $$\begin{cases} x_1+2x_2-200=4 \\ 2x_1-4x_2=5 \end{cases}\Rightarrow x_2=\frac{403}{8}, x_1=\frac{413}{4}$$

Nevertheless, just by doing naive computations, I realised that we could also "express $$x_1$$, $$x_3$$ as a function of $$x_2$$, while setting $$x_2$$ equal to any number". For example, I can set $$x_2=7$$. In turn, $$\begin{cases} x_1+14-x_3=4\\ 2x_1-28=5 \end{cases}\Rightarrow x_1=\frac{33}{2}, x_3=\frac{53}{2}$$ Alternatively, we could also "express $$x_2$$, $$x_3$$ as a function of $$x_1$$, while setting $$x_1$$ equal to any number". For example, I can set $$x_1=0$$. In turn, $$\begin{cases} 2x_2-x_3=4\\ -4x_2=5 \end{cases}\Rightarrow x_2=-\frac{13}{2}, x_3=-\frac{5}{4}$$

Hence, my question: what is the relation between

• the fact that $$x_3$$ is a free variable and $$x_1,x_2$$ are basic variables (from the row-reduced matrix)

• and the fact that in the system above I am free to fix the value of anyone among $$x_1,x_2,x_3$$ and I will always be able to solve for the other two unrestricted variables

• Note that the order of the columns is arbitrary in a way. Feb 5 at 13:22

You've got three column vectors, and any pair of them are independent and thus span $$\Bbb R^2$$. The solution set is one-dimensional (affine) (aka a line) so of the form $$v+tw$$, where $$v,w$$ are some vectors and $$t$$ is a free real variable. We can choose this to be $$x_3$$, or any of the other variables if you prefer.

You can then transform the equations to \begin{align}x_1+2x_2 &= 4+x_3\\ x_1 - 4x_2 &= 5\end{align}

and solve these in terms of $$x_3$$ uniquely. We get (add twice the first equation to the second and we get $$3x_1 = 13+2x_3$$ so $$x_1 = \frac{1}{3}(13+2x_3)$$, and subtraction the equations: $$6x_2 = x_3-1$$ and hence $$x_2 = \frac{1}{6}(x_3 - 1)$$ (we could also do another elimination step on this new system) so we can write the solution line as

$$\begin{pmatrix} \frac{13}{3}\\ -\frac16\end{pmatrix} + x_3 \begin{pmatrix} \frac{2}{3}\\ \frac16\end{pmatrix}$$

• Thanks. (1) Can we say that in this specific case "$x_3$ is a free variable, $x_1,x_2$ are basic variables" or, equivalently, "$x_1$ is a free variable, $x_3,x_2$ are basic variables" or, equivalently, "$x_2$ is a free variable, $x_1,x_3$ are basic variables"? (2) Is there any way to realise from the echelon form that we are "free" to permute the columns? This somehow relates to my other question math.stackexchange.com/questions/4012647/… if you can help.
– TEX
Feb 5 at 13:27
• @TEX They can all be free, that's the point. It doesn't matter. Feb 5 at 13:28
• Thanks. My doubt is: is there a formal way to understand from the echelon form (or any other form) when any variable can be set as the free variable? This is not always the case.
– TEX
Feb 5 at 13:29
• @TEX you just use $x_3$ and get the full solution set. Why do more? Feb 5 at 13:31
• Because this relates to a more general result that I need to prove. I want to understand when I'm free to set the free variables and how I can detect that.
– TEX
Feb 5 at 13:32


# Group tables for a group of four elements.

I should consider group tables obtained by renaming elements as essentially the same and then show that there are only two essentially different groups of order 4.

There seems to be so many different possible group tables for all the different binary operations - which is why I'm confused. I was thinking about using Cayley's table to show their commutativity but I'm really not too sure. Any help please!

Edit: After all of your help, I completely understand how to show that there are only two different groups of order four. Thank you. The only thing which I am still unclear of is how to note down the 'other tables' before stating that they are essentially the same as one of the other tables - meaning that there are just two different ones. It's pointless work but I think it's what the question requires.

Answer: After a bit of playing around - I've realised that there are four different tables, however, 3 tables are the same as each other, just with different values (the Klein 4 Group with 3 different generators). Hence there are two tables.

• It might seem like there are many possibilities, but try to start writing out some of them and note that there are certain properties they must have in order to satisfy the group axioms. – Tobias Kildetoft Feb 11 '13 at 17:47
• Re: duplicate vote. It's the same question, but the discussion in the older one is left unfinished. – user53153 Feb 11 '13 at 19:12

Hints: we are talking about groups of order 4, which narrows the possible binary relations and elements available:

• each must contain the identity,
• each must be associative,
• each must be closed under inverses (if $a \in G,\;a^{-1} \in G$)
• (and of course, closed under the group operation).

There are essentially (up to isomorphism) only two groups of order 4:

• one of course will be the additive cyclic group $\mathbb{Z}_4$, and
• the other will be the Klein 4-group, which is indeed abelian.

