# Showing a function is not uniformly continuous

I am looking at uniform continuity (for my exam) at the moment and I'm fine with showing that a function is uniformly continuous but I'm having a bit more trouble showing that it is not uniformly continuous, for example:

show that $x^4$ is not uniformly continuous on $\mathbb{R}$, so my solution would be something like:

Assume that it is uniformly continuous then:

$$\forall\epsilon\geq0\exists\delta>0:\forall{x,y}\in\mathbb{R}\ \mbox{if}\ |x-y|<\delta \mbox{then} |x^4-y^4|<\epsilon$$

Take $x=\frac{\delta}{2}+\frac{1}{\delta}$ and $y=\frac{1}{\delta}$ then we have that $|x-y|=|\frac{\delta}{2}+\frac{1}{\delta}-\frac{1}{\delta}|=|\frac{\delta}{2}|<\delta$ however $$|f(x)-f(y)|=|\frac{\delta^3}{8}+3\frac{\delta}{4}+\frac{3}{2\delta}|$$

Now if $\delta\leq 1$ then $|f(x)-f(y)|>\frac{3}{4}$ and if $\delta\geq 1$ then $|f(x)-f(y)|>\frac{3}{4}$ so there exists not $\delta$ for $\epsilon < \frac{3}{4}$ and we have a contradiction.

So I was wondering if this was ok (I think it's fine) but also if this was the general way to go about showing that some function is not uniformly continuous? Or if there was any other ways of doing this that are not from the definition?

Thanks very much for any help

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So, is this an exam question? –  user21436 Apr 22 '12 at 12:17
No, I'm just practicing for my exam where questions like this (not this one though) will come it. This is from one of the past papers that are for revision –  hmmmm Apr 22 '12 at 12:29
Just trying to ensure we are not taken for a ride. Hope you don't mind. :) –  user21436 Apr 22 '12 at 12:33
@KannappanSampath no its fine- it annoys me when I see people posting assessment questions on forums :) –  hmmmm Apr 22 '12 at 12:36

To show that it is not uniformly continuous on the whole line, there are two usual (and similar) ways to do it:

1. Show that for every $\delta > 0$ there exist $x$ and $y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|$ is greater than some positive constant (usually this is even arbitrarily large).
2. Fix the $\varepsilon$ and show that for $|f(x)-f(y)|<\varepsilon$ we need $\delta = 0$.

First way:

Fix $\delta > 0$, set $y = x+\delta$ and check $$\lim_{x\to\infty}|x^4 - (x+\delta)^4| = \lim_{x\to\infty} 4x^3\delta + o(x^3) = +\infty.$$

Second way:

Fix $\epsilon > 0$, thus $$|x^4-y^4| < \epsilon$$ $$|(x-y)(x+y)(x^2+y^2)| < \epsilon$$ $$|x-y|\cdot|x+y|\cdot|x^2+y^2| < \epsilon$$ $$|x-y| < \frac{\epsilon}{|x+y|\cdot|x^2+y^2|}$$

but this describes a necessary condition, so $\delta$ has to be at least as small as the right side, i.e.

$$|x-y| < \delta \leq \frac{\epsilon}{|x+y|\cdot|x^2+y^2|}$$

so if either of $x$ or $y$ tends to infinity then $\delta$ tends to $0$.

Hope that helps ;-)

Edit: after explanation and calculation fixes, I don't disagree with your proof.

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Thanks for the reply, I think that I use that we are considering all of $\mathbb{R}$ when i choose $x=\delta+\frac{1}{\delta}$ and $y=\frac{1}{\delta}$ as these would not be valid for small $\delta$ in bounded interval? –  hmmmm Apr 22 '12 at 13:38
@hmmmm Ok, I misunderstood what you were saying there. If the calculations are alright, then your proof is fine. –  dtldarek Apr 22 '12 at 13:44

I will comment on your solution after writing another approach. For any $x,y\in\mathbb{R}$ we have: \begin{align*} |x^{4}-y^{4}|=|(x^{2}-y^{2})(x^{2}+y^{2})|=|(x-y)(x+y)(x^{2}+y^{2})|=|x-y|\cdot |x+y|\cdot |x^{2}+y^{2}| \end{align*}

So what you can see is that even if you take arbitrarily close $x$ and $y$, you can grow the distance of $x^{4}$ and $y^{4}$ as much as you want by taking them far enough away from zero. You can easily conclude from here that the function is not uniformly continuous by a contraposition for example.

Alright, then to your solution. If the calculations would be correct, then it would be fine. You could assume at first that such $\delta>0$ exists for $0<\varepsilon<3$ and conclude with a contradiction. However, I got a bit different calculations than you. Using the above equation we see that: \begin{align*} |f(\frac{\delta}{2}+\frac{1}{\delta})-f(\frac{1}{\delta})|&=|(\frac{\delta}{2}+\frac{1}{\delta})^{4}-\frac{1}{\delta^{4}}|=|\frac{\delta}{2}(\frac{\delta}{2}+\frac{2}{\delta})((\frac{\delta}{2}+\frac{1}{\delta})^{2}+\frac{1}{\delta^{2}})| \\ &= |(\frac{\delta^{2}}{4}+1)(\frac{\delta^{2}}{4}+2\cdot \frac{\delta}{2}\cdot \frac{1}{\delta}+\frac{1}{\delta^{2}}+\frac{1}{\delta^{2}})| \\ &=|(\frac{\delta^{2}}{4}+1)(\frac{\delta^{2}}{4}+1+\frac{2}{\delta^{2}})| \\ &= |\frac{\delta^{4}}{16}+\frac{\delta^{2}}{4}+\frac{1}{2}+\frac{\delta^{2}}{4}+1+\frac{2}{\delta^{2}}| \\ &= |\frac{\delta^{4}}{16}+\frac{\delta^{2}}{2}+\frac{2}{\delta^{2}}+\frac{3}{2}|\\ &= \frac{\delta^{4}}{16}+\frac{\delta^{2}}{2}+\frac{2}{\delta^{2}}+\frac{3}{2} \end{align*} If you're able to find a lower bound for this (which is quite easy) as you did previously, then by choosing an epsilon smaller than that fixed number you may conclude as you did in your original post by contradiction.

