# Math Help - Divisibility by 7

1. ## Divisibility by 7

Prove that $2222^{5555}+5555^{2222}$ is divisible by 7.

I set about trying to show that this expression is equal to $0 \mod 7$.

I have that:
$2222=3 \mod7=2 \mod 6$
$5555=4 \mod 7=5 \mod 6$

The question also has Fermat's little theorem above it so I think I should be using that somewhere.

Anyone have any ideas?

2. Hello,

Let's take for example $2222^{5555}$
It's congruent to $3^{5555}(\bmod 7)$

Now, you know by Fermat's little theorem, that $3^{6}\equiv 1(\bmod 7)$

As you noticed, $5555\equiv 5(\bmod 6)$

It can be proved that hence $2222^{5555}\equiv 3^5 (\bmod 7)$

Why ? Because $5555=6k+5$

And then $3^{6k+5}=(3^6)^k\cdot 3^5 \dots$

3. Originally Posted by Showcase_22
Prove that $2222^{5555}+5555^{2222}$ is divisible by 7.

I set about trying to show that this expression is equal to $0 \mod 7$.

I have that:
$2222=3 \mod7=2 \mod 6$
$5555=4 \mod 7=5 \mod 6$

The question also has Fermat's little theorem above it so I think I should be using that somewhere.

Anyone have any ideas?
Hi Showcase_22.

Fermat’s little theorem is a little cumbersome for this problem. Here is a shorter way.

$2222\equiv3\,\bmod7\ \implies\ 2222^3\equiv-1\,\bmod7$ $\implies\ 2222^{5555}=2222^{3\cdot1851+2}\equiv-3^2\,\bmod7\equiv5\,\bmod7$

$5555\equiv4\,\bmod7\ \implies\ 5555^3\equiv1\,\bmod7$ $\implies\ 5555^{2222}=5555^{3\cdot740+2}\equiv4^2\,\bmod7\eq uiv2\,\bmod7$

Hence $2222^{5555}+5555^{2222}\equiv7\,\bmod7\equiv0\,\bm od7.$

The reason I prefer $2222^3$ and $5555^3$ rather than $2222^6$ and $5555^6$ is that their powers can be brought closer this way to 5555 and 2222 respectively.

4. Hello,
Hello!

I finally got it to work out using TheAbstractionist's method (I looked at what you did, did it for 2222 whilst looking at yours and then did it for 5555 separately).

I'll have a go using your method now Moo since it seems to be the one in the mark scheme.

5. Okay, here's my way of doing it. I think it's both of your methods combined (or I may just be copying one of your methods, i'm not that good at number theory!):

$2222^{5555}=3^{5555} \bmod 7$

$3^6=1 \bmod 7$ by Fermat's little theorem

$3^{5555} \bmod 7=3^{6(925)+5}=3^{6(925)}3^5 \bmod 7$

$=(1 \bmod 7)(3^2. 3^2.3 \bmod 7)=(1 \bmod 7 )(5 \bmod 7)=5 \bmod 7$

$\Rightarrow 2222^{5555}=5 \bmod 7$

$5555^{2222}=(5 \bmod 6)^{2222}$

$4^6=1 \bmod 7$ by Fermat's little theorem.

$5555^{2222}=(4 \bmod 7)^{2222}$

$=4^{370(6)+2} \bmod 7=4^{370(6)}.4^2 \bmod 7=(1 \bmod 7)(2 \bmod 7)=2 \bmod 7$

$\Rightarrow 5555^{2222}=2 \bmod 7$

$\Rightarrow 2222^{5555}+5555^{2222}=2 \bmod 7+5 \mod 7=7 \bmod 7= 0 \bmod 7$

**WIN**

6. Well, actually that was exactly the way I wanted you to do

Good job.

Though your notations are very strange...
$=(1 \bmod 7)(3^2. 3^2.3 \bmod 7)$
Just write $1\cdot 3^2 \cdot 3^2 \cdot 3 \bmod 7$

$5555^{2222}=(5 \bmod {\color{red}7})^{2222}
$
Write $5555^{2222}=5^{2222} \bmod 7$

7. Ah, sorry about my notations!

I didn't realise there was a command for $"\cdot"$. The more I know!

That second one was just plain wrong, sorry I wrote it!

On the plus side:

Spoiler:
At least I know how to use these spoiler boxes now!


# How many ways to select $k$ vertices of an $n$-gon?

I have a regular $n$-gon, of which I have to select $k$ vertices. The selections must be rotationally distinct; two selections would be considered equivalent if one is a rotation of the other. For example, if I have a square, and I want to select 2 vertices, there are only 2 possible ways to do that according to the constraint. One is "x - x -", another is "x x - -".

If we denote the function by $CR(n,k)$, then these are the trivial cases:

1. $CR(n, 1) = 1$

2. $CR(n, 2) = \lfloor\frac{n}{2}\rfloor$

3. $CR(n, k) = CR(n, n - k)$

I am quite short of ideas on how to find the recurrence or closed formula of this problem, or if this problem has any closed form / recurrence solution at all. Any help with a bit detailed walk through would be much appreciated.

• What do you mean by selection must be rotationally invariant? – Anurag A May 28 '15 at 19:07
• Supposing, I have a 4-gon and I have to select 2 vertices. Hence, these two selections are the same. x-x-, -x-x. These selections are the same, x--x, xx--, -xx-, --xx – sarker306 May 28 '15 at 19:09
• The following MSE meta link has a section on enumerating necklaces and bracelets. – Marko Riedel May 28 '15 at 19:51
• One way to think about this (perhaps the best way) is the group action of cyclic rotations on the subsets of vertices. Rotations preserve the number of vertices in a subset (obviously), so the counting comes down to counting orbits of the $k$-subsets of $n$-vertices under those rotations. – hardmath May 28 '15 at 19:53

As the OP seems interested in fixing $k$ and letting $n$ vary we will do an example to show how this might work.

By the Polya Enumeration Theorem what we have here is $$[z^k] Z(C_n)(1+z)$$ with $Z(C_n)$ the cycle index of the cyclic group which is $$Z(C_n) = \frac{1}{n}\sum_{d|n} \varphi(d) a_d^{n/d}.$$

The desired quantity is thus given by $$[z^k] Z(C_n)(1+z) = [z^k] \frac{1}{n}\sum_{d|n} \varphi(d) (1+z^d)^{n/d}.$$

This is $$\frac{1}{n}\sum_{d|n, d|k} \varphi(d) [z^k] (1+z^d)^{n/d}$$ or $$\frac{1}{n}\sum_{d|n, d|k} \varphi(d) {n/d \choose k/d} = \frac{1}{n}\sum_{d|\gcd(n,k)} \varphi(d) {n/d \choose k/d}.$$

For $k=4$ starting at $n=1$ we obtain the sequence $$0, 0, 0, 1, 1, 3, 5, 10, 14, 22, 30, 43, 55, 73,\ldots$$ which points us to OEIS A008610 where we find confirmation.

Another interesting one is $k=6$ which yields $$0, 0, 0, 0, 0, 1, 1, 4, 10, 22, 42, 80, 132, 217, 335, 504,\ldots$$ which points to OEIS A032191.

A Maple session with these looks like this:

> with(numtheory):
> Q := (n,k) -> 1/n*add(phi(d)*binomial(n/d,k/d), d in divisors(gcd(n,k)));
Q := (n, k) -> add(numtheory:-phi(d) binomial(n/d, k/d),

d  in  numtheory:-divisors(gcd(n, k)))/n

> seq(Q(n,8), n=1..18);
0, 0, 0, 0, 0, 0, 0, 1, 1, 5, 15, 43, 99, 217, 429, 810, 1430, 2438

which incidentally is OEIS A032193.

• It'll be a sin by my side not to refer you to my similiar question here: math.stackexchange.com/questions/1294224/… . Perhaps you'll be able to help :) I apologize, OP! Really am! – Matan May 28 '15 at 20:40
• I have been following your question for three hours. @Matan , felt really intrigued. – sarker306 May 28 '15 at 20:45
• How wonderful to hear. Thanks! I update it on regular basis and work on it when I have time. Keep track of it :) @sarker306 – Matan May 28 '15 at 20:49

Taking the necklace proof for Fermat's Little Theorem as inspiration:

For $n,k$ coprime, every possible selection has $n$ rotations. So

$$CR(n,k) = \frac 1n{n \choose k}$$

Otherwise letting $d=\gcd(n,k)$, we need to account for those choices where there is a repeat pattern smaller than $n$, which will be all possible choices over the shorter interval with the reduced number of choices (but repeated through the entire set):

\begin{align} CR(n,k) &= \frac 1n\left[{n \choose k} - {n/d \choose k/d }\right] + CR\left(\frac nd, \frac kd \right) \\ &=\frac 1n\left[{n \choose k} - {n/d \choose k/d }\right] + \frac 1d {n/d \choose k/d } \end{align}

since $\gcd(\frac nd, \frac kd)=1$

And also note that $CR(n,0)=1$ and $CR(n,n)=1$.

• Any explanation? Or any source where the derivation is explained? – sarker306 May 28 '15 at 19:25
• Joffan, if you could please refer here math.stackexchange.com/questions/1294224/… for a related problem I asked I will be grateful. Perhaps you'll be able to help. – Matan May 28 '15 at 19:26
• @sarker306 If you review the necklace proof of Fermat's Little theorem I think you'll understand where this comes from. If I have time I'll add to it. – Joffan May 28 '15 at 19:34
• I get CR(6,3) = 25/6 when I apply your solution. – TokenToucan May 28 '15 at 20:17
• CR(n, 2) has its correct value when n is odd, not even. But really thanks for bringing in this gcd issue, it seems to be a right step to the right direction, I didn't suspect it before. I'll think it through the night. – sarker306 May 28 '15 at 20:25


# How many distinct ways to climb stairs in 1 or 2 steps at a time?

I came across an interesting puzzle:

You are climbing a stair case. It takes $n$ steps to reach to the top. Each time you can either climb $1$ or $2$ steps. In how many distinct ways can you climb to the top?