If you follow the suggestions for solving the problem, you'll find, indeed, that any possible GROUP of order 4 can be shown isomorphic to one of the two groups mentioned simply by a renaming elements.

Yes: use of the Cayley table will be of great importance:

• for a group: no element can appear twice in any column,
• no element can appear twice in any row.

You'll find that the only ways of completing a table satisfying these criteria are limited, and then you show that a simple renaming of elements will reveal the group is isomorphic to $\mathbb{Z}_4$ or else the Klein-4 group.

• Yes, this has been very helpful, thank you. Your answer is so neatly written too! The only question I have is that I believe I should write down all the group tables... Is this just 2 group tables (of what binary operation???) OR do I list more but eliminate the rest. With Cayley's table, to show commutativity, is it sufficient to use the symmetry of the diagonal line? Thanks again. PS, I have encountered Lagrange - just struggling with Algebra! – user61854 Feb 11 '13 at 21:00
• If you have encountered Lagrange, than it suffices to argue that the only groups of order 4 have are the cyclic group $\mathbb{Z}$ (generated by one element), and a group whose non-identity elements have order 2 (since two and four divide 4). Completing any table that satisfies the properties of a group (one and only one element in each column and each row will limit the number of Cayley tables to construct. You can then show that of those, each is isomorphic to $Z_4$ or the Klein 4-group. And yes, you can show that those two non-isomorphic groups are indeed commutative by the Cayley table. – amWhy Feb 12 '13 at 1:47

There aren't that many. One of the elements will be the identity, let's call it $1$. (I am writing the operation as multiplication.) So the table looks like $$\begin{matrix} 1 & a & b & c\\ a\\ b\\ c\\ \end{matrix}$$ (I'm not writing the row and column labels, as they are the same as the first column and row). Now what could $ab$ be? Not $a$, not $b$... And $ba$? Proceed with $ac$, $ca$, $bc$, $cb$, and then you are left with the squares. They might all be $1$, or...

This simplifies it:

If a group has 4 elements, we separate two cases:

Case1: all the elements have order 2, then the group is abelian, and $a^2=1\Rightarrow a=a^{-1}$ for all $a\in G$: $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$ the table should be easy to make in this case: there is only one.

Case2: there is an element $a\not =1$ that has order different from 2, then by LAgrange's theorem, the order must be 4, and so $G=\langle a\rangle$, and so the group is cyclic, and therefore it's abelian. We have proved that a 4-element group is always abelian. The table shouldn't be difficult in this case either.

• You're assuming Lagrange, and chances are, at this stage, the OP has not yet encountered Lagrange. See my comment below Michael Hardy's answer. – amWhy Feb 11 '13 at 19:17
• @amWhy I would have never imagined, Lagrange always comes really soon. I guess that then he will have to do every possible table to get to the same conclusion. – MyUserIsThis Feb 11 '13 at 21:25
• so b^2 = a, c^2 = a in one table and in the other, b^2 = 1, c^2 = 1. – user61854 Feb 12 '13 at 16:45
• @user61854 Yes, in my case 1, you have that: $a^2=b^2=c^2=1$, so: $ab=c, bc=a$, it's abelian, so the whole table is there. In my case 2, $a$ is the generator, so $a=a,a^2=b,a^3=c$, and you can compose to get the rest of the combinations. – MyUserIsThis Feb 12 '13 at 17:00

The order of an element must divide $$4$$, so it must be either $$1$$, $$2$$, or $$4$$. Only the identity element can have order $$1$$. If one element has order $$4$$, then it generates the whole group and you have a cyclic (hence abelian) group.

Otherwise the three non-identity elements each have order $$2$$.

$$e=\text{the identity}$$.

$$a,b,c=\text{the other three}$$.

$$a^2=b^2=c^2=e$$, and each of these is its own inverse.

So what is $$ab$$? It can't be $$e$$ since if $$ab=e$$ then $$(ab)b^{-1} = b^{-1}=b$$, and hence $$a=b$$. It can't be $$a$$ or $$b$$ since if $$ab=b$$ then $$(ab)b^{-1}=bb^{-1}=e$$ and so $$a=e$$, and similar reasoning shows it can't be $$a$$ (but you have to multiply on the left be the inverse).

So $$ab=c$$.

Similar reasoning shows $$ab=ba=c$$, and $$ac=ca=b$$ and $$bc=cb=a$$.

So the group is abelian.

• Chances are, at this stage of the game, the OP hasn't encountered Lagrange. I know Fraleigh includes this exercise long before introducing Lagrange... – amWhy Feb 11 '13 at 19:15
• @amWhy Because this is such a toy example, you don't need Lagrange; it's trivial to see that only the identity can have order 1, or that if one element has order 4 then the group must be cyclic. The only other possibility aside from the all-order-2 case is that some element (and therefore two elements) has order 3, and it's easy to eliminate this possibility. (Suppose that $a$ and $b$ have order 3, with $a^2=b$. $ca$ can't be $a$ and can't be $c$ by cancellation; it can't be $b$ because $ca=aa\implies c=a$, and it can't be the identity because then $ca=1=ba$ and again by cancellation $c=b$.) – Steven Stadnicki Feb 12 '13 at 23:33


# What does ceil mean ?

1. Apr 7, 2006

### momentum

Its diifcult to express my question....so, i am posting this

ceil(4.5) =?
ceil(4.1)=?
ceil(4.6)=?