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Hey sorry about that, I edited it-hopefully it's right now? –  hmmmm Apr 22 '12 at 13:39
I also edited now my calculation which still differs abit from your new one. Could you show the steps of how you got this answer for $|f(x)-f(y)|$? –  Thomas E. Apr 22 '12 at 13:47
yeah I messed that up quite a bit sorry (I had the wrong power and the wrong delta's) –  hmmmm Apr 22 '12 at 13:50
It should be $|(\frac{\delta}{2}+\frac{1}{\delta})^4-\frac{1}{\delta^4}|$ which would give $|\frac{\delta^4}{16}+\frac{\delta^2}{2}+\frac{2}{\delta}+\frac{3}{2}|$ I think I could conclude a similar thing from here? –  hmmmm Apr 22 '12 at 13:52
Except that is the last $-\frac{1}{\delta^{4}}$ missing from there? Otherwise it looks close to mine. –  Thomas E. Apr 22 '12 at 13:57
I think you should make this a little simpler (and for uniform continuity in general) All you need to do to show $f:X \to Y$ is not uniformly continuous on $X$ (let's suppose there both subsets of $\Bbb R$), then just give me a SINGLE epsilon such that, NO MATTER HOW SMALL delta is chosen, there will be x and y closer than delta for which the difference in function values exceed epsilon. Thus for instance $|(N+\theta)^4- N^4| \ge 4\theta N^3$ so if you choose $x$ really big (with respect to $\delta, x=N+\theta\,\,\,\ \text{and}\,\,\, y = N,$ then if $0 < \delta/2 < \theta < \delta$ you have $|x-y| < \delta$ yet you still have the variable N to play with to make the difference in function values as large as you like (in particular the difference in function values can always be made bigger than 3 regardless of how small $\delta$ is). Nevertheless, I think your proof is an accurate job!


# TrigonometryMore than one equation for a given Trig Graph?

#### m3dicat3d

##### New member
Hi all.. another Trig question here...

Let's say I'm given a graph of a sinusoidal function and asked to find its equation, but I'm not told whether this is a sine or cosine function and I'm left to determine that myself.

I understand that evaluating where the graph intersects the y axis is the straight-forward, easiest approach. For instance, take this graph where the y interval is .5 and the x interval is pi/2

I can say that it's a sine graph easily by sight, but also b/c it intersects the y axis at y=0. And given the phase shift and no vertical shift, the equation is f(x) = sin [(2/3)x].

BUT, couldn't this also be f(x) = cos [{(2/3)x} - (pi/2)] since sin(x) and cos(x) are seperated only by a phase shift of pi/2?

This is meant for my own edification and not to make this kind of exercise more confusing than be. I'm simply interseted if this is in fact mathematically accurate that you could have more than one equation (a sine or a cosine "version") for a given sinusoidal curve.

My calculator returns coincidental curves when I graph both the sine and cosine "versions" of this given graph, but I know my calculator isn't really a mathematician either haha, so I thought I'd ask some real mathematicians instead

Thanks

#### Ackbach

##### Indicium Physicus
Staff member
Note that by the addition of angles identity, that
$$\cos(2x/3- \pi/2)= \cos(2x/3) \cos( \pi/2)+ \sin(2x/3) \sin( \pi/2) = \cos(2x/3) \cdot 0+ \sin(2x/3) \cdot 1= \sin(2x/3).$$
So yes, you can definitely have more than one representation of the same graph, as you have seen on your calculator.

#### m3dicat3d

##### New member
Thank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!!

Excellent answer, and as I am still reviewing my Trig, I hadn't even considered the identity perspective of it... I can't stress enough how much having that perspective helps me even more with this...

Again, Thank you... man this place rocks!

#### MarkFL

Staff member
As sine and cosine are complementary or co-functions, this just means they are out of phase by $\displaystyle \frac{\pi}{2}$ radians, or 90°.

You may have noticed that a sine curve, if moved 1/4 period to the left, becomes a cosine curve, or conversely, a cosine curve moved 1/4 period to the right becomes a sine curve.

You are doing well to see this, it shows you are trying to understand it on a deeper level rather than just plugging into formulas. Both the sine function and the cosine function, and linear combinations of the two (with equal amplitudes) are called sinusoidal functions.