Is there a closed-form solution to the problem? One can compute it by creating a 'tree' of possibilities of each step. That is, I can either take 1 or 2 steps at each stage and terminate a branch once it sums to $n$. But this is would get really unwieldy very quickly since the maximum number of nodes in a binary tree is $2^{n+1}-1$, i.e., exponential. Is there an easier way to solving this puzzle?

Let $F_n$ be the number of ways to climb $n$ stairs taking only $1$ or $2$ steps. We know that $F_1 = 1$ and $F_2 = 2$. Now, consider $F_n$ for $n\ge 3$. The final step will be of size $1$ or $2$, so $F_n$ = $F_{n-1} + F_{n-2}$. This is the Fibonacci recurrence.

• Ha! Some simple and yet a convoluted formation of the sequence. That explains it. Can't believe I couldn't see it :) – PhD May 11 '14 at 1:53
• Is it possible to generalize it? Such as what if your steps are 2, 3, and 5? Also, we can compute fibonacci in matrix form which is faster. In the case of 2,3 and 5 steps, what would be the closed form solution? – Bagus Trihatmaja Dec 5 '18 at 8:53
• Yes, this falls into the more general framework of solving linear recurrence equations. They always admit fast evaluations via matrix exponentiation and often have closed-form solutions, especially when the characteristic equation has low degree since you need to find all of its roots. – fahrbach Dec 6 '18 at 22:19
• Can someone help me understand a tweak to this problem where the order of the steps doesn't matter? How to come up to a closed form equation for it? – mankadnandan May 26 '19 at 7:38

The solution to this problem indeed corresponds to the Fibonacci numbers, as mentioned by @fahrbach.

Here is an illustration of what you are trying to solve for the case of $n=4$ steps (taken from this website, which also gives a combinatorial solution)

Any staircase with $n$ steps allowing paths with increments of 1 or 2 steps at a time will end up in one of two states before the last path is taken: either we've climbed $(n-1)$ steps already and have $\color{red}{one}$ more step to take, or we've climbed $(n-2)$ steps already and we have $\color{blue}{two}$ more steps to take (if we took only one step here then we'd end up in an arrangement from the first state).

Thus, to get the total number of possible ways to climb $n$ steps, we just add the number of possible ways we can climb $(n-1)$ steps and the number of possible ways we can climb $(n-2)$ steps, giving the familiar recurrence relation:

\begin{equation*} F_n = \left\{ \begin{array}{l@{}l@{}l} 1 & n = 0,1\\ \color{red}{F_{n-1}} + \color{blue}{F_{n-2}} & n \ge 2 \end{array} \right. \end{equation*}

• Can this answer be expanded to answer the question "Given a set of possible step sizes S, calculate the number of possible ways you can climb n steps using step sizes in S)"? – Ephraim Nov 4 '15 at 20:21
• @Ephraim That's simple, it's just $F_n=\sum_{i\in I} F_i$ where $I$ is the set of steps that can jump to step $n$. – Stella Biderman Jan 31 '17 at 16:37
• This is the explanation that clicked to me, the colored figure makes a lot of sense, thanks. – neevek Oct 20 '20 at 16:47

This is the Fibonacci numbers - One interpretation of the n-th Fibonacci number is the number of ways to compose $n$ with parts in $\{1,2\}$ (that is, the n-th fibonacci number counts the number of sequences whose values are in $\{1,2\}$ whose sums are $n$). This is often called "Golfs and Cadillacs" or something similar.

• Could you please elaborate as to "how" it helps solve the problem? – PhD May 11 '14 at 1:46
• If there are $n$ steps, you have $F_n$ ways to reach the top of the stairs? The sequences map directly to the amount of stairs you climb with each step. I don't see how it can be any clearer. This interpretation can be used as a definition of the Fibonacci numbers and can be shown to be equivalent to the recurrence by using a straightforward combinatorial argument. – Batman May 11 '14 at 1:47
• I see what you mean. Perhaps what I'm confused by is by this: that is, the n-th fibonacci number counts the number of sequences whose values are in {1,2} whose sums are n. Could you please provide an example to explain this? – PhD May 11 '14 at 1:49
• You form a sequence made up of 1's and 2's whose sum is $n$. For example, for $n=3$, the sequences are (1,1,1), (1,2) and (2,1) giving you 3 ways to climb the stairs. Note that this is the common combintorics use of Fibonacci numbers, which is offset by 1 from the kind you learn in elementary school (this Fibonacci sequence is $1,2,3,5,...$ instead of $1,1,2,3,5,...$). – Batman May 11 '14 at 1:51

We can change this question into: In how many ways is possible to write a number as the ordered sum of 1s and 2s?

We can prove this using direct proof.

Hypothesis: For Fibbonaci number $F_1=1, F_2=1, F_{n}=F_{n-1}+F_{n-2}$, $Q(k)=F_{k+1}$

We say that we have constructed every order of $Q(n)$. Then, we construct $Q(n+1)$ this way:

1. For all orders, we add $+1$ to change $n$ to $n+1$. (There are now $Q(n)$ numbers.) Note: all numbers that are constructed here ends in $+1$.
2. For all orders that end with $+1$, we change it to $+2$. We shall prove that the amount will be $Q(n-1)$ to complete the proof. Note: all numbers that are constructed here ends in $+2$.

It can be seen that by constructing in this way there contains all orders.

Proof of step 2: If we constructed $Q(n)$ in the way above, then all numbers ended in $+1$ will be the amount $Q(n-1)$ by step 1. above, and we are done.


#### Problem

The positive integers, $x$, $y$, and $z$ are consecutive terms in an arithmetic progression. Given that $n$ is also a positive integer, for how many values of $n$ below one-thousand does the equation $x^2 - y^2 - z^2 = n$ have no solutions?

#### Solution

Let $x = a + d$, $y = a$, and $z = a - d$.

$\therefore (a + d)^2 - a^2 - (a - d)^2 = n$
$a^2 + 2ad + d^2 - a^2 - a^2 + 2ad - d^2 = n$

$\therefore 4ad - a^2 = a(4d - a) = n$.

Let $u = a$ and $v = 4d - a \implies u + v = 4d \equiv 0 \mod 4$. In other words, for a solution to exist the factors of $n$ must add to a multiple of four.

We shall deal with $n$ being of the form $2^m r$, where $r$ is odd, and for increasing values of $m$.

• $m = 0$ ($n$ is odd):
If $n = r$ then the factors $u$ and $v$ must both be odd. But if they are both congruent with 1 or both congruent with -1 modulo 4 then $u + v \equiv 2 \mod 4$, and there will be no solution; if they are different then $u + v \equiv 0 mod 4$, and there will always be a solution.
Hence if they are the same then $n = uv \equiv 1 \mod 4$, or $n$ being of the form 4$k$ + 1, will have no solutions.

• $m = 1 \implies n = 2r = a(4d - a)$:
If $a = 2r$, $4d - a = 1 \implies 4d = 2r + 1$. But as the RHS is odd, this is impossible.
If $a = r$, $4d - a = 2 \implies 4d = r + 2$. Impossible, as RHS is odd.
In other words if $n = 2(2k + 1) = 4k + 2$ then there will be no solutions.

• $m = 2 \implies n = 4r$:
If $a = 2r$, $4d - a = 2 \implies 4d = 2r + 2 = 2(r + 1)$.
And as $r + 1$ is even, we will always have at least one solution if $n = 4r$.

• $m = 3 \implies n = 8r$:
If $a = 8r$, $4d - a = 1 \implies 4d = 8r + 1$. Impossible.
If $a = 4r$, $4d - a = 2 \implies 4d = 4r + 2$. Impossible.
If $a = 2r$, $4d - a = 4 \implies 4d = 2r + 4$. Impossible.
If $a = r$, $4d - a = 8 \implies 4d = r + 8$. Impossible.
Hence if $n = 8(2k + 1) = 16k + 8$ then there will be no solutions.

• $m \ge 4$:
If $a = 4r$, $4d - a = 2^{m-2} \implies 4d = 4(r+2^{m-4})$.
Hence for $m \ge 4$ there will always be at least one solution.

Thus there will be no solutions for numbers of the form $4k + 1$, $4k + 2$, and $16k + 8$. As the first and second cases are odd and even respectively, they are mutually exclusive, and although the the second and third are both even, the third is divisible by 4, whereas the second is not divisible by 4. Hence all three forms are mutually exclusive.

As $4 \times 249 + 1 = 997$, $4 \times 249 + 2 = 998$, and $16 \times 61 + 8 = 984$, there are exactly $249 + 249 + 61 = 559$ values of $n$ below one-thousand that have no solution.

For which values of $n$ will there be exactly one solution?

Problem ID: 295 (26 Nov 2006)     Difficulty: 4 Star

Only Show Problem


## Sunday, May 23, 2010

### GR9768.63: Intersections

63. At how many points in the xy-plane do the graphs of $y = x^{12}$ and $y=2^x$ intersect?
(A)  None
(B)  One
(C)  Two
(D)  Three
(E)  Four

Solution:
We know what the graph of $2^x$ looks like: always positive, the $y$-intercept is at $(0,1)$. For $y = x^{12}$, it passes through the vertex $(0,0)$ and it concaves up. There are at most 2 intersections. Now, the two graph intersect exactly once on $(-\infty, 0)$. On the other half, $[0, \infty)$, we'll look at $f(x) = 2^x - x^{12}$ and note that $f(1) = 2^1 - 1^{12} = 1 > 0$, where as $f(2) = 2^2 - 2^{12} < 0$. Thus, by the Rolle's Theorem, we see that $f$ has a zero on $(1,2)$. Thus, $x^{12}$ intersects $2^x$ on this interval. So there are two points of intersection.
The answer is C.
-sg-
Note: the answer provided in the Answer Sheet is D, 3 points of intersection.
result from wolframalpha

#### 13 comments:

1. By the same Rolle's theorem, there *is* one more point of intersection. YOu exained x=2. Now, examine x=L, where L is very large. Clearly, 2^x wins here. So, again by ROlle's theorem, there is yet another point of intersection between 2 and L. The answer is indeed, D. - Deego.