2. Apr 7, 2006

### AKG

ceil(x) is the smallest integer which is greater than or equal to x. In particular, if x is an integer, then ceil(x) = x, and if x is not an integer, then ceil(x) > x.

3. Apr 7, 2006

### momentum

ceil(4.5)=4 // is it ok ?
ceil(4.1)=4 //is it ok ?
ceil(4.6)=5 //is it ok ?

4. Apr 7, 2006

### Integral

Staff Emeritus
Nope, read the defintion given above, then try again.

5. Apr 7, 2006

### momentum

ah...i see, all of them should be 5 .....i had confusion on fractional part .5.
but i see ..it does not care for .5 which we use for round-off.

6. Apr 7, 2006

Yes, all 5.

7. Apr 7, 2006

### momentum

thank you for the clarifcation

8. Apr 7, 2006

### bomba923

Recall that

$$\begin{gathered} \forall x \in \left( {a,a + 1} \right)\;{\text{where }}a \in \mathbb{Z}, \hfill \\ {\text{floor}}\left( x \right) = \left\lfloor x \right\rfloor = a \hfill \\ {\text{ceil}}\left( x \right) = \left\lceil x \right\rceil = a + 1 \hfill \\ \end{gathered}$$

$$\forall x \in \mathbb{Z},\;\left\lfloor x \right\rfloor = \left\lceil x \right\rceil = x$$

Last edited: Apr 7, 2006
9. Apr 7, 2006

### Staff: Mentor

ceil -> "goes up" if it needs to, in order reach an integer
floor -> "goes down" as it needs to, in order to reach an integer
What happens with negative numbers:

$$floor( -1.1 ) = -2 \; ceil( -1.1 ) = -1$$
$$floor( -0.1 ) = -1 \; ceil( -0.1 ) = 0$$
$$floor( 0.9 ) = 0 \; ceil( 0.9 ) = 1$$


# How to translate this statement into a mathematical one(using appropriate quantifiers)?

The statement I'd like to translate into a mathematical one is

"Every American has a dream".

Let $$A$$ and $$D$$ denote the set of all Americans and the set of all dreams, respectively, and $$P(a,d)$$ denote the proposition "American $$a$$ has a dream $$d$$". The mathematically equivalent statement I've deduced is $$\forall a\in A.\exists d\in D.P(a, d)$$

However, I suspect the above statement implies that for every American there exists a common dream $$d$$ such that $$P(a,d)$$ holds true. I would like to know how to rectify this error(if there is one).

• Correct: "forall a there is a d..." does not mean that the d is the same for all a. To state that the d is the same, you have to write "there is a d for all a...". Compare $\forall n \exists m (n < m)$ and $\exists m \forall n (n < m)$. May 25, 2020 at 10:01

1. The verifier of a sentence of the form "$$∀a{∈}A\ ( P(a) )$$" must let the refuter first choose any arbitrary $$a∈A$$ and then verify $$P(a)$$ no matter what $$a∈A$$ was chosen.
2. The verifier of a sentence of the form "$$∃d{∈}D\ ( Q(d) )$$" must first choose some $$d∈D$$ and then verify $$Q(d)$$ for that chosen $$d∈D$$.
In your example, the verifier of "$$∀a{∈}A\ ∃d{∈}D\ ( P(a,d) )$$" must let the refuter make the first move in choosing an $$a∈A$$, and then verify "$$∃d{∈}D\ ( P(a,d) )$$" no matter what $$a$$ was chosen. But since the verifier makes the second move in choosing some $$d∈D$$, the verifier can choose this $$d$$ based on the refuter's first move (i.e. based on $$a$$). That is why "Every American has a dream." corresponds to this sentence.
In contrast, the verifier of "$$∃d{∈}D\ ∀a{∈}A\ ( P(a,d) )$$" must make the first move in choosing some $$d∈D$$, before the refuter makes the second move in choosing an $$a∈A$$. You can see easily that the verifier can win only if there is a single choice of $$d∈D$$ that defeats every possible choice of $$a∈A$$. That is why "All Americans have a common dream." corresponds to this sentence and not the other one.