#### m3dicat3d

##### New member
Thanks MarkFL, I appreciate the encouraging words. I'm studying for my State certification exam to teach HS math here in TX, and I tutor HS students in the meantime. I'm no math genius by far, so when I ask some of my questions I sometimes feel they might be dumb (which no one here has made me feel like I'm glad to say). I'm trying to see those "nuances" in the math in case they might help my students, and again, I appreciate your words, and the full out decency of the community here. It's a great place to learn

#### Ackbach

##### Indicium Physicus
Staff member
Thank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!!

Excellent answer, and as I am still reviewing my Trig, I hadn't even considered the identity perspective of it... I can't stress enough how much having that perspective helps me even more with this...

Again, Thank you... man this place rocks!
You're quite welcome! Glad to be of help.


# Thread: Roots of unity and the length from one root to the other roots

1. ## Roots of unity and the length from one root to the other roots

Hello,

I had an assignment that required me to solve for the roots of unity for various equations of the form $z^n -1 = 0$. Then , I was asked to represent the roots of unity for each equation on an argand diagram in the form of a regular polygon.

I did all of that , however, i have a question:

is there a relation ship between the power n and the length from one root to the other roots?

When n = 3,
The roots are $-0.5 \pm 0.8660i$ and 1. The length from one root to the other roots are both 1.7321 (sqrt of 3)

When n = 4,

The roots are $\pm 1, 0 \pm 1i$. The length from one root to the other roots are (sqrt 2, sqrt and 2)

I am wondering , for the equation $z^3 - 1 = 0$ , is there a formula relating the power of z to the length from one roots to the other roots?
Thanks.

2. ## Re: Roots of unity and the length from one root to the other roots

The nth roots of unity lie at the vertices of a polygon with n sides, each vertex having distance from the center of 1. Drawing a line from the center to each vertex divides the polygon into n isosceles triangles, each having two sides of length 1, vertex angle of 360/n degrees and you want to find the length of the third side. If you draw a line from the vertex of such a triangle to the center of the base, you get two right triangles with hypotenuse length 1 and one angle of 180/n. The opposite side of that triangle has length 1(sin(180)/n) so the side of the polygon is twice that :
2sin(180/n).

3. ## Re: Roots of unity and the length from one root to the other roots

Hi, thank you very much for answering the question first.

However, i have already got the conjecture that the length of the polygon would be 2 sin (180/n), but this conjecture cannot be proven algebraically or MI that this statement would be true for any value of n, can it?

I am just wondering is there any other relationships between the length from one root to the other roots? (Lets say, if you draw a line from one root (any root) to the other roots (not only the adjacent root)

n = 3, length is sqrt 3 and sqrt 3
n = 4, length is sqrt 2 , 2 and sqrt 2
n = 5, length is not an exact value.. but what i have got is 1.1756, 1.9021, 1.9021 and 1.1756
n = 6, length is 1, sqrt 3, 2, sqrt 3, 1
n = 7, length is 0.86, 1.56, 1.94, 1.94, 1.56, 0.86

seems like theres a pattern here, but i can't seem to figure it out... would help a lot if someone could tell me if theres a conjecture for this or not, thank you very much.


+0

# help

0
62
4

Which is larger, the blue area or the orange area?

Nov 14, 2019

#1
+105411
+2

Let the radius  of the orange circle   = r

So......its area  =  pi * r^2

We can find the side length,s, of the  equilateral triangle thusly

tan (30°)  =  r / [(1/2)s]

1/ √3 = r / [ (1/2)s ]

(1/ √3) ( 1/2)s  = r

s =  (2√3)  r   = [ √12  ] r

And the area of the equilateral  triangle  is   (  √3/ 4 ) ([√12]r)^2  = 3√3 r^2

So....the blue area inside the equilateral triangle  =

[area of equailateral triangle - area of small circle  ]  /  3

[√3 r^2  - (1/3) pi r^2 ]  =   r^2 [ √3 - pi/3]      (1)

The radius  of the larger circle  can be found as

√[ [√3r ]^2  + r^2 ]   =  √ [ 3r^2 + r^2 ]   =  2r

So....the area  of the  larger circle =  pi (2r)^2  =  4pi r^2

So   the area between the side of the equilateral triangle and the larger circle   is

[Area of larger circle  - area of equilateral triangle  ]  / 3  =

[ 4pi r^2  - 3√3r^2 ] / 3  =  r^2 [ (4/3)pi   - √3]    (2)

So the sum of  (1) and (2)   is  the sum of the blue areas

r^2  [ √3 - pi/3  + (4/3) pi - √3 ]   = pi r^2

So.....the orange and blue areas are equal   !!!!

Nov 14, 2019
#2
+70
+2

Here is a different approach.The arc AC has measure $$\frac{1}{3}\cdot2\pi$$ and the angle AOB has measure half of the arc or $$\frac {\pi}{3}$$. If the radius of the larger circle is $$R$$, then the measure of OB, the radius of the smaller circle, is $$Rcos(\frac{\pi}{3})$$and so the orange region has area $$\pi(Rcos(\frac{\pi}{3}))^2=\frac{1}{4}\pi R^2$$ . The blue region, however, has area equal to $$\frac{1}{3}$$of the difference between area of the larger circle and the area of the smaller circle, i.e. $$\frac{1}{3}(\pi R^2-\frac{1}{4}\pi R^2) =\frac{1}{4}\pi R^2$$. So the two regions have the same area, each being $$\frac{1}{4}$$ of the area of the larger circle.

Nov 14, 2019
#3
+105411
+1

Thanks, Gadfly....I like that approach  !!!