2. Only 17 percent of the actual GRE test takers in 1997 got this right, worse than a random selection strategy!

3. Proper result from wolfram:

http://www.wolframalpha.com/input/?i=Plot[{2^x,+x^12},+{x,+-100.,+100.}]+

4. one vote for D

5. shouldn't it be Intermediate Value Theorem instead of Rolle's Theorem?

---

yodogyo@gmail.com

6. answer is (d) ...

basic property is that 2^x increases faster than x^12 and in fact for any a^x for which a>1 we would have a^x > x^n for any n as x--> infinity.

Hence all we need to find is whether the graph intersects twice or once for smaller values of x. e^x and x^2 I believe intersect only once.

2^x and x^12 intersects twice for x<2 and also for some big value of x as 2^x would be greater than x^12 for some value eventually. Hence this function intersects thrice.

7. I agree with Shantanu.

X>=0: We know that 0^12< 2^0 (x=0) and 2^12>2^2 (X=2).
Thus there is at least one intersection between 0 and 2.
At the same time, we know that 2^x is NOT O(x^12) as x->infinity. Also, they are both positive functions, so we know that there is some x>2 such that 2^x exceeds x^12 forever.

x<0. Its clear there is a cross in this region.

This gives us at least 3 crosses. My gut tells me that this is all because I am familiar with the graphs of both functions. So I would definitely choose D.

Rigorously, I do not know why there is only 3. Examining the derivatives would probably be the place to look.

8. Sorry for my grammar, english is not my native.
First 2 solutions are clearly. Lets prove that there is the 3-th solution. If we prove that lim(2^x/x^12)>1 for x->+oo, we will prove that the curves intersect once more time (because if x2 is x of 2-nd intersection and x near the x2 and x>x2 than 2^x1 for x->+oo. Just use L'Hospital's rule 12 times and we have: lim(ln2^12*2^x/(x^12))=+oo for x->+oo. So, because +oo>1 than we have the 3-th point of intersection.

9. Why are there not more than 3 intersections?

10. There are precisely 3 intersections. At x=0, 2^x = 2 > 0 = x^12. At x=2, 2^x = 4 < 2^12 = x^12, hence there is one intersection point in the interval (0,2). It is a well-known fact that as x grows to infinity 2^x exceeds x^12 (quite rapidly, in fact). Hence there is a second intersection point for x>2. By drawing the graphs of these functions it is easy to see that these are all intersection points for positive values of x.

A similar argument shows that there is a third and last intersection point between -1 and 0, and these are all.

11. What is the value of 3rd point of intersection? How large is it?

12. The third point of intersection occurs approximately at (74.66932553, 3.00404713 X 10^22)


# Open sets in $\mathbb R\times \mathbb R$ with the dictionary order

In the example below, I wonder why Munkres depicted the first type of intervals in this way?

More specifically, why doesn't the picture look like this:

where all points below, above, and on the red segment are removed?

This is my justification why I think this should be the right picture:

$$(a\times b,c\times d)=\{x|a \times b <_{dict} x <_{dict} c\times d\}\\=\{(p,q)|a <_{R} p <_{R} c\}\cup \{(r,s)| a=r=c \text{ and } c <_{R} s <_{R} d \}$$ (where $<_{dict}$ is the dictionary order and $<_R$ is the standard order on $\mathbb R$). The first set in the union is the set of the first type, the second set in the union is the set of the second type.

Now, any point above, below, or on the red line has the first coordinate $a$, which cannot lie in a set of the first type.

Update: I defined the sets of the first type in a way that differs from Munkres, but I don't see what's wrong with my description of open sets (in the display), and no open set in my description can be a set of the first type in Munkres.

• Um, removing the red segments on your second image produces exactly the first one, doesn't it? – Henning Makholm Sep 1 '18 at 19:12
• @HenningMakholm I'm not removing just the red segment, I'm removing everything above and below it as well. – user531587 Sep 1 '18 at 19:21
• What then happen to points such as $a\times(b+1)$, which is definitely both $>_{\rm dict} a\times b$ and $<_{\rm dict} c\times d$? – Henning Makholm Sep 1 '18 at 19:29
• @HenningMakholm okay, this makes sense... I was confused (and still confused) by the description of the basis in the text. – user531587 Sep 1 '18 at 19:51
• Actually this answer made things more clear: math.stackexchange.com/a/1129061/531587 – user531587 Sep 1 '18 at 20:02

Suppose a point $p = (x,y) \in (a \times b , c \times d)$ has $a$ as the first coordinate. Since $(a,b) < (x,y) = (a,y)$, this necessarily implies that $b < y$. In the same vein, if $x = c$ and since $(x,y) = (c,y) < (c,d)$, necessarily we must have that $y < d$. This justifies why the interval only contains an open ray of $\{x = a\}$ and $\{x = c\}$ instead of the whole line.
Your characterization of the interval is not quite right. Note that $r$ cannot be $a$ and $c$ at the same time, so you should separate in two cases (which is basically what I wrote) depending on the value that the first coordinate takes. This should clarify the picture.


# Principal submatrix and Principal minor of a matrix

A submatrix of a matrix is obtained by deleting any collection of rows and/or columns.

For example, from the following 3-by-4 matrix, we can construct a 2-by-3 submatrix by removing row 3 and column 2:

The minors and cofactors of a matrix are found by computing the determinant of certain submatrices.

A principal submatrix is a square submatrix obtained by removing certain rows and columns. The definition varies from author to author. According to some authors, a principal submatrix is a submatrix in which the set of row indices that remain is the same as the set of column indices that remain.

For a general 3 × 3 matrix in Mathematica,

(mat = Array[Subscript[a, ##] &, {3, 3}]) // MatrixForm

$mat=\left( \begin{array}{ccc} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{array} \right)$

there is one third order principal submatrix, namely mat. There are three second order principal submatrix:

$mat_{33}=\left( \begin{array}{ccc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \\ \end{array} \right)$, formed by deleting column 3 and row 3;

$mat_{22}=\left( \begin{array}{ccc} a_{1,1} & a_{1,3} \\ a_{3,1} & a_{3,3} \\ \end{array} \right)$, formed by deleting column 2 and row 2;

$mat_{11}=\left( \begin{array}{ccc} a_{2,2} & a_{2,3} \\ a_{3,2} & a_{3,3} \\ \end{array} \right)$, formed by deleting column 1 and row 1;

And there are three first order principal submatrix:

$mat=\left( \begin{array}{ccc} a_{1,1} \\ \end{array} \right)$, formed by deleting column 2,3 and row 2,3;

$mat=\left( \begin{array}{ccc} a_{2,2} \\ \end{array} \right)$, formed by deleting column 1, 3 and row 1, 3;

$mat=\left( \begin{array}{ccc} a_{3,3} \\ \end{array} \right)$, formed by deleting column 1,2 and row 1,2;

Do you have a way to find all principal submatrix in Mathematica ?

or find the determine of all principal submatrix (principal minor) in Mathematica ?

By @klgr comment.

Diagonal@Map[Reverse, Minors[mat, k, Identity], {0, 1}] //MatrixForm /@ # &

• Perhaps Diagonal @ Minors[mat, k]? – Carl Woll Aug 18 '17 at 17:01
• Diagonal@Map[Reverse, Minors[mat, k, Identity], {0, 1}] // MatrixForm /@ # & – kglr Aug 18 '17 at 17:04
• @kglr Thanks for the answer but is this in general. What if the matrix is a 4X4 measurement . – Emad kareem Aug 18 '17 at 17:14
• @kglr Thank you very much. This gives the exact result in general. – Emad kareem Aug 18 '17 at 17:24

To get the principal submatrices:

Diagonal[Map[Reverse, Minors[mat, #, Identity], {0, 1}]] & /@ {1, 2}

For the principal minors

Diagonal[Map[Reverse, Minors[mat, #], {0, 1}]] & /@ {1, 2}

• What is the & /@ {1, 2} for? – AzJ Jan 25 '18 at 19:46
• @AzJ, the & belongs to the left part, that is, the expression is actually (Diagonal[Map[Reverse, Minors[mat, #], {0, 1}]] & ) /@ {1,2}. For a pure function (a function with unnamed arguments) foo[#]& (where foo is an instruction on what to do with the stuff #; for example, multiply it by 2 and add 1 , (1+ 3 #)&) , the expression foo[#]&/@{a,b,c} Maps foo over a list of arguments ({a, b}) , that is, it is a short hand for  {foo[a], foo[b], foo[c]}. ... – kglr Jan 25 '18 at 20:34
• ... Please see the doc pages on for details on of esoteric symbols such as Function (&), Slot (#) and Map – kglr Jan 25 '18 at 20:35

I arrived three years later but I think that the code of @klgr can be significantly improved as it computes first all the minors to take at the end just those on the diagonal. I think that the next routine works faster for the submatrices with the obvious change making it work also for the minors, of course. I also suppose for convenience that the matrix is square, the obvious change make this also work for non-square matrices.

PrincipalSubmatrices[mat_, size_] :=
Module[{choices = Subsets[Table[i, {i, 1, Length[mat]}], {size}], count,
rc, symsubmatrix, symsubmatrices}, count = choices;
symsubmatrices = {};
While[count != {}, rc = count[[1]]; symsubmatrix = mat[[rc, rc]];
symsubmatrices = Append[symsubmatrices, symsubmatrix];
count = Delete[count, 1]]; symsubmatrices]


How many classes does the equivalence relation partition the set?

Looking for help on B) How many classes does the equivalence relation partition set X

Consider the relations $R$ and $S$, defined on the set $X = \{1, 2, . . . , 99\}$ as follows.

$xRy \iff x + y$ is a multiple of $11$,

$xSy \iff x − y$ is a multiple of $11$.

A) One of $R$ and $S$ is an equivalence relation, the other is not. Determine which is which and justify your answers.

B) Into how many classes does the equivalence relation partition set $X$?

So far I've determined that S is the equivalence relation as R isn't reflexive or transitive. My only attempt at B is that there is 11-1=10 non-zero congruence classes, so does each one correspond to a non-zero equivalence class?