A right triangle may be isosceles … Input 3 triangle side lengths (A, B and C), then click "ENTER". Constructing an isosceles triangle Constructing an isosceles triangle also known as drawing an isosceles triangle using only a straightedge and a compass is what I will show you here. B - isosceles triangle. Yes, because it has an obtuse angle. Please update your bookmarks accordingly. Obtuse isosceles triangle with an irrational but algebraic ratio between the lengths of its sides and its base. This concept will teach students the properties of isosceles triangles and how to apply them to different types of problems. An isosceles right triangle therefore has angles of 45 degrees, 45 degrees, and 90 degrees. Apply the properties of isosceles triangles. An isosceles triangle is a special case of a triangle where 2 sides, a and c, are equal and 2 angles, A and C, are equal. Given below are the properties of isosceles and acute triangles. Parts of an Isosceles Triangle. The angles opposite the equal sides are also equal. Find information related to equilateral triangles, isosceles triangles, scalene triangles, obtuse triangles, acute triangles, right angle triangles, the hypotenuse, angles of a triangle and more. C - obtuse triangle. Let the coordinates of the right-angled isosceles triangle be O(0, 0), A(a, 0) and B(0, a). Area of the acute triangle is $$A = {1 \over 2} \times b \times h$$ The Perimeter of the Acute triangle is: $$P=a+b+c$$ The types of acute triangles are: a) Acute Equilateral Triangle b) Acute Isosceles Triangle c) Acute Scalene Triangle All three angles are different implies that all three sides of an acute scalene triangle are also different. The two shorter sides measure x cm and 3x cm. Without Using The Calculator When given 3 triangle sides, to determine if the triangle is acute… An acute angle triangle (or acute-angled triangle) is a triangle in which all the interior angles are acute angles. Step-by-step explanation: Correct statements and reasons are in bold. A - acute triangle. For an acute triangle, all three angles of the triangle are from $$0^o$$ to $$90^o$$. This calculator will determine whether those 3 sides will form an equilateral, isoceles, acute, right or obtuse triangle or no triangle at all. In a right triangle, one of the angles is a right angle—an angle of 90 degrees. For example, if we know a and b we know c since c = a. Addtionally, like all triangles, the three angles will sum to . All the three angles situated within the isosceles triangle are acute, which signifies that the angles are less than 90°. We have moved all content for this concept to for better organization. Step 1: One of the angles of the given triangle is a right angle. Add these two angles together and subtract the answer from 180° to find the remaining third angle. Answer : D Explanation. the isosceles triangle can be acute, right or obtuse, but it depends only on the vertex angle (base angles are always acute) The equilateral triangle is a special case of a isosceles triangle. PROPERTIES OF ISOSCELES TRIANGLE ABC Let ABC be an isosceles triangle with
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# C F Gauss, on how to add all numbers of 1 to 100?

#### CaptainBlack

##### Well-known member

Carl gauss 1777 1855 , on how to add all numbers of 1 to 100?

I can only add from 1 to 100 in order, apparently he could add lighteningly fast, and backwards, can anyone explain how in the confines of one message on here please
Allegedly his method was something like:

$(1+100)+(2+99)+ ... +(99+2)+(100+1)=2(1+2+...+100)$

But the left hand side is $101\times 100$, so $1+2+..100=101\times 100/2$

(Like Marlow's Dr Faustus he could sum them forwards and backwards - but had no need of anagramatisation)

CB

Last edited:

#### Bacterius

##### Well-known member
MHB Math Helper
This also works for any sum:

$\displaystyle \sum_{k = 1}^{n} k = 1 + 2 + \cdots + n$

This sum can be reorganized in this fashion:

$\displaystyle \left ( 1 + \left ( n \right ) \right ) + \left ( 2 + \left ( n - 1 \right ) \right ) + \cdots + \left (n + \left (1 \right ) \right )$

This produces $n$ sums of $n + 1$, so the total sum is $n(n + 1)$. But note that this goes through the list twice, from the left and from the right (for instance, $1$ is added to $n$ at the beginning and at the end). So this is actually twice the sum we need, we conclude:

$\displaystyle \sum_{k = 1}^{n} k = \frac{n(n + 1)}{2}$

$n = 100$ can then be easily calculated from the formula, just as can $n = 4885$ or $n = 6$

#### QuestForInsight

##### Member
I would like to share two more ways of doing this.

Proof #0 (Notice that what Gauss did is basically this):

\begin{aligned} \displaystyle & \sum_{0 \le k \le n}k = \sum_{0 \le k \le n}(n-k) = n\sum_{0 \le k \le n}-\sum_{0 \le k \le n}k \\& \implies 2\sum_{0 \le k \le n}k = n(n+1) \implies \sum_{0 \le k \le n}k = \frac{1}{2}n(n+1).\end{aligned}

Proof #1 (Now try that with $\sum_{0 \le k \le n}k^2$ and you will be pleasantly surprised):

\begin{aligned} \displaystyle & \sum_{0 \le k \le n}k^2 = \sum_{0 \le k \le n}(n-k)^2 = n^2\sum_{0 \le k \le n}-2n\sum_{0 \le k \le n}k+\sum_{0 \le k \le n}k^2 \\& \implies 2n\sum_{0 \le k \le n}k = n^2(n+1) \implies \sum_{0 \le k \le n}k = \frac{1}{2}n(n+1).\end{aligned}