Nov 14, 2019
#4
+23575
+2

Which is larger, the blue area or the orange area?

$$\begin{array}{|rcll|} \hline {\color{orange}\text{orange}} +3\times {\color{blue}\text{blue}} &=& \pi r_{\text{circumcircle}}^2 \quad | \quad : {\color{orange}\text{orange}} \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{ \pi r_{\text{circumcircle}}^2 } { {\color{orange}\text{orange}} } \quad | \quad {\color{orange}\text{orange}} = \pi r_{\text{incircle}}^2 \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{ \pi r_{\text{circumcircle}}^2 } { \pi r_{\text{incircle}}^2 } \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 \\ 3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 -1 \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\left( \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 -1 \right) \\ \\ && \boxed{\text{here, if triangle is equilateral: }\\ \mathbf{2\times r_{\text{incircle}} = r_{\text{circumcircle}}!!!} } \\\\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\left( \left( \dfrac{ 2\times r_{\text{incircle}} } { r_{\text{incircle}} } \right)^2 -1 \right)\\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\times(2^2 -1) \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\times(3) \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& 1 \\ \mathbf{ {\color{blue}\text{blue}} } &=& \mathbf{ {\color{orange}\text{orange}} } \\ \hline \end{array}$$

Nov 15, 2019
edited by heureka  Nov 15, 2019


A while ago @davidphys1 asked why nobody had made animations of the shunting yard algorithm with cutesy trains.
There is no surer way to summon me!
I've spent some of my spare time over the bank holidays making exactly that: somethingorotherwhatever.com/s

· · Web · · ·

The shunting yard algorithm neatly solves the problem of translating a mathematical expression written in infix notation (operators go between the numbers/letters) to postfix notation (operators go after the things they act on).

The core problem is that you need to work out what just what an operator applies to: with the order of operations, it might be just one number, or it might a large sub-expression.

The algorithm solves this by holding operators on a separate stack until they're needed

Here are some animations to illustrate. In the first, the operations happen left-to-right, so they appear in the same order in the output as in the input.
In the second, the addition must happen after the multiplication, so it's held back.
In the third, brackets ensure that the addition happens first.

The last wrinkle for standard arithmetic is that exponentiation is right-associative: while for the other operations you work left-to-right:
1 − 2 − 3 = (1 − 2) − 3,
the order goes the other way for exponentiation:
1 ^ 2 ^ 3 = 1 ^ (2 ^ 3)

@christianp @davidphys1 Wow that's a very cute animation!

@christianp
Doesn't that depend on the domain?

I mean, some rules aren't valid anymore for, say, Quaternions or Octaves…

@RyunoKi *leans heavily on the word "standard"*

@RyunoKi but no, I think that if I wrote an expression $$a \times b \times c$$, where a, b and c are octonions and so multiplication isn't associative, I'd expect you to interpret it as $$a \times b \times c$$, by convention.
I might include some brackets, to avoid relying on convention.

@christianp
So… the algorithm isn't considering this constraint. Fine. Space for further research :)

@RyunoKi no, it is. Or I don't understand what you mean.

@RyunoKi just noticed I missed the brackets in my second-last tweet! I had trouble LaTeXing, then didn't check how it looked!
I'd expect you to interpret $$a \times b \times c$$ as $$(a \times b) \times c$$

@christianp
I even switched to browser view to check whether something got lost in transmission 😅

@christianp
„The core problem is that you need to work out what just what an operator applies to: with the order of operations, it might be just one number, or it might a large sub-expression.

The algorithm solves this by holding operators on a separate stack until they're needed“

Is that assuming „standard" arithmetic like associative multiplication? Or the lack thereof?

Like, once „computer does that“ many people rely on it without questioning the result.

@RyunoKi right, the algorithm says that multiplication is left-associative. For real numbers, it doesn't matter.

@christianp
Thanks. Important fact in certain circumstances.

@christianp
Especially the order and arguments can change depending on how you put the brackets.

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# Is $\frac{200!}{(10!)^{20} \cdot 19!}$ an integer or not?

A friend of mine asked me to prove that $$\frac{200!}{(10!)^{20}}$$ is an integer

I used a basic example in which I assumed that there are $200$ objects places in $20$ boxes (which means that effectively there are $10$ objects in one box). One more condition that I adopted was that the boxes are distinguishable but the items within each box are not. Now the number of permutations possible for such an arrangement are : $$\frac{200!}{\underbrace{10! \cdot 10! \cdot 10!\cdots 10!}_{\text{20 times}}}$$ $$\Rightarrow \frac{200!}{(10!) ^{20}}$$

Since these are just ways of arranging, we can be pretty sure that this number is an integer.

Then he made the problem more complex by adding a $19!$ in the denominator, thus making the problem: Is $$\frac{200!}{(10!)^{20} \cdot 19!}$$ an integer or not?

The $19!$ in the denominator seemed to be pretty odd and hence I couldn't find any intuitive way to determine the thing. Can anybody please help me with the problem?