Any help is appreciated!

• Where does this "non-zero" stuff come from? – ancientmathematician Apr 14 '18 at 16:53
• What is a "non-zero" equivalence class? – Morgan Rodgers Apr 14 '18 at 16:54
• @CyclotomicField: Sorry, but "a zero equivalence class" is a bit of a nonsense on a set that doesn't even contain zero, such that the given set $X$. – zipirovich Apr 14 '18 at 18:29
• @CyclotomicField It's unambiguous, but it's also meaningless in the normal context of equivalence relations. And is a term that doesn't actually show up in the problem – Morgan Rodgers Apr 15 '18 at 17:11
• What's wrong with a "zero equivalence class? Doe $11 -11=0$ make $11$ any less equivalent? Where did you get this weird idea that you aren't supposed to count the the equivalence class to zero? – fleablood Apr 15 '18 at 17:20

Just count them.

$1 R 12 R 23 R 34 R .... R 90$

$2 R 13 R 24 R 35 R .... R 91$

....

$10 R 21 R 32 R 43 R..... R 98$

$11 R 22 R 33 R 44 R ..... R 99$

That's 11.

It doesn't make any sense to subtract the "zero" equivalence class for 2 reasons.

1) The "zero" equivalence class is STILL an equivalence class so why they heck would you omit it? NOWHERE in the question does it say ANYTHING about how many non-zero equivalence classes; it ask how many equivalence classes. And $\{11,22,33,44,55,66,77,88,99\}$ is certainly an equivalence class, isn't it?

2) Since no method of "addition" has been defined or discussed, there is no meaning to defining any one of the equivalence classes as be a "zero" equivalence class. IF we were to define $\{x|x R c\} + \{x|x R d\} = \{x| x R (d+c\pm 11k \text{ for some integer } k)\}$ and define $[0]$ as the equivelence clas so that $[0] + \{x|x R c\} = \{x|x R c\}$ then, yes, $\{11,22,33,44,55,66,77,88,99\} = [0]$. But again, so what, it's still an equivalence class, isn't it?

To show $S$ is an equivalence relation you can either verify directly that it is reflexive, transitive, and symmetric or demonstrate that the relation partitions the set and use the fundamental theorem of equivalence relations. If you can demonstrate the existence of the congruence classes you have such a partition and that's sufficient to demonstrate that you have an equivalence relation. As you've deduced there are eleven such classes, and ten of them do not contain zero. I would also encourage you to consider the first method of directly verifying the three defining properties as it's both instructive and relatively easy.

• So is the answer 10 for B) correct? – RealGib Apr 14 '18 at 18:23
• @RealGib: No, it's not. – zipirovich Apr 14 '18 at 18:30
• None of the equivalence classes on $X$ contain 0, since $X$ does not contain 0. – Morgan Rodgers Apr 15 '18 at 17:17
• "Morgan Rodgers continued insistence on ignoring this obvious interpretation of the OPs comments". But it goes both ways. I agree that "$\{11,...99\}$ do not contain $0$ as so can't be 'zero' modulo class" is a bit too oblique and inaccurate and $11\equiv 0 \mod 11$ does indeed mean $\{11,...99\}$ is the zero element. But on another, and very important level A "zero" (additive identity) element can only be defined if addition on the classes (and NOT on the elements of the classes) is defined. So it is not unjustified. – fleablood Apr 16 '18 at 16:10
• ... so I mostly agree with Morgan Rodgers although I believe saying "it isn't a zero module as it doesn't contain zero" evades the issue that it actually isn't a zero module because the means for which it is "zero" are not stated. Of more concern is... why leave out the zero modules. These seem to much of a "monkey see; monkey do" approach to math that is doomed to failure. You don't just do everything one way despite whether that is what a question asks. If was asked how many elments does $\{0,1,6,8\}$ you don't answer 3 because zero doesn't count and you don't answer "it goes to 8". – fleablood Apr 16 '18 at 16:15

The question is not asking you about "nonzero" equivalence classes, it is asking for the total number of equivalence classes. So the equivalence class that contains 0 counts the same as any other equivalence class (however you want to interpret this, your set $X$ does not contain 0 so the interpretation of which equivalence class is the "zero" involves concepts that go beyond the scope of what this problem is asking).


# A circle touches two sides of a triangle and two of its medians

A circle touches two sides of a triangle and two of its medians. Prove that the triangle is isosceles.

This problem came from the Mathematical Digest issue 62 (Jan 1986) which in turn cited a Russian mag called KVANT (meaning "Quantum").

• Assume it's not true. Then you wouldn't ask the question. A contradiction. Therefore it's true. QED. :) Mar 13 '21 at 1:02

As the triangles ADC and BEC have the same incircle and their areas are equal (half that of ABC), so are their perimeters
DC+DA+AC = EC+EB+BC
or, subtracting from both sides CD+CE+DA+EB
AE-EB = BD-DA.
This means that D and E lie on the same pair of hyperbolas with foci A and B. Since they also have the same distance to the base AB (half that of C) the triangle must be isosceles by symmetry.

• It took me a moment to see why they have the same distance to the base, but that should have been obvious. This is a really clever proof. Mar 13 '21 at 8:38
• @JaapScherphuis It's short ;-) I've added a half sentence explaining (if you can call it that) the distance to the base. Mar 13 '21 at 9:50
• That is clever. I don't think I've ever seen anyone use a hyperbola locus in a proof like this! Mar 13 '21 at 17:34
• Elegant, indeed. Mar 16 '21 at 6:16

Alternative Proof

By Pitot Theorem we have $$|CE|+|MD| = |CD|+|ME|$$ and since the centroid divides each median in the ratio 2:1, this means that $$\frac{1}{2}|AC| + \frac{1}{3}|AD| = \frac{1}{2}|BC|+\frac{1}{3}|BE|$$ Now let $$|AB| = c, |BC| = a, |CA|=b, |AD|=m_a, |BE|=m_b$$.
Then Apollonius' Theorem tells us that $$m_a = \sqrt{\frac{2b^2+2c^2-a^2}{4}}\,\,\,\,,\,\,\,\,m_b = \sqrt{\frac{2a^2+2c^2-b^2}{4}}$$ Substituting this in above and multiplying across by $$6$$ yields $$3a + \sqrt{2b^2+2c^2-a^2} = 3b + \sqrt{2a^2+2c^2-b^2}$$ $$\Rightarrow 3(a-b) + \sqrt{2b^2+2c^2-a^2} = \sqrt{2a^2+2c^2-b^2}$$ $$\Rightarrow 9(a-b)^2 + 6(a-b)\sqrt{2b^2+2c^2-a^2} + 2b^2+2c^2-a^2 = 2a^2+2c^2-b^2$$ $$\Rightarrow (a-b)\left(9(a-b) + 6\sqrt{2b^2+2c^2-a^2} - 3(a+b)\right) = 0$$ which means either $$a=b$$ or $$\sqrt{2b^2+2c^2-a^2} = 2b-a$$.
If it is the latter then $$2b^2+2c^2-a^2 = 4b^2-4ab+a^2$$ $$\Rightarrow c^2 = b^2-2ab+a^2 = (b-a)^2$$ $$\Rightarrow c = \pm(b-a)$$ which only happens for a degenerate triangle (does not satisfy the triangle inequality).
Hence $$a=b$$

@loopwalt and @hexomino both gave excellent answers. I wanted to share the answer from the Mathematical Digest, because it's also elegant in its own way:

It starts the same way as @hexomino's proof (I didn't know it was called Pitot's Theorem!):

Since $$CEMD$$ circumscribes a circle, $$CE+MD = CD + ME$$ (Pitot's Theorem, but fun to prove). So (as in hexomino's proof):
$$\frac{1}{2}AC+\frac{1}{3}AD = \frac{1}{2}BC+\frac{1}{3}BE \tag1$$
Next, note that $$\triangle ADC$$ and $$\triangle BEC$$ have the same area ($$\frac{1}{2}\triangle ABC$$) and share a common incircle. This means that their perimeters are equal (another fun to prove and left as an exercise!). So $$AD+\frac{1}{2}BC+AC = BE+\frac{1}{2}AC+BC$$, which gives:
$$\frac{1}{2}AC+AD = \frac{1}{2}BC+BE \tag2$$
Now, $$3\times(1)-(2)$$ gives $$AC=BC$$


### Probability of k collisions

, Friday, 30 January 2015
While I was reading about hash tables I stumbled upon this deceptively simple question. I have rephrased it to make it easier to understand.
Say we have $m$ buckets. We select a random bucket and put a ball in it, we repeat this (select-put) $n$ times. In the end what is the probability of having at least one bucket with $k$ balls?
My initial attempts to solve it failed, and I posted the question on math.stackexchange. An approximation for large values of $m$ and $n$ was posted as an answer but I was interested in exact results. Once I realized I had no way of finding someone else's solution, I was able to solve it myself.

## Simulation

First, to check if my solution is correct, I have written a small program to simulate the select-put procedure million times and report the experimental probability for different values of $m$,$n$ and $k$. If you are interested, you can find the java code at the end of my post.

## Solution

First we will start with writing the probability of 1 box having $k$ balls. Then using inclusion-exclusion principle we will generalize. Let's write the probability of putting first $k$ balls into the first box and remaining balls to the other boxes: $$\left(\frac1{m}\right)^k \left(1-\frac1{m}\right)^{n-k}$$ We have $\binom{n}{k}$ ways of putting $k$ balls into a particular box (e.g. first box), and we have $m$ such boxes. So the probability becomes: $$m \binom{n}{k} \left(\frac1{m}\right)^k \left(1-\frac1{m}\right)^{n-k}$$ But the expression above suffers from double counting, i.e. it counts configurations that contains multiple boxes with $k$ balls more than once. So we need to subtract the value for 2 boxes having $k$ balls, add the value for 3 boxes having $k$ balls, etc... Let's write the probability of putting first $k$ balls into the 1st box, next $k$ balls into the 2nd box, and remaining balls to the other boxes. $$\left(\frac1{m}\right)^k \left(\frac1{m}\right)^k \left(1-\frac1{m}\right)^{n-2k}$$ We have $\binom{n}{k}\binom{n-k}{k}$ ways of putting $2k$ balls into two particular boxes (e.g. first and second box), and we have $\binom{m}{2}$ such boxes. So the probability becomes: $$\binom{m}{2}\binom{n}{k}\binom{n-k}{k}\left(\frac1{m}\right)^k \left(\frac1{m}\right)^k \left(1-\frac1{m}\right)^{n-2k}$$ At this point you should be able to see the pattern that is emerging.