Proof #2 (Writing it as a double sum and switching the order of summation):

\begin{aligned}\displaystyle & \begin{aligned}\sum_{1 \le k \le n}k & = \sum_{1 \le k \le n}~\sum_{1 \le r \le k} = \sum_{1 \le r \le n} ~\sum_{r \le k \le n} = \sum_{1 \le r \le n}\bigg(\sum_{1 \le k \le n}-\sum_{1 \le k \le r-1}\bigg) \\& =\sum_{1 \le r \le n}\bigg(n-r+1\bigg) = n\sum_{1 \le k \le n}-\sum_{1 \le k \le n}k+\sum_{1 \le k \le n}\end{aligned} \\& \implies 2\sum_{1 \le k \le n}k = n^2+n \implies \sum_{1 \le k \le n}k = \frac{1}{2}n(n+1), ~ \mathbb{Q. E. D.}\end{aligned}

Last edited:

#### MarkFL

Staff member

Here are two closely related methods to derive the formulas for summations of sequences of natural number powers of integers from 0 - n.

Method 1: A "bottom-up" approach...

We may state:

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

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

$\displaystyle0=\sum_{k=0}^n(1)-(n+1)$

$\displaystyle\sum_{k=0}^n(1)=n+1$

Now we may compute:

$\displaystyle\sum_{k=0}^n(k)$

We may state:

$\displaystyle\sum_{k=0}^n\left(k^2 \right)=\sum_{k=0}^n\left((k+1)^2 \right)-(n+1)^2$

$\displaystyle\sum_{k=0}^n\left(k^2 \right)=\sum_{k=0}^n\left(k^2+2k+1 \right)-(n+1)^2$

$\displaystyle\sum_{k=0}^n\left(k^2 \right)=\sum_{k=0}^n\left(k^2 \right)+2\sum_{k=0}^n(k)+\sum_{k=0}^n(1)-(n+1)^2$

$\displaystyle2\sum_{k=0}^n(k)=-\sum_{k=0}^n(1)+(n+1)^2$

Now, using our previous result, we have:

$\displaystyle2\sum_{k=0}^n(k)=-(n+1)+(n+1)^2$

$\displaystyle2\sum_{k=0}^n(k)=(n+1)\left(-1+(n+1) \right)$

$\displaystyle2\sum_{k=0}^n(k)=(n+1)(n)$

$\displaystyle\sum_{k=0}^n(k)=\frac{n(n+1)}{2}$

Now we may continue and find:

$\displaystyle\sum_{k=0}^n\left(k^2 \right)$

$\displaystyle\sum_{k=0}^n\left(k^3 \right)=\sum_{k=0}^n\left((k+1)^3 \right)-(n+1)^3$

$\displaystyle\sum_{k=0}^n\left(k^3 \right)=\sum_{k=0}^n\left(k^3+3k^2+3k+1 \right)-(n+1)^3$

$\displaystyle\sum_{k=0}^n\left(k^3 \right)=\sum_{k=0}^n\left(k^3 \right)+3\sum_{k=0}^n\left(k^2 \right)+3\sum_{k=0}^n(k)+\sum_{k=0}^n(1)-(n+1)^3$

$\displaystyle3\sum_{k=0}^n\left(k^2 \right)=-3\sum_{k=0}^n(k)-\sum_{k=0}^n(1)+(n+1)^3$

Using our previous results, we have:

$\displaystyle3\sum_{k=0}^n\left(k^2 \right)=-3\left(\frac{n(n+1)}{2} \right)-(n+1)+(n+1)^3$

$\displaystyle3\sum_{k=0}^n\left(k^2 \right)=(n+1)\left(-\frac{3n}{2}-1+(n+1)^2 \right)$

$\displaystyle3\sum_{k=0}^n\left(k^2 \right)=(n+1)\left(-\frac{3n}{2}-1+n^2+2n+1 \right)$

$\displaystyle3\sum_{k=0}^n\left(k^2 \right)=(n+1)\left(n^2+\frac{n}{2} \right)$

$\displaystyle3\sum_{k=0}^n\left(k^2 \right)=\frac{n(n+1)(2n+1)}{2}$

$\displaystyle\sum_{k=0}^n\left(k^2 \right)=\frac{n(n+1)(2n+1)}{6}$

We may continue this process, to find further power summations.

Method 2: A recursion approach...

Suppose we want to find:

$\displaystyle S_n=\sum_{k=0}^n\left(k^3 \right)$

We may use the inhomogeneous recursion to state:

$\displaystyle S_n=S_{n-1}+n^3$

Now, we may use symbolic differencing to ultimately derive a homogeneous recursion.