• There might be some clever way, but this is where I would start counting primes. Are there enough $2$'s in the numerator? $3$'s? – Arthur Aug 15 '17 at 7:23
• Use the fact that every set of $k$ integers will have one integer divisible by $k$. Show that that means $(m+1)(m+2).... (m+k)$ will be divisible by $k!$. And that means $(1*2....10)(11*... 20)...... (191*....200)$ is divisible by $10!*10!*....10!$. – fleablood Aug 15 '17 at 7:26
• That's an interesting friend you have there. – uniquesolution Aug 15 '17 at 7:30
• To eliminate 19! Not that all k divide 10*k. Then each 10k+1 to 10k+9 is divisible by 9! And 10k/k = 10 so it is divisible by 10 as well. – fleablood Aug 15 '17 at 7:34
• An interesting question (at least I think it would be) is what is the largest integer $k$ so that $k!*10^{20}$ divides $200!$. – fleablood Aug 15 '17 at 16:27

You assumed the boxes were distinguishable, leading to $\frac{200!}{(10!)^{20}}$, ways to fill the boxes. If you make them indistinguishable, you merge the $20!$ ways of reordering the boxes into one, so that previous answer overcounts each way of filling indistinguishable boxes by a factor of $20!$. Therefore you are left with $\frac{200!}{(10!)^{20}}/20!$ ways to fill 20 indistinguishable boxes, which then must be an integer. After multiplying by $20$ it is of course still an integer.

We know that $\dfrac{(mn)!}{n!(m!)^n}$ is an integer for $m,n \in \Bbb N$ $^{(*)}$ . Let $n = 20$ and $m = 10$, then $\dfrac{(200)!}{20!(10!)^{20}}$ is an integer.

Multiply by $20$, $\dfrac{(200)!}{19!(10!)^{20}}$ is an integer.

Using induction, this answer says that $$\frac{(mn)!}{(m!)^nn!}=\prod_{k=1}^n\binom{mk-1}{m-1}$$ Plug in $m=10$ and $n=20$ to get $$\frac{200!}{10!^{20}\,20!}=\prod_{k=1}^{20}\binom{10k-1}{9}$$ Multiply by $20$ to get $$\frac{200!}{10!^{20}\,19!}=20\,\prod_{k=1}^{20}\binom{10k-1}{9}$$

Another Approach

Note that \begin{align} \binom{kn}{n} &=\frac{(kn-n+1)(kn-n+2)\cdots(kn-1)\,kn}{1\cdot2\cdots(n-1)\,n}\\ &=\frac{(kn-n+1)(kn-n+2)\cdots(kn-1)\,k}{1\cdot2\cdots(n-1)}\\ &=\binom{kn-1}{n-1}\,k \end{align} Therefore, since we can write a multinomial as a product of binomials, \begin{align} \frac{(mn)!}{n!^m} &=\prod_{k=1}^m\binom{kn}{n}\\ &=\prod_{k=1}^m\binom{kn-1}{n-1}\,k\\ &=m!\,\prod_{k=1}^m\binom{kn-1}{n-1} \end{align} and so $$\frac{(mn)!}{n!^m\,m!}=\prod_{k=1}^m\binom{kn-1}{n-1}$$ Plug in $m=20$ and $n=10$ and multiply by $20$ to get $$\frac{200!}{10!^{20}\,19!}=20\,\prod_{k=1}^{20}\binom{10k-1}{9}$$

• But this is my answer ... – user8277998 Aug 16 '17 at 8:49
• @123: I see... I cited an answer that I wrote and didn't see that you had parenthetically cited the question to which that was an answer. However, without the connection given by the citations, the answers do not look the same. If this bothers you, I will delete my answer. – robjohn Aug 16 '17 at 13:01
• @123: I have added another proof of the same identity to differentiate our answers. I will still delete this answer if you think they are too close. – robjohn Aug 16 '17 at 13:28
• No, I have no problem with your answer, you can keep it as it is. – user8277998 Aug 16 '17 at 14:36

A long version: $$\frac{200!}{10!^{20} \cdot 19!}=\frac{30\cdot31\cdot .. \cdot200}{10!^{19}}\cdot \frac{29!}{10!\cdot(29-10)!}=...$$ which is $$...=\frac{30\cdot31\cdot .. \cdot200}{10!^{19}}\cdot \binom{29}{10}=\\ \frac{\color{red}{30} ..\color{red}{40} ..\color{red}{50} ..\color{red}{60}..\color{red}{70}..\color{red}{80}..\color{red}{90}..\color{red}{10^2}..\color{red}{110}..\color{red}{120}..\color{red}{130}..\color{red}{140}..\color{red}{150}..\color{red}{160}..\color{red}{170}..\color{red}{180}..\color{red}{190}..\color{red}{2\cdot10^{2}}}{10!^{19}}\cdot \binom{29}{10}=...$$ $20$ numbers divisible by 10, or $$3\cdot4\cdot5\cdot..\cdot9\cdot11\cdot..\cdot19\cdot20\cdot\frac{31..39\cdot41..49\cdot51..59\cdot..\cdot191..199}{9!^{19}}\cdot \binom{29}{10}=\\ 10\cdot\frac{2..9\cdot11..19\cdot31..39\cdot41..49\cdot51..59\cdot..\cdot191..199}{9!^{19}}\cdot \binom{29}{10}=...$$ cardinality of $\{31,41,51,61,71,81,91,101,111,121,131,141,151,161,171,181,191\}$ is 17 $$...=10\cdot \frac{1..9}{9!}\cdot\frac{11..19}{9!}\cdot\frac{31..39}{9!}\cdot..\cdot\frac{191..199}{9!}\cdot \binom{29}{10}=\\ 10\cdot \binom{9}{9} \cdot \binom{19}{9} \cdot \binom{39}{9}\cdot .. \cdot \binom{199}{9} \cdot \binom{29}{10}$$

• I would appreciate the down-voters to at least comment ... – rtybase Aug 15 '17 at 8:29

Consider $V=(10k+1)*....*(10k+9)$.