## Generalization

If we put together and generalize what we have done so far we get the following expression as probability of having a box with $k$ balls: $$\sum^{\lfloor \frac{n}{k} \rfloor}_{i=1} (-1)^{i+1} \binom{m}{i} \left[ \prod^{i-1}_{j=0} \binom{n-jk}{k} \right] \left( \frac1{m} \right)^{ik} \left( \frac{m-i}{m} \right)^{n-ik}$$ The powers of $-1$ at the beginning enables adding values for odd number of balls and subtracting values for even number of balls, i.e. inclusion-exclusion principle.

## The code

  private static double calculateKcollisions(int m, int n, int k) {
double probability = 0;

int sign = 1;

for (int i = 1; n - i * k >= 0 && i <= m; i++) {
double p = binomial(m, i) * pow(1.0 / m, i * k);
p *= pow((double) (m - i) / m, n - i * k);

for (int j = 0; j < i; j++) {
p *= binomial(n - j * k, k);
}

probability += sign * p;
sign *= -1;
}

return probability;
}

## Simulation code

  private static final Random RNG = new Random();

public static void main(String[] args) {
int sampleCount = 1000000;

for (int experimentNo = 0; experimentNo < 10; experimentNo++) {
int m = RNG.nextInt(7) + 3;
int n = RNG.nextInt(m) + RNG.nextInt(10) + 1;
int k = RNG.nextInt(n + 1);

System.out.println(format("m:%d  n:%d  k:%d", m, n, k));

int successCount = 0;
for (int sampleNo = 0; sampleNo < sampleCount; sampleNo++) {

int[] boxes = new int[m];

for (int i = 0; i < n; i++)
boxes[RNG.nextInt(m)]++;

for (int i = 0; i < m; i++) {
if (boxes[i] == k) {
successCount++;
break;
}
}
}

double experimental = (double) successCount / sampleCount;

double calculated = calculateKcollisions(m, n, k);

System.out.println(format("xperimental: %f  calculated: %f", experimental, calculated));
System.out.println(format("diff: %.2f\n", abs(experimental - calculated)));
System.out.println();
}
}

#### 1 comment:

1. Excellent blog thanks for sharing the valuable information..it becomes easy to read and easily understand the information.
Useful article which was very helpful. also interesting and contains good information.


# Math Help - Finding point on sphere furthest from point in space

1. ## Finding point on sphere furthest from point in space

Find the point on the sphere $(x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99$ which is furthest from the point C(1,3,4).

Now I know that you suppose to find an equation of a line relating the centre of the sphere (2,2,7) and C(1,3,4) e.g. x = (1,3,4) + t(1,-1,3) where t can be any real number. But I dont understand what you would do to find the furthest point on the sphere?

2. Originally Posted by SyNtHeSiS
Find the point on the sphere $(x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99$ which is furthest from the point C(1,3,4).

Now I know that you suppose to find an equation of a line relating the centre of the sphere (2,2,7) and C(1,3,4) e.g. x = (1,3,4) + t(1,-1,3) where t can be any real number. But I dont understand what you would do to find the furthest point on the sphere?
Think about it geometrically. To get from C to the furthest point on the sphere, you have to go from C to the centre of the sphere and then continue in the same direction for a distance equal to the radius of the sphere, which is $\sqrt{99} = 3\sqrt{11}$. You have already identified the direction of travel as the vector (1,–1,3), which has length $\sqrt{11}$.

Those $\sqrt{11}$s can't be coincidental, can they?

3. Hello, SyNtHeSiS!

$\text{Find the point on the sphere }\,(x - 2)^2 + (y - 2)^2 + (z-7)^2 \:= \:99$

$\text{which is furthest from the point }\,C(1,3,4).$
Code:
              * * *         o C
*           *   *
*               ♥ P
*              *  *
*
*           *       *
*         o         *
*       * A         *
*
*  *              *
Q ♥               *
*   *           *
*         * * *

Construct a line through $\,C$ and the center of the sphere $\,A.$

It will intersect the sphere in two points, $\,P$ and $Q.$
. . We want $Q$, the point further from $\,C.$

The line from $C(1,3,4)$ to $A(2,2,7)$ has direction $\langle 1,-1,3\rangle$

The line has parametric equations: . $\begin{Bmatrix}x &=& 1 + t \\ y &=& 3 - t \\ z &=& 4+3t \end{Bmatrix}$

Substitute into the equation of the sphere:

. . $\left[(1-t) - 2\right]^2 + \left[(3+t)-2\right]^2 + \left[([4-3t)-7\right]^2 \:=\:99$

. . $11t^2 + 22t + 11 \:=\:99 \quad\Rightarrow\quad t^2 + 2t + 1 \:=\:9$

. . $t^2 + 2t - 8 \:=\:0 \quad\Rightarrow\quad (t-2)(t+4)\:=\:0 \quad\Rightarrow\quad t \:=\:2,\:-4$

$\begin{array}{ccccccc}t = 2\!: & (x,y,z) &=& (\text{-}1,5,\text{-}2) & \text{point J} \\
t = \text{-}4\!: & (x,y,z) &=& (5,\text{-}1,16) & \text{point K}\end{array}$

I suspect that point $\,K$ is further from $\,C$, but let's make sure.

Distance from $C(1,3,4)$ to $J(\text{-}1,5,\text{-}2)\!:$

. . $\overline{CJ} \;=\;\sqrt{(\text{-}1-1)^2 + (5-3)^3 + (\text{-}2-4)^2} \;=\;\sqrt{44}$

Distance from $C(1,3,4)$ to $K(5,\text{-}1,16)\!:$

. . $CK \;=\;\sqrt{(5-1)^2 + (\text{-}1-3)^2 + (16-4)^2} \;=\;\sqrt{176}$

Hence, $K$ is the point we are seeking.

Therefore: . $Q(5,\text{-}1,16)$

4. Originally Posted by Soroban
The line from $C(1,3,4)$ to $A(2,2,7)$ has direction $\langle 1,-1,3\rangle$
Thanks for the great post. I was just curious why is it that choosing a line from A(2,2,7) to C(1,3,4) with direction <-1,1,-3> equivalent to the quote above? I thought this vector be the same magnitude but opposite, since points A and C is switched around.

5. Just to point out that the makings of a much simpler solution were already in the original post, where the equation of the line from C to the centre of the sphere is given as $x = (1,3,4) + t(1,-1,3)$. If you put t=1 in that equation then you get the point (2,2,7) (the centre of the sphere). To get to the furthest point of the sphere you have to travel a further distance along that same line. The extra distance that you need to travel is $\sqrt{99}$ units, which is three times the length of the vector (1,–1,3).

So to get to the furthest point of the sphere you need to take t = 4 (one unit of t to take you to the centre of the sphere, and another three units of t to take you to the furthest point). The furthest point is therefore at (1,3,4) + 4(1,–1,3) = (5,–1,16).


# When to use mean vs median

I'm new to data science and stats, so this might seems like a beginner question.

I'm working on a dataset where I've user's Twitter followers gain per day. I want to measure the average growth he had over a period of time, which I did by finding the mean of growth. But someone is suggesting me to use median for this.

Can anyone explains, in which use-case we should use mean and when to use median?

The arithmetic mean is denoted as $$\bar{x}$$

$$\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$$

where each $$x_i$$ represent an unique observation. The arithmetic mean measures the average value for a given set of numbers.

In contrast to this, the median is the value which falls directly in the middle of your dataset. The median is especially useful when you are dealing with a wide range or when there is an outlier (a very high or low number compared to the rest) which would skew the mean.

For example, salaries are usually discussed using medians. This due to the large disparity between the majority of people and a very few people with a lot of money (with the few people with a lot of money being the outliers). Thus, looking at the 50% percentile individual will give a more representative value than the mean in this circumstance.

Alternatively, grades are usually described using the mean (average) because most students should be near the average and few will be far below or far above.

• That's a great answer. So, If I think it like this, I can plot my data and see if it values are continuous, then we can use mean and if they're more clustered (some high and some low), then median would be better, right? – Mukul Jain Mar 6 '19 at 4:48
• @MukulJain, Yes it depends on the distribution of the data as you mentioned. Plotting is always my go to way to get a sense of my data. Easy to spot anomalies and get a sense of its spread. – JahKnows Mar 6 '19 at 5:48
• I think you could explain this better using the term "outlier" – MilkyWay90 Mar 7 '19 at 1:11
• So, if data has lots of outliers, is it good to use median right? Outliers can be calculated using z-score (<3 or >-3) – Mukul Jain Mar 7 '19 at 4:19
• @MukulJain, correct, and you can also calculate outliers using p-value, – JahKnows Mar 7 '19 at 5:52

It depends what question you are trying to answer. You are looking at the rate of change of a time series, and it sounds like you are trying to show how that changed over time. The mean gives the reader one intuitive insight: they can trivially estimate the number of followers at any date $$d$$ days since the start by multiplying by the mean rate of change.

The downside to this single metric is that it doesn't illustrate something which is very common in series such as this: the rate of change is not fixed over time. One reasonable metric for giving readers an idea of whether the rate of change is static is giving them the median. If they know the minimum of the series (presumably zero in your case), the current value, the mean and the median, they can in many cases get a "feel for" how close to linear the increase has been.

There is a great cautionary tale in Anscombe's quartet - four completely different time series which all share several important statistical measures. Basically it always comes back to what you are trying to answer. Are you trying to find users which are likely to become prominent soon? Users which are steadily accruing followers year by year? One hit wonders? Botnets?

As you've probably guessed, this means it's not possible to universally call mean or median "better" than the other.