We begin by replacing n with n + 1:

$\displaystyle S_{n+1}=S_{n}+(n+1)^3$

Subtracting the former from the latter, there results:

$\displaystyle S_{n+1}=2S_{n}-S_{n-1}+3n^2+3n+1$

$\displaystyle S_{n+2}=2S_{n+1}-S_{n}+3(n+1)^2+3(n+1)+1$

Subtracting again:

$\displaystyle S_{n+2}=3S_{n+1}-3S_{n}+S_{n-1}+6n+6$

$\displaystyle S_{n+3}=3S_{n+2}-3S_{n+1}+S_{n}+6(n+1)+6$

Subtracting again:

$\displaystyle S_{n+3}=4S_{n+2}-6S_{n+1}+4S_{n}-S_{n-1}+6$

$\displaystyle S_{n+4}=4S_{n+3}-6S_{n+2}+4S_{n+1}-S_{n}+6$

Subtracting again:

$\displaystyle S_{n+4}=5S_{n+3}-10S_{n+2}+10S_{n+1}-5S_{n}+S_{n-1}$

Now we have a homogeneous recursion whose associated characteristic equation is:

$\displaystyle (r-1)^5=0$

Since we have the root $\displaystyle r=1$ of multiplicity 5, we know the closed form for the sum will take the form:

$\displaystyle S_n=k_0+k_1n+k_2n^2+k_3n^3+k_4n^4$

We may use known initial values to determine the constants $\displaystyle k_i$.

$\displaystyle S_0=k_0=0$

This means, we are left with the 4X4 system:

$\displaystyle k_1+k_2+k_3+k_4=1$

$\displaystyle 2k_1+4k_2+8k_3+16k_4=9$

$\displaystyle 3k_1+9k_2+27k_3+81k_4=36$

$\displaystyle 4k_1+16k_2+64k_3+256k_4=100$

Did you notice we get the squares of successive triangular numbers? Anyway, solving this system, we find:

$\displaystyle k_1=0,k_2=\frac{1}{4},k_3=\frac{1}{2},k_4=\frac{1}{4}$

And so, we have:

$\displaystyle S_n=\frac{1}{4}n^2+\frac{1}{2}n^3+\frac{1}{4}n^4= \frac{n^4+2n^3+n^2}{4}=$

$\displaystyle\frac{n^2(n+1)^2}{4}=\left( \frac{n(n+1)}{2} \right)^2$

#### melese

##### Member
a nice combinatorial proof

To each yellow circle we can associate a pair of green circles in the way suggested by the drawing. This shows a bijection between (and hence the same number of) all the yellow circles and all the pairs of green circles.

Here, there are $1+2+3+4+5+6+7+8$ yellow circles, and the number of green-circle pairs is $\binom{9}{2}$; so $1+2+3+4+5+6+7+8=\binom{9}{2}$.

(I saw this 'proof without words' in a lecture by Gil Kalai)

#### HallsofIvy

##### Well-known member
MHB Math Helper
$(1+100)+(2+99)+ ... +(99+2)+(100+1)=2(1+2+...+100)$
But the left hand side is $101\times 100$, so $1+2+..100=101\times 100/2$


math, probability

13. Exercise: Convergence in probability:

a) Suppose that Xn is an exponential random variable with parameter lambda = n. Does the sequence {Xn} converge in probability?

b) Suppose that Xn is an exponential random variable with parameter lambda = 1/n. Does the sequence {Xn} converge in probability?

c) Suppose that the random variable in the sequence {Xn} are independent, and that the sequence converges to some number a, in probability.
Let {Yn} be another sequence of random variables that are dependent, but where each Yn has the same distribution (CDF) as Xn. Is it necessarily true that the sequence {Yn} converges to a in probability?

1. 👍
2. 👎
3. 👁
1. a) yes
b) no
c) yes

1. 👍
2. 👎
2. y
n
n

1. 👍
2. 👎
3. 142n=47 find n

1. 👍
2. 👎

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Accelerating the pace of engineering and science

# istriu

Determine if matrix is upper triangular

## Description

example

tf = istriu(A) returns logical 1 (true) if A is an upper triangular matrix; otherwise, it returns logical 0 (false).

## Examples

expand all

### Test Upper Triangular Matrix

Create a 5-by-5 matrix.

`A = triu(magic(5))`
```A =

17    24     1     8    15
0     5     7    14    16
0     0    13    20    22
0     0     0    21     3
0     0     0     0     9```

Test A to see if it is upper triangular.

`istriu(A)`
```ans =

1
```

The result is logical 1 (true) because all elements below the main diagonal are zero.

### Test Matrix of Zeros

Create a 5-by-5 matrix of zeros.

`Z = zeros(5);`

Test Z to see if it is upper triangular.

`istriu(Z)`
```ans =