By your reasoning, ${10k+9 \choose 9}=(10k+1)*....*(10k+9)/9!$ is an integer.

And $10(k+1)/10(k+1)$ is an integer.

So $(10k+1)*....*(10 (k+1))$ is divisible by $9!*10*(k+1)=10!*(k+1)$.

So $200!$ is divisible by $10!*1*10!*2*10!*3*.....*10!*19=(10!)^{20}*19!$

• Don't you mean it is divisible by $10!*(k+1)$ ? – Jaap Scherphuis Aug 15 '17 at 7:55
• Yeah, I guess I did. – fleablood Aug 15 '17 at 16:22

I computed the answer just for fun using Java, and it's indeed an integer!

41355508127520659545494261323391337886154686759988983912363570790033502473625361601944917427369977161391866491251801111884812210789772970682172860398969828337097889527312353089859289462934116034461288917394623420753412096000000

import java.math.BigDecimal;
import java.math.RoundingMode;

public class JustForFun{
public static void main(String []args){
BigDecimal thFact = new BigDecimal("1");
BigDecimal tenFact = null, ntFact = null, tenFactPow20 = null;
/* Computes 200! and stores it in a */
for (int i = 1; i <= 200; i++) {
thFact = thFact.multiply(new BigDecimal(i + ""));
/* stores 10! in b */
if (i == 10)
tenFact = thFact;
/* stores 19! in c */
if (i == 19)
ntFact = thFact;
}
tenFactPow20 = tenFact.pow(20);
tenFactPow20 = tenFactPow20.multiply(ntFact);
thFact = thFact.divide(tenFactPow20);
System.out.println(thFact);
}
}

Since I am pushing 70 yrs old, it seems appropriate to dinosaur-excerpt "Elementary Number Theory" 1938 (Uspensky and Heaslett).

For real # $$r$$, let $$\lfloor r\rfloor \equiv$$ the floor of $$r$$.

Let $$p$$ be any prime #.
Let $$V_p(n) : n ~\in ~\mathbb{Z^+} ~\equiv~$$ the largest exponent $$\alpha$$ such that $$p^{\alpha} | n$$.
That is, if $$\alpha = V_p(n),$$ then $$p^{(\alpha + 1)} \not | ~n.$$

From Uspensky and Heaslett, $$V_p(n!) = \left\lfloor\frac{n}{p^1}\right\rfloor ~+~ \left\lfloor\frac{n}{p^2}\right\rfloor ~+~ \left\lfloor\frac{n}{p^3}\right\rfloor ~+~ \left\lfloor\frac{n}{p^4}\right\rfloor \cdots$$

Clearly, given two positive integers $$A,B$$, $$~\frac{A}{B}$$ will be an integer $$\iff$$
for every prime # $$p$$ that occurs in the prime factorization of $$B$$,
$$V_p(B) \leq V_p(A).$$

It is immediate, that given the OP's original question, the only prime #'s that need to be checked are those prime #'s that are $$\leq 19.$$ Further, you can see at a glance the prime #'s 11, 13, 17, and 19 can not pose a problem.

Therefore, the problem reduces to manually applying Uspenky and Heaslett's formula to the numerator and denominator of the OP's original query with respect to the prime #'s 2,3,5,7.

Empirically, they each check out okay. Therefore, the fraction is an integer.

In your face $$21^{\text{st}}$$ century!


# Coprimality Relation is not Antisymmetric

## Theorem

Consider the coprimality relation on the set of integers:

$\forall x, y \in \Z: x \perp y \iff \gcd \set {x, y} = 1$

where $\gcd \set {x, y}$ denotes the greatest common divisor of $x$ and $y$.

Then:

$\perp$ is not antisymmetric.

## Proof

We have:

$\gcd \set {3, 5} = 1 = \gcd \set {5, 3}$

and so:

$3 \perp 5$ and $5 \perp 3$

However, it is not the case that $3 = 5$.

The result follows by definition of antisymmetric relation.

$\blacksquare$


# Limit of powers of $3\times3$ matrix

Consider the matrix

$$A = \begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$$

What is $$\lim_{n→\infty}A^n$$ ?

A)$$\begin{bmatrix} 0 & 0 & 0\\ 0& 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$$ B)$$\begin{bmatrix} \frac{1}{4} &\frac{1}{2} & \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2} & \frac{1}{2}\\ \frac{1}{4}& \frac{1}{2} & \frac{1}{2}\end{bmatrix}$$ C)$$\begin{bmatrix} \frac{1}{2} &\frac{1}{4} & \frac{1}{4}\\ \frac{1}{2}& \frac{1}{4} & \frac{1}{4}\\ \frac{1}{2}& \frac{1}{4} & \frac{1}{4}\end{bmatrix}$$ D)$$\begin{bmatrix} 0 &\frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}$$ E) The limit exists, but it is none of the above

The given answer is D). How does one arrive at this result?