Simply to say, If your data is corrupted with noise or say erroneous no.of twitter followers as in your case, Taking mean as a metric could be detrimental as the model will perform badly. In this case, If you take the median of the values, It will take care of outliers in the data. Hope it helps

Often median is more robust to extreme value to mean. Try to think it as a minimization task. Median corresponds to absolute loss while mean corresponds to square loss.

I find myself explaining this a lot and the example I use is the famous Bill Gates version. Bill Gates is in your data science class. Your instructor asks you: what is the average income or net worth of this class? Bill Gates sheepishly obliges and tells you what his income is. Now when you say the average income of your group is a zillion dollars - technically correct but does not describe the reality - that Bill Gates is an outlier skewing everything.

So you line up all the people in your group in ascending or descending order - whatever the person in the middle is making - that is your median. In this example, everybody but Bill Gates is likely to be in spitting distance of that median, and Bill Gates will be the only one making anything close to the mean.

Now say buddy Bill Gates is hiring a money manager. Based on the returns they produced so far. Should he look at their average returns over a 10 year period or their median return or a combination of the two? Did they outperform the market each year? Some years? How does portfolio size factor in? In the case of Twitter followers, Obama would have a different growth compared to someone with say 500K-1MM followers. As @l0b0 alludes to in his excellent answer - it all depends. Are you measuring follower growth or the rate of change of follower growth and what is the question you are trying to answer, strategy/product you are trying to develop - accordingly you pick mean or median. Getting the mean and median is always the easy part. It's always better to never ever have the average of 2.1 kids. Have a whole number of kids. But what can you say about population growth rates if mean number of kids is 2.1 and median is 1 or 2? Or median is 3 or more? Is growth accelerating or decelerating? What is mode doing? Compute all the basics first - and then ask the reason why you are using mean versus median.


# Calculate $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}k\sin\left(\frac{a}{k}\right)$

I'm trying to calculate $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}k\sin\left(\frac{a}{k}\right)$. Intuitively the answer is $a$, but I can't see any way to show this. Can anyone help? Thanks!

-
Is there anything special about $a$? I'm assuming it is an arbitrary real number? –  JavaMan Feb 22 '12 at 2:14
Welcome to math.SE! Do you have learned Riemann integration? –  leo Feb 22 '12 at 2:14
You may prove a more general statement: *If $a_n \to a$, then $\frac{1}{n}\sum_{k=1}^{n}a_k \to a$*. This can be proved by $\epsilon-\delta$ argument. –  sos440 Feb 22 '12 at 2:18
a is an arbitrary real number, yeah. And I've learned Riemann integration! –  ro44 Feb 22 '12 at 2:27
@sos440: Thanks, I thought I could try something like this. But I'm wondering if there's perhaps a 'special' trick with this particular limit? leo mentions Riemann integration. –  ro44 Feb 22 '12 at 2:29

I’m not going to work out all of the details; rather, I’ll suggest in some detail a way to approach the problem.

First, it suffices to prove the result for $a>0$, since the sine is an odd function. For $a>0$ we have $k\sin\left(\frac{a}k\right)<k\left(\frac{a}k\right)=a$, so $$\frac1n\sum_{k=1}^nk\sin\left(\frac{a}k\right)<\frac1n\sum_{k=1}^na=a\;;$$ this gives you an upper bound of $a$ on any possible limit.

You know that $\lim\limits_{x\to 0}\frac{\sin x}x=1$, so there is a $c>0$ such that $\sin x>\frac{x}2$ whenever $0<x<c$. This means that $$k\sin\left(\frac{a}k\right)>\frac{a}2$$ whenever $\frac{a}k<c$, i.e., whenever $k>\frac{a}c$. Now suppose that $n$ is very large compared with $\frac{a}c$; then ‘most’ of the terms of $$\frac1n\sum_{k=1}^nk\sin\left(\frac{a}k\right)\tag{1}$$ will be greater than $\frac{a}2$, and hence so will $(1)$ itself. You may have to do a little fiddling to say just how big $n$ should be taken relative to $\frac{a}c$, but it should be clear that this idea works to show that the limit of $(1)$ as $n\to\infty$ must be at least $\frac{a}2$.

But what I did with $\frac12$ can clearly be done with any positive fraction less than $1$: if $0<\epsilon<1$, there is a $c>0$ such that $\sin x>\epsilon x$ whenever $0<x<c$. If you’ve filled in the missing details for the previous paragraph, you shouldn’t have too much trouble generalizing to show that the limit of $(1)$ must be at least $\epsilon a$ for any $\epsilon <1$ and hence must be at least $a$.

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Based on Brian and sos's arguments, I was wondering if I'm correctly generalizing this situation: Say we have $a_n->a$, and let's take some arbitrary $l>0$ (using 'l' instead of epsilon). Then $|(\frac{1}{n}\sum_{k=1}^{n}a_n)-a|=|\frac{1}{n}\sum_{k=1}^{n}(a_n-a)|\leq \frac{1}{n}\sum_{k=1}^{n}|a_n-a|$. For almost every n, we have that $|a_n-a|<l$, yielding that $\frac{1}{n}\sum_{k=1}^{n}|a_n-a|<l$ for almost every n. So $\frac{1}{n}\sum_{k=1}^{n}a_n \rightarrow a$. Does this sound correct? Thanks! –  ro44 Feb 22 '12 at 2:49
@ro44: Not quite. You want $|(\frac1n\sum_{k=1}^na_k)-a|=|\frac1n\sum_{k=1}^n(a_k-a)|\le\frac1n\sum_{k=1}^n|a_k-a|$. Then you want to say that for sufficiently large $n$, ‘most’ terms of this sum are less than $l$, so the sum is less than (say) $2l$. There’s still a bit of work to be done show that if $n$ is large enough, the early ‘bad’ terms are a small enough fraction of the total not to mess things up. –  Brian M. Scott Feb 22 '12 at 3:37

$\displaystyle \sin x \leq x$ for $x \geq 0.$ Integrating this over $[0,t]$ gives $$-\cos t +1 \leq \frac{t^2}{2} .$$

Integrating both sides again from $[0,x]$ gives $$-\sin x + x \leq \frac{x^3}{6} .$$

Thus, $$x - \frac{x^3}{6} \leq \sin x \leq x.$$

Hence, $$\frac{1}{n} \sum_{k=1}^{n} k \left( \frac{a}{k} - \frac{a^3}{6k^3} \right ) \leq \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \leq \frac{1}{n} \sum_{k=1}^{n} k \left( \frac{a}{k} \right).$$

Since $\displaystyle \sum_{k=1}^n \frac{a^3}{6k^3}$ is convergent, the Squeeze theorem shows that $\displaystyle \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \to a.$

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# math

The volume V of a sphere of radius r is given by the formula V (r) = (4/3)πr^3. A balloon in the shape of a sphere is being inflated with gas. Assume that the radius of the balloon is increasing at the constant rate of 2 inches per second, and is zero whent = 0.
(a) Find a formula for the volume V of the balloon as a function of time t.
(b) Determine the volume of the balloon after 5 seconds.
(c) Starting with an empty balloon, suppose that the balloon will burst when
its volume is 10, 000 cubic inches. At what time will the balloon burst?
(d) Find a formula for the surface area S of the balloon as a function of time
t; recall the surface area formula for a sphere of radius r is S(r) = 4πr^2.
(e) Determine the surface area of the balloon after 6 seconds.
(f) What will be the surface area of the balloon when it bursts?

1. 👍 0
2. 👎 0
3. 👁 233
1. V = 4/3 πr^3
(a) dV = 4πr^2 dr
V = ∫[0,t] 4πr^2 dr
(b) now plug in t=5
(c) find when (a) = 10000
(d) dA = 8πr dr
work it as in (a)
(e) work as in (b)
(f) plug in t as in (c)

1. 👍 0
2. 👎 0
👨🏫
oobleck
2. since r = 2t, let's replace r in the equation

V = (4/3)π r^3
= (4/3)π (2t)^3
= (32/3)π t^3

b) replace t with 5 in my formula.
c) (32/3)π t^3 = 10000
solve for r

d) S(r) = 4πr^2 , let do the same thing: r = 2t
= 4π(2t)^2 = 16πt^2

e) let t = 6 and evaluate using the new formula from d)

f) sub in the value of t you got in c) into the new S(r) formula

let me know what your answers are so I can check them

1. 👍 0
2. 👎 0
👨🏫
Reiny
3. yep i got the same answers as you did :) thank you for your help

1. 👍 0
2. 👎 0

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# Relationship between the coefficients of a polynomial equation and its roots

My question is- Solve: $x^3 - 6x^2 + 3x + 10= 0$ given that the roots are in arithmetic progression.

Any help would be greatly appreciated.

Let the three roots be $a-d$, $a$, and $a+d$.

The sum of the roots is the negative of the coefficient of $x^2$, so $(a-d)+a+(a+d)=6$, and therefore $a=2$.

The product of the roots is the negative of the constant term, and therefore $(a-d)(a)(a+d)=-10$. Since we know $a$, we can now find $d$. We get $a^2-d^2=-5$, and therefore $d^2=9$.

Remarks: $1.$ In general, if we have a cubic equation $px^3+qx^2+rx+s=0$, where $p\ne 0$, and the roots are $\alpha$, $\beta$, and $\gamma$, then $$\alpha+\beta+\gamma=-\frac{q}{p}, \qquad \alpha\beta+\beta\gamma+\gamma\alpha=\frac{r}{p},\qquad \alpha\beta\gamma=-\frac{s}{p}.\tag{1}$$ There are analogous relations between coefficients and roots for polynomials of any degree.

$2$. Note that we called the roots $a-d, a, a+d$. We could have called them $a$, $a+d$, and $a+2d$, but the algebra would be a little more complicated. Symmetry is your friend.

$3.$ If the fact we used about roots is not available, there are a couple of things we can do. We can hope that the roots are rational. then they must all divide the constant term $10$. the only candidates are $\pm 1$, $\pm 2$, $\pm 5$, and $\pm 10$, and we quickly get to an answer.