1
```

The result is logical 1 (true) because an upper triangular matrix can have any number of zeros on the main diagonal.

## Input Arguments

expand all

### A — Input arraynumeric array

Input array, specified as a numeric array. istriu returns logical 0 (false) if A has more than two dimensions.

Data Types: single | double
Complex Number Support: Yes

expand all

### Upper Triangular Matrix

A matrix is upper triangular if all elements below the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.

For example, the matrix

$A=\left(\begin{array}{cccc}1& -1& -1& -1\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -2& -2\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -3\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\right)$

is upper triangular. A diagonal matrix is both upper and lower triangular.

### Tips

• Use the triu function to produce upper triangular matrices for which istriu returns logical 1 (true).

• The functions isdiag, istriu, and istril are special cases of the function isbanded, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, istriu(A) == isbanded(A,0,size(A,2)).


# Generalized Fibonacci sequences

Why Fibonacci sequence start at $0$, Tribonacci sequence with $0,0$, Tetranacci with $0,0,0$, etc. [ref OEIS] Has any good reasons for that?

These sequences arise in generalization of Pascal Triangle as diagonal sums and there they start at $1$.

Pascal triangle:

$$\begin{array}{} \color{red}1& \color{blue}0& \color{green}0& \color{cyan}0&\dots\\ \color{blue}1& \color{green}1& \color{cyan}0& \color{magenta}0&\dots\\ \color{green}1& \color{cyan}2& \color{magenta}1& 0&\dots\\ \color{cyan}1& \color{magenta}3& 3& \color{red}1&\dots\\ \vdots&\vdots&\vdots&\vdots&\ddots \end{array}$$

diagonal sum gives $\color{red}1,\color{blue}1,\color{green}2,\color{cyan}3,\color{magenta}5,8,\color{red}{13}\dots$ Fibonacci sequence

First generalization:

$$\begin{array}{r} \color{red}1& \color{blue}0& \color{green}0& \color{cyan}0& \color{magenta}0& 0&\color{red}0&\dots\\ \color{blue}1& \color{green}1& \color{cyan}1& \color{magenta}0& 0& \color{red}0&\color{blue}0&\dots\\ \color{green}1& \color{cyan}2& \color{magenta}3& 2& \color{red}1& \color{blue}0&\color{green}0&\dots\\ \color{cyan}1& \color{magenta}3& 6& \color{red}7& \color{blue}6& \color{green}3&\color{cyan}1&\dots\\ \color{magenta}1&4&\color{red}{10}&\color{blue}{16}&\color{green}{19}&\color{cyan}{16}&\color{magenta}{10}&\dots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots \end{array}$$

gives sequence of diagonal sum $\color{red}1,\color{blue}1,\color{green}2,\color{cyan}4,\color{magenta}7,\dots$, Tribonacci sequence, etc.

-
Sorry, I'm slow: how are you computing your diagonal sums? –  Guess who it is. Dec 18 '11 at 1:29
Also, if you look here, I'd say it's pretty convenient to have the generalized Fibonacci numbers take on positive values for positive index. –  Guess who it is. Dec 18 '11 at 1:33
Thanks @Robert! I found this in the meantime. –  Guess who it is. Dec 18 '11 at 1:37
@J.M.: I added some color to make it more obvious. –  Brian M. Scott Dec 18 '11 at 2:19
@Peter: If $a(r,c)$ is the entry in row $r$, column $c$, $$a(r+1,c)=a(r,c)+a(r,c-1)+a(r,c-2)\;.$$ There’s an error in the table, which I’ve now fixed. –  Brian M. Scott Dec 18 '11 at 10:04

The essential definition of a $k$-nacci sequence is that the earliest positive term is $1$ and all subsequent terms are the sum of the previous $k$ terms.

So to find the second earliest positive term (which is also $1$) you need at least $k-1$ zeros at the beginning of the sequence to do the sum. In your tables there are at least $k-1$ columns of implicit zeroes to the left of the columns you show.

Since the Fibonacci sequence conventionally starts with $F(1)=1$ and $F(0)=0$, it is a plausible but not necessary convention to start the $k$-nacci sequence with the initial leading zero also at a zero index, and this is what the OEIS has done.

You could do something different, but then you would need to translate between your indices and those used by others. That is the effect of conventions.

-

While it looks natural to start Fibonacci at $f_1=1$, the Fibonacci actually goes both ways. Indeed

$$f_{n+1}=f_n+f_{n-1} \,,$$ implies

$$f_{n-1}= f_{n+1}-f_n \,.$$

Thus you can calculate $f_0$ and $f_{-n}$.

Given a recurrence which goes in both directions, any "start" point is somewhat artificial.

I find the start $f_0=0$ useful when I work with the matrices

$$F:=\left( \begin{array}{rr} 1 & 1\\ 1 & 0 \end{array}\right)=\left(\begin{array}{rr} f_2 & f_1\\ f_1 & f_0 \end{array}\right) \,.$$

Note that without starting with $f_0=0$, the standard formula

$$F^n= \left(\begin{array}{rr} f_{n+1} & f_n\\ f_n & f_{n-1} \end{array}\right)$$

would not work for $n=1$. This would make some consequences weaker, since you'd need always the eliminate the case when the parameter(s) are 1. But this would be a not needed elimination.