• Did you try doing for small values of $n$. Take $n=2, 3,4$ and post your observations (if any) as well. – Vizag May 24 '19 at 11:12

By this question, we know that

$$$$A^n= \begin{pmatrix} 2^{-n} & n\cdot 2^{-n-1} - 2^{-n-1} + \frac12 & {1-\frac{n+1}{2^n}\over2}\\ 0 & {2^{-n}+1\over2} & {1-2^{-n}\over2} \\ 0 & {1-2^{-n}\over2} & {2^{-n}+1\over2} \end{pmatrix}.$$$$

It is thus clear that $$\lim_{n\to\infty} A^n = \begin{pmatrix} 0 &\frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\\ 0 & \frac{1}{2} & \frac{1}{2}\end{pmatrix}$$.

• $2^{-n}$ tends to 0, when $n->\infty$.....right?? – Srestha May 25 '19 at 19:27
• @Srestha that is correct – Maximilian Janisch May 25 '19 at 19:40

If you are in $$1$$, you have same probability to stay there or to pass to $$2$$, but no way to get back from there. Thus you are finally drifting to $$2$$.

States $$2$$ and $$3$$ are symmetrical: at long they will tend to be equally populated, independently of the starting conditions.

Therefore also starting from $$1$$ you will at long be split between $$2$$ and $$3$$.

It’s often worth examining a matrix for obvious eigenvectors and eigenvalues, especially in artificial exercises, before plunging into computing and solving the characteristic equation. From the first column of $$A$$, we see that $$(1,0,0)^T$$ is an eigenvector with eigenvalue $$\frac12$$. The rows of $$A$$ all sum to $$1$$, so $$(1,1,1)$$ is an eigenvector with eigenvalue $$1$$. The remaining eigenvalue $$\frac12$$ can be found by examining the trace.
$$A$$ is therefore similar to a matrix of the form $$J=D+N$$, where $$D=\operatorname{diag}\left(1,\frac12,\frac12\right)$$ and $$N$$ is nilpotent of order no greater than 2. (If $$A$$ is diagonalizable, then $$N=0$$.) $$D$$ and $$N$$ commute, so expanding via the Binomial Theorem, $$(D+N)^n=D^n+nND^{n-1}$$. In the limit, $$D^n=\operatorname{diag}(1,0,0)$$ and the first column of $$N$$ is zero, so the second term vanishes. Thus, if $$A=PJP^{-1}$$, then $$\lim_{n\to\infty}A^n=P\operatorname{diag}(1,0,0)P^{-1}$$, but the right-hand side is just the projector onto the eigenspace of $$1$$. Informally, repeatedly multiplying a vector by $$A$$ leaves that vector’s component in the direction of $$(1,1,1)^T$$ fixed, while the remainder of the vector eventually dwindles away to nothing.
Since $$1$$ is a simple eigenvalue, there’s a shortcut for computing this projector that doesn’t require computing the change-of-basis matrix $$P$$: if $$\mathbf u^T$$ is a left eigenvector of $$1$$ and $$\mathbf v$$ a right eigenvector, then the projector onto the right eigenspace of $$1$$ is $${\mathbf v\mathbf u^T\over\mathbf u^T\mathbf v}.$$ (This formula is related to the fact that left and right eigenvectors with different eigenvalues are orthogonal.) We already have a right eigenvector, and a left eigenvector is easily found by inspection: the last two columns both sum to $$1$$, so $$(0,1,1)$$ is a left eigenvector of $$1$$. This gives us $$\lim_{n\to\infty}A^n = \frac12\begin{bmatrix}1\\1\\1\end{bmatrix}\begin{bmatrix}0&1&1\end{bmatrix} = \begin{bmatrix}0&\frac12&\frac12\\0&\frac12&\frac12\\0&\frac12&\frac12\end{bmatrix}.$$


How to draw Fractal images of iteration functions on the Riemann sphere?

Prof. McClure, in the work "M. McClure, Newton's method for complex polynomials. A preprint version of a “Mathematical graphics” column from Mathematica in Education and Research, pp. 1–15 (2006)", discusses how Mathematica can be applied to iteration functions for obtaining the basins of attraction (or their fractal images). Below, I provide his code for the fractal image of the polynomial $p(z)=z^3-1$:

p[z_] := z^3 - 1;
theRoots = z /. NSolve[p[z] == 0, z]
cp = Compile[{{z, _Complex}}, Evaluate[p[z]]];
n = Compile[{{z, _Complex}}, Evaluate[Simplify[z - p[z]/p'[z]]]];
bail = 150;
orbitData = Table[
NestWhileList[n, x + I y, Abs[cp[#]] > 0.01 &, 1, bail],
{y, -1, 1, 0.01}, {x, -1, 1, 0.01}
];
numRoots = Length[Union[theRoots]];
sameRootFunc = Compile[{{z, _Complex}}, Evaluate[Abs[3 p[z]/p'[z]]]];
whichRoot[orbit_] :=
Module[{i, z},
z = Last[orbit]; i = 1;
Scan[If[Abs[z - #] < sameRootFunc[z], Return[i], i++] &, theRoots];
If[i <= numRoots, {i, Length[orbit]}, None]
];
rootData = Map[whichRoot, orbitData, {2}];
colorList = {{cc, 0, 0}, {cc, cc, 0}, {0, 0, cc}};
cols = rootData /. {
{k_Integer, l_Integer} :> (colorList[[k]] /. cc -> (1 - l/(bail + 1))^8),
None -> {0, 0, 0}
};
Graphics[{Raster[cols]}]

My main question is here. He nicely obtained the fractal images on the complex plane, while it would be an interesting challenge to obtain these images on the Riemann sphere, e.g.