Or else we can let the roots be $\alpha$, $\beta$, and $\gamma$, and see that the polynomial must be $(x-\alpha)(x-\beta)(x-\gamma)$. Expanding, we find in essence the relationship $(1)$.

Or else (but this is ugly) we can substitute $a-d$, $a$, and $a+d$ in our polynomial, and expand. Then solve for $a$ and $d$. The equations we get have a fair bit of symmetry, so this is not as difficult as it seems.

The roots are in arithmetic progression, so we can denote them as $a-d, a, a+d$ where $d$ is the common difference. The sum of these roots is then $3a$ and Viete's formulas shows that the sum of the roots must be $6$, so $a=2.$ So we have found a root, and can get the others by dividing the polynomial by $x-2$ and solving the remaining quadratic.

A three degree polynomial would have max 3 roots say p,q,r

The relations between the roots and coefficients in $ax^3 + bx^2 + cx + d$ are:

$\frac{-b}{a} = p+q+r$

$\frac{c}{a} = pq + qr + pr$

$\frac{-d}{a} = pqr$

Note: where $a \neq 0$. Being 0 simply implies that the order of equation isn't actually 3.

Three equations can solve three unknowns (To use the fact that they are in AP, assume $p,q,r$ to be $q-d$, $q$, $q+d$ where $q$ is the mean and $d$ is the common difference of AP).

Note: The pattern of the roots and coefficients can be the same for n degrees of polynomial as well.

• where $a\neq 0$ – Belgi Jun 14 '12 at 14:39
• Thanks. I have edited the answer – Saurabh Agarwal Jun 14 '12 at 14:39

Hint: Note that $2$ is a root of the polynomial, how can you get the other two roots ?

• I did not use the fact that the roots are in arithmetic progression – Belgi Jun 14 '12 at 14:36
• That fact could be used to ascertain that 6/3 = 2 is one of the roots. – hardmath Feb 1 '14 at 4:40

Hint $\rm\ \ f(x) = (x+a\!-\!b)(x+a)(x+a\!+\!b)\iff f(x\!-\!a) = x^3 - b^2 x.\:$ But the only shift killing the $\rm\:-6\,x^2\:$ in your cubic is $\rm\:x \to x+2,\:$ yielding $\rm\:f(x\!+\!2) = x^3 - 9\,x,\:$ so $\rm\: b = \pm 3.$

Note $\rm\,\ \ f(x\!-\!a) = x^3\! -\! b^2x\:\Rightarrow\:b^2 = -f'(x\!-\!a)|_{x=0} =\, -f'(-a)\ \ [\, = -(3\cdot 2^2\! -\! 12\cdot 2\! +\! 3) = 9\ ].\:$

Therefore $\rm\ \ f(x) = x^3 + c\: x^2 +\ldots\: \Rightarrow\ a = c/3,\ \ b = \sqrt{-f'(-a)}$

First of all the easiest mtd is

$$1)~$$ Get one of the roots by hit and trial mtd in your question clearly $$~-1~$$ is one of the roots i.e when $$~x =-1~$$ the whole equation results $$~0~$$ therefore $$~X=-1 ,~~~ X+1=0~$$

$$2)~$$ apply division method to reduce it to quadratic and then $$~u~$$ can easily have your roots that are in A. P


# complex number to rectangular form

The x is the real number of the expression and the y represents the imaginary number … We move 2 units along the horizontal axis, followed by 1 unit up on the vertical axis. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Complex Numbers in Rectangular and Polar Form To represent complex numbers x yi geometrically, we use the rectangular coordinate system with the horizontal axis representing the real part and the vertical axis representing the imaginary part of the complex number. Post was not sent - check your email addresses! If necessary, round to the nearest tenth. An identification of the copyright claimed to have been infringed; Convert The Following Complex Numbers To Rectangular Form. In other words, there are two ways to describe a complex number written in the form a+bi: To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. Khan Academy is a 501(c)(3) nonprofit organization. The absolute value of a complex number is the same as its magnitude. Find more Mathematics widgets in Wolfram|Alpha. information described below to the designated agent listed below. Use i or j to represent the imaginary number −1 . St. Louis, MO 63105. Related Topics: Common Core (The Complex Number System) Common Core for Mathematics. Varsity Tutors LLC your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the ( Log Out /  For this reason the rectangular form used to plot complex numbers is also sometimes called the Cartesian Form of complex numbers. Change ), You are commenting using your Twitter account. Recall that the complex plane has a horizontal real axis running from left to right to represent the real component (a) of a complex number, and a vertical imaginary axis running from bottom to top to represent the imaginary part (b) of a complex number. A. Zi = 6e 25" B. Z2 = 5e-145 3. The rectangular form of the equation appears as , and can be found by finding the trigonometric values of the cosine and sine equations. A complex number can be expressed in standard form by writing it as a+bi. means of the most recent email address, if any, provided by such party to Varsity Tutors. ChillingEffects.org. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). But complex numbers, just like vectors, can also be expressed in polar coordinate form, r ∠ θ . Change ), You are commenting using your Google account. Theorem 11.16. Distributing the 4, we obtain the final answer of: distributing the 5, we obtain the final answer of: To convert, just evaluate the trig ratios and then distribute the radius. Your name, address, telephone number and email address; and Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. Yes, you guessed it, that is why (a+bi) is also called the rectangular form of a complex number. Get the free "Convert Complex Numbers to Polar Form" widget for your website, blog, Wordpress, Blogger, or iGoogle. The rectangular form of the equation appears as , and can be found by finding the trigonometric values of the cosine and sine equations. Complex numbers in the form a+bi\displaystyle a+bia+bi are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. To convert to rectangular form, just evaluate the trig functions and then distribute the radius: To convert, evaluate the trig ratios and then distribute the radius: If you've found an issue with this question, please let us know. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. Polar to Rectangular Online Calculator Below is an interactive calculator that allows you to easily convert complex numbers in polar form to rectangular form, and vice-versa. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing Varsity Tutors. If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one This can be a helpful reminder that if you know how to plot (x, y) points on the Cartesian Plane, then you know how to plot (a, b) points on the Complex Plane. Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to email this to a friend (Opens in new window), put it into the standard form of a complex number by writing it as, How To Write A Complex Number In Standard Form (a+bi), The Multiplicative Inverse (Reciprocal) Of A Complex Number, Simplifying A Number Using The Imaginary Unit i, The Multiplicative Inverse (Reciprocal) Of A Complex Number. Add The Two Complex Numbers In Rectangular Form Zı = 10-j6 Z2 = 3 +j16 4. Solution for Write the complex number 4(cos 240° + i sin 240°) in rectangular form. To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are To plot a complex number a+bi on the complex plane: For example, to plot 2 + i we first note that the complex number is in rectangular (a+bi) form. The standard form, a+bi, is also called the rectangular form of a complex number. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Convert The Following Complex Numbers To Polar Form. In order to work with complex numbers without drawing vectors, we first need some kind of standard mathematical notation. If you were to represent a complex number according to its Cartesian Coordinates, it would be in the form: (a, b); where a, the real part, lies along the x axis and the imaginary part, b, along the y axis. ( Log Out /  The multiplication of complex numbers in the rectangular form follows more or less the same rules as for normal algebra along with some additional rules for the successive multiplication of the j-operator where: j2 = -1. Explanation: . Complex Numbers in Polar Coordinate Form The form a + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a and height b, as shown in the graph in the previous section. You can input only integer numbers or fractions in this online calculator. Distributing the 4, we obtain the final answer of: A description of the nature and exact location of the content that you claim to infringe your copyright, in \ Therefore the correct answer is (4) with a=7, and b=4. A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. Change ), You are commenting using your Facebook account. … Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; The Complex Hub aims to make learning about complex numbers easy and fun. The calculator will simplify any complex expression, with steps shown. an The angle must be converted to radians when entering numbers in complex exponential form: In other words, given z = r(cosθ + isinθ), first evaluate the trigonometric functions cosθ and sinθ. This point is at the co-ordinate (2, 1) on the complex plane. The rectangular form of a complex number is given by z = a + bi. Additional features of complex numbers convert. Zi = 3 + J5 Z2 = 9-14 Well, rectangular form relates to the complex plane and it describes the ability to plot a complex number on the complex plane once it is in rectangular form. ; The absolute value of a complex number is the same as its magnitude. Show Instructions. > 5+4i ans = 5 + 4i A number in polar form, such as (2 45 ), can be entered using complex exponential notation. Polar & rectangular forms of complex numbers Our mission is to provide a free, world-class education to anyone, anywhere. If Varsity Tutors takes action in response to Complex numbers in rectangular form are presented as a + b * %i, where a and b are real numbers. In other words, given z = r(cosθ + isinθ), first evaluate the trigonometric functions cosθ and sinθ. Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially There's also a graph which shows you the meaning of what you've found. Example 1 – Determine which of the following is the rectangular form of a complex number. Notice the rectangle that is formed between the two axes and the move across and then up? Substitute the values of a and b. z = a + bi = rcosθ + (rsinθ)i = r(cosθ + isinθ) distributing the 3, we obtain the final answer of: we find that the value of  and the value of . Label the x-axis as the real axis and the y-axis as the imaginary axis. That’s right – it kinda looks like the the Cartesian plane which you have previously used to plot (x, y) points and functions before. In rectangular form a complex number take the form, a + bi. improve our educational resources. The rectangular representation of a complex number is in the form z = a + bi. Z = 6+j8 B. Z2 = 25-12 2. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). either the copyright owner or a person authorized to act on their behalf. The rectangular from of a complex number is written as a single real number a combined with a single imaginary term bi in the form a+bi. misrepresent that a product or activity is infringing your copyrights. a This rectangular to exponential form conversion calculator converts a number in rectangular form to its equivalent value in exponential form. Unlike the polar form, which is expressed in unit degrees, a complex exponential number is expressed in unit radians. Then, multiply through by r. Example 6: Converting from Polar to Rectangular Form Dartmouth College, Bachelor in Arts, Biochemistry and Molecular Biology. 101 S. Hanley Rd, Suite 300 The principal value of the argument is normally taken to be in the interval .However, this creates a discontinuity as moves across the negative real axis. Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. Key Concepts. By … Examples, solutions, and lessons to help High School students learn how to represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. Question: 1) Convert The Complex Number From Polar To Rectangular Form. Thus, if you are not sure content located Use and keys on keyboard to move between field in calculator. Rectangular form, on the other hand, is where a complex number is denoted by its respective horizontal and vertical components. The correct answer is therefore (2). In essence, the angled vector is taken to be the hypotenuse of a right triangle, described by the lengths of the adjacent and opposite sides. Z = 3 Cis (7/6) - 2) Find The Power Of The Complex Number In Polar Form. sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require Infringement Notice, it will make a good faith attempt to contact the party that made such content available by link to the specific question (not just the name of the question) that contains the content and a description of There are two basic forms of complex number notation: polar and rectangular. Rectangular, polar and exponential forms of complex numbers When in rectangular form, the real and imaginary parts of the complex number are co-ordinates on the complex plane, and the way you plot them gives rise to the term “Rectangular Form”. © 2007-2021 All Rights Reserved, Express Complex Numbers In Rectangular Form, Calculus Tutors in San Francisco-Bay Area. In polar form the modulus and argument are used to rewrite the complex number in the form: z = |z|(cos(θ) + i sin (θ)) where θ = arg(z) The steps to converting a complex number into polar form. Your Infringement Notice may be forwarded to the party that made the content available or to third parties such bi+a instead of a+bi). (This is spoken as “r at angle θ ”.) A complex number in rectangular form means it can be represented as a point on the complex plane. The rectangular form of a complex number is written as a+bi where a and b are both real numbers. 3. With the help of the community we can continue to Rectangular forms of numbers take on the format, rectangular number= x + jy, where x and y are numbers. Change ). Convert the following to rectangular form: Distribute the coefficient 2, and evaluate each term: Using the general form of a polar equation: we find that the value of  is   and the value of  is . which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Send your complaint to our designated agent at: Charles Cohn Columbia University in the City of New York, Master ... New Jersey Institute of Technology, Bachelor of Science, Mechanical Engineering. Example 2 – Determine which of the following is the rectangular form of a complex number. Label the x-axis as the real axis and the y-axis as the imaginary axis. Using the general form of a polar equation: we find that the value of and the value of .The rectangular form of the equation appears as , and can be found by finding the trigonometric values of the cosine and sine equations. A. Entering data into the complex number convert. Divide The Two Complex Numbers, ZJZı. There are two basic forms of complex number notation: polar and rectangular. I'm not fimular with MATLAB keywords but need to use this to prove my answers. Find the absolute value of z= 5 −i. 1. Find Z3 When Z = 4 Cis(85°). Rules. New Jersey Institute of Technology, Master o... Princeton University, Bachelor in Arts, Public Policy Analysis. Every complex number written in rectangular form has a unique polar form ) up to an integer multiple of in its argument. 2. Polar form is where a complex number is denoted by the length (otherwise known as the magnitude, absolute value, or modulus) and the angle of its vector (usually denoted by an angle symbol that looks like this: ∠). Show Hide all comments. But then why are there two terms for the form a+bi? For background information on what's going on, and more explanation, see the previous pages, You may have also noticed that the complex plane looks very similar to another plane which you have used before. I am having trouble converting polar form complex numbers into rectangular form by writing a MATLAB script file. Track your scores, create tests, and take your learning to the next level! -1.92 -1.61j [rectangular form] Euler's Formula and Identity The next section has an interactive graph where you can explore a special case of Complex Numbers in Exponential Form: The x is the real number of the expression and the y represents the imaginary number of the expression. ( Log Out /  Finding the Absolute Value of a Complex Number with a Radical. Products, Powers and Quotients Complex Numbers in Polar Form: Suppose z and w are complex numbers with polar forms z = | z | cis(α) and w = | w | cis(β). You can use rad function to convert from degrees to radians: r * exp (rad (d) * %i). Sorry, your blog cannot share posts by email. More in-depth information read at these rules. See . Complex Number Calculator. Although the complex numbers (4) and (3) are equivalent, (3) is not in standard form since the imaginary term is written first (i.e. It is the distance from the origin to the point: See and . the Rectangular forms of numbers take on the format, rectangular number= x + jy, where x and y are numbers. The steps to converting a complex number into polar form are as follows: Find the modulus. Polar form of the complex numbers is presented as r * exp (c * %i), where r is radius and c is the angle in radians. Z = 0.5 angle( pi / 4 ) or. It is the distance from the origin to the point: ∣z∣=a2+b2\displaystyle |z|=\sqrt{{a}^{2}+{b}^{2}}∣z∣=√a2+b2. Z = 0.5 angle( - 45 degrees ) 2 Comments. Then, multiply through by r. Example 10.5.6A: Converting from Polar to Rectangular Form 11.7 Polar Form of Complex Numbers 997 The following theorem summarizes the advantages of working with complex numbers in polar form. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). Note that all the complex number expressions are equivalent since they can all ultimately be reduced to -6 + 2i by adding the real and imaginary terms together. as We sketch a vector with initial point 0,0 and terminal point P x,y. By default, MATLAB/Octave accepts complex numbers only in rectangular form. ( Log Out /  Script file not sent - check your email addresses i am having trouble converting form... '' B. Z2 = 5e-145 3 Our educational resources 1 unit up on the vertical.., we first need some kind of standard mathematical notation cosθ + )... 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# What does directional derivative zeros imply when directional vector is not zero?