It is somehow similar to the following situation: You have a function $f$ in the complex plane, and you can prove it is entire, but you only need the fact that it is Analytical in some region $R$. What do you do? Do you prove that it is Analytic in the region you need for the applications you are interested, or prove it is entire?

Who knows, at some point in the future you might need that the function it is entire. You also might need to work with $f_0$, and I explained above why $f_n$ can be extended in an unique way below 0. Same can be said about the other sequences.

-

As for "Why Fibonacci sequence start at $0$?" one compelling reason is that this indexing is the natural one that reveals their interesting divisibility properties, namely that they form a strong divisibility sequence, i.e.

$$\rm\ gcd(f_m,f_n)\ =\ f_{\:gcd(m,n)}$$

hence $\rm\ m\ |\ n\ \Rightarrow\ f_m\ |\ f_n\$ etc. Analogous results hold for the more general class of Lucas-Lehmer sequences, which prove convenient when studying elementary number theory of quadratic fields, e.g. generalizations of Fermat's little theorem, Euler's $\phi$ totient function, etc. If one changed the indexing then many of these results would be greatly obfuscated.

That said, it should be emphasized that such definitions are merely conventions that prove useful in the context at hand. In other contexts - where such divisibility properties play no role - another indexing might prove more convenient.

-
Can you be more specific about Lucas-Lehmer sequences as a general notion? The only somewhat authoritative mention I can find is in OEIS, but is is only one sequence, even though its title enigmatically uses the indefinite article. As far as I know the only thing really called after Lucas and Lehmer is the primality test (in which the above sequence occurs, and no other). Possibly you meant Lucas sequences? –  Marc van Leeuwen Dec 18 '11 at 8:02
@Marc D.H. Lehmer developed an extended theory of Lucas sequences in his PhD thesis, which was published in Annals of Math., 1930 Due to such, many authors refer to the general class as Lucas-Lehmer sequences, e.g. see Ribenboim's The New Book of Prime Number Records, 1996. –  Bill Dubuque Dec 18 '11 at 15:07

Once you have also an expression, which allows interpolating to noninteger n (thus putting the sequence into a greater framework), then this might provide an argument to select a specific n as beginning of the sequence. For the Fibonacci-numbers I think the Binet-formula is
$$\small fib(n) = {((1+\sqrt 5 )^n - (1-\sqrt 5)^n) \over 2^n \sqrt 5 }$$ and then $\small fib(0) = 0$. Surely one can insert an arbitrary offset for n into the defining expression, but the given definition may be taken as the most natural.

-


# Why there are no constant functions on $\mathbb{R}$ with compact support?

For some the question might seem trivial but the concept is new to me and I have been wondering why there are no constant functions on $$\mathbb{R}$$ with compact support?

Following wiki:

Def.1) Functions with compact support on a topological space $$X$$ are those whose support is a compact subset of $$X$$.

Def.2) $$\operatorname{supp}f=:\overline{\{x\in X\;,f(x)\neq 0\}}$$

What if we take such function $$f:\mathbb{R}\to\mathbb{R}$$ defined as $$f(x)=0\;,\forall x\in\mathbb{R}$$ then $$\operatorname{supp}f=\overline{\{x\in\mathbb{R}\;,f(x)\neq 0\}}=\overline{\emptyset}=\emptyset$$ but $$\emptyset$$ is a compact subset of $$\mathbb{R}$$ I think....

Could someone enlighten me? Thank you.

• Where were you told that no constant functions on $\mathbb{R}$ have compact support? – Hayden May 31 '14 at 18:28
• well, anyway that's the only one there is, $f \equiv 0$ – mm-aops May 31 '14 at 18:30
• @Hayden: here for example www-personal.umich.edu/~wangzuoq/437W13/Notes/Lec%2030.pdf (12th line) – user124471 May 31 '14 at 18:32
• Well, I'd beg to differ with the claim, for precisely the example you gave. But as mm-aops pointed out, it's the only example. – Hayden May 31 '14 at 18:35
• the statement is very clear : ''there are NO functions'', (such functions don't exists). The real question here is if my counter-example is really correct (i.e that I followed all definitions correctly and so on). – user124471 May 31 '14 at 18:37

## 1 Answer

The author in the article you link to (www-personal.umich.edu/~wangzuoq/437W13/Notes/Lec%2030.pdf) is just being sloppy. He really should have said there are no non-zero constant functions with compact support on $\mathbb R$ (and thus $H_c^0(\mathbb R)$.

So yes, your example is correct. The empty set is (tautologically) compact.

• Hi Fredrik. Thank you. Unfortunately it is not the first time I have seen it....Exactly the same statement I found in Bott, Tu, ''Differential forms in algebraic topology''. Cheers – user124471 May 31 '14 at 18:44
• @user124471 Yep, I've seen it as well. In fact, I was just reading in the book by Bott & Tu a week ago, and had to think about the exact same question as you asked. – Fredrik Meyer May 31 '14 at 18:47
• @user124471 I was wondering the same thing! Now I know. Can you include Bott Tu in your question? – user636532 May 29 '19 at 11:27