It seems the complex plane in this case has been replaced by a sphere, but how? I will be thankful if someone could revise the code given above for obtaining such beautiful fractal images on the Riemann sphere. Any tips and tricks will be fully appreciated as well.

-
Would you care to share where your example spherical projection came from? –  Mark McClure Dec 24 '12 at 13:04

As the other answers have shown, it's fairly easy to map an image onto a parametrized surface using textures. It can be a bit tricky, though, getting the image to mesh well with the transformation. J.M. hit on the crucial issue, namely that we compute the image using points that map to the sphere with minimal distortion. This answer is largely an expansion on his, although there are some differences and other ideas as well.

First, the article that Fazlollah refers to is some years old now, and the code can be improved in light of the many changes since V5, so let's start by showing how to generate regular Newton iteration images for general polynomials. Given a polynomial function $f(z)$, the following code computes the corresponding Newton's method iteration function $n$. It then defines the command limitInfo that iterates $n$ up to $50$ times from a starting point $z_0$ terminating when $|f(z)|$ is small and returning the last iterate and the number of iterates required for $|f(z)|$ to get small. It's compiled, listable and set to run in parallel, so it should be pretty fast. In addition to the function f, there are two numeric parameters to set, bail and r.

f = Function[z, z^3 - 1]; (* A very standard example *)
n = Function[z, Evaluate[Simplify[z - f[z]/f'[z]]]];
bail = 50;
(* bail is the number of iterates before bailing. *)
(* Doesn't have to be particularly large, *)
(* if there are only simple roots. *)
r = 0.01;
(* We assume that if |z-z0|<r, then we've *)
(* converged to the root z0. *)
limitInfo = With[{bail = bail, r = r, f = f, n = n},
Compile[{{z0, _Complex}},
Module[{z, cnt},
cnt = 0; z = z0;
While[Abs[f[z]] > r && cnt < bail,
z = n[z];
cnt = cnt + 1
];
{z, cnt}],
RuntimeAttributes -> {Listable},
Parallelization -> True,
RuntimeOptions -> "Speed"
]];

Since the function is listable and runs in parallel, we can simply apply it to a table of data on one fell swoop.

step = 4/801; (* The denominator is essentially the resolution. *)
limitData = limitInfo[
Table[x + I*y, {y, 2, -2, -step}, {x, -2, 2, step}]]; // AbsoluteTiming

(* Out: {1.492716, Null} *)

Each element is a pair that indicates the limiting behavior and how long it took to get there.

limitData[[1, 1]]

(* Out: {-0.499835 + 0.866044 I, 5. + 0. I} *)

I guess we need a function that takes something like that and turns it into a color.

roots = z /. NSolve[f[z] == 0, z];
preColors = List @@@ Table[ColorData[61, k], {k, 1, Length[roots]}];
preColors = Append[preColors, {0.0, 0.0, 0.0}];
color = With[{bail = bail, roots = roots, preColors = preColors},
Compile[{{z, _Complex}, {cnt, _Complex}},
Module[{arg, time, i},
arg = Arg[z];
time = Abs[cnt];
i = 1;
Scan[If[Abs[z - #] < 0.1, Return[i], i++] &, roots];
Abs[preColors[[i]]*(cnt/bail)^(0.2)]
(* The exponent 0.2 adjusts the brightness of the image. *)]]
];

Now, we apply that function and generate the image.

colors = Apply[color, limitData, {2}];
Image[colors, ImageSize -> 2/step]

To map onto a sphere nicely, we'll discard the rectangular grid of points that we used above in favor of a collection of points that looks something like the following (although, we'll want higher resolution, of course):

step = Pi/12;
pts = Table[Cot[phi/2] Exp[I*theta],
{phi, step, Pi - step, step}, {theta, -Pi, Pi, step}];
ListPlot[{Re[#], Im[#]} & /@ Flatten[pts, 1],
AspectRatio -> Automatic, PlotRange -> All,
Epilog -> {Red, Circle[]}]

The expression $\cot(\phi/2) e^{i\theta}$ is the stereographic projection of a point expressed in spherical coordinates $(1,\phi,\theta)$ onto the plane. As a result, the corresponding points on the sphere are nicely distributed. Note, for example, that the number of points inside and outside of the unit circle are the same.

Graphics3D[{{Opacity[0.8], Sphere[]},
Point[Flatten[Table[{Cos[theta] Sin[phi], Sin[theta] Sin[phi], Cos[phi]},
{phi, step, Pi - step, step}, {theta, -Pi, Pi, step}], 1]]}]

Now, we increase the resolution and use the same limitInfo and color functions as before.

step = Pi/500;
limitData = limitInfo[Table[Cot[phi/2] Exp[I*theta],
{phi, step, Pi - step, step}, {theta, -Pi, Pi, step}]];
colors = Apply[color, limitData, {2}];
rect = Image[colors, ImageSize -> 4/step]

The image looks a bit different, but it's perfect for use as a spherical texture.

ParametricPlot3D[{Cos[theta] Sin[phi], Sin[theta] Sin[phi], Cos[phi]} ,
{theta, -Pi, Pi}, {phi, 0, Pi}, Mesh -> None, PlotPoints -> 100,
Boxed -> False, PlotStyle -> Texture[Show[rect]],
Lighting -> "Neutral", Axes -