This question might sound stupid but I want to confirm an answer from it.

I saw somewhere online that it means that when the directional derivative of function $f$ along the none zero vector $v$ at certain point is equal to $0$, it means that the function $f$ is constant in that direction. But what does "constant in direction" mean? can anyone give me an example of it such as $f(x,y)$ to explain this?

Thanks!

• I think it simply means that the value of the function does not change as you move along the direction given by that vector. I sounds redundant, I guess. Hope this helps – Vishesh Sep 13 '13 at 12:38
• As an afterthought, this happens when you look at tangent vectors to level curves of the given function. Also if you know $D_{v}f = \nabla f . v$, where $v$ is your tangent vector. The gradient is always normal to a level curve of a function. – Vishesh Sep 13 '13 at 12:43
• But if the value long curve of that direction doesn't change, doesn't it imply that the curve is a constant? But hardly can I imagine a concrete function like this. – Cancan Sep 13 '13 at 12:54
• Other better answers have already been given. Anyway for the sake of completion, just take a look at $f(x,y)= 2e^{x}+3e^{y}$ at $(0,0)$ Then use the vector $(-3,2)$. Cheers – Vishesh Sep 13 '13 at 13:05
• What do you mean by saying a curve is a constant??? – Vishesh Sep 13 '13 at 13:06

If $D_v f(x_0)=0$, then $f$ is constant to first order in the direction $v$ - that is, if you consider the values of $f$ along the line in the direction $v$, you find that there is no linear-order term:

$$f(x_0+tv) = f(x_0) + t D_vf(x_0) + o(t) = f(x_0) + o(t).$$

(Here $o(t)$ is some function such that $o(t)/t \to 0$ as $t \to 0$, which is what we mean by zero to first order.)

Instead of considering a line, you can consider the level set $\{x : f(x) = f(x_0)\}$. So long as $x_0$ isn't a critical point of $f$, this level set will (at least in some neighbourhood of $x_0$) be a curve. Thus $f$ is constant along a curve in the direction $v$.

Let's break our understanding in several pieces and glue them together to understand the puzzle:

1. $f(x,y)$ is a number.
2. $\nabla f(x,y)$ is a vector.
3. $\vec{v}$ is also a vector.
4. The directional derivative is a number that measures increase or decrease if you consider points in the direction given by $\vec{v}$.
5. Therefore if $\nabla f(x,y) \cdot \vec{v} = 0$ then nothing happens. The function does not increase (nor decrease) when you consider points in the direction of $\vec{v}$.

Let's start with the simplest case first. $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^2$. Then the directional derivative

$$\frac{\partial f}{\partial x} = \frac{df}{dx} = 2x$$

which is zero at the origin. If you look at a graph of a parabola, you see that the closer you zoom in on the origin, the more flat the graph looks (check this yourself on an online graph calculator).

Now take $g: \mathbb{R}^2 \to \mathbb{R}$ to be $g(x,y) = x^2 + y$, which is a parabolic cylinder, as can be seen here: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427empntaacb4q.

If you take the directional derivative along $y$ (also known as the partial derivative with respect to $y$), you get $1$. Thus, this function is increasing as you fix $x$ and increase $y$. However, the directional derivative along $x$ is $2x$, which is zero at the origin. Thus, if you zoom in really close to the origin, what you will see is something that looks like a tilted plane. As you run along the $x$ axis, the plane stays at the same height (constant), but it changes as you run along the $y$ axis.

Suppose that $$f$$ is differentiable and that the directional derivative of $$f$$ along the vector $$v=(a_1,a_2,...,a_n)$$ is zero i.e. $$\nabla f\cdot v=0$$ or $$a_1f_1+a_2f_2+...+a_nf_n=0$$.

Claim: $$f$$ is constant along any line having direction $$v$$

$$L$$ be any such line which passes through $$p=(b_1,b_2,...,b_n)$$. Now the parametric equations of $$L$$ are: $$x_1(s)=b_1+a_1s,x_2(s)=b_2+a_2s,...,x_n(s)=b_n+a_ns$$. Keeping this in mind, we define $$F(s)=f(x_1(s),x_2(s),...,x_n(s))$$. The derivative $$F'(s)=f_1\dot x_1(s)+f_2\dot x_2(s)+...+f_n\dot x_n(s)=f_1a_1+f_2a_2+...+f_na_n=0$$. So $$F$$ must be a constant function. This shows that $$f$$ is constant along $$L$$.

• Why does $F'(s) = 0$ implies that $F$ is constant? – John Mars Apr 7 at 20:49
• @JohnMars By Lagrange's mean value theorem, $F(x)-F(y)=F'(z)(x-y)=0$ for all $x,y$. So $F$ is constant. – Hrit Roy Apr 8 at 23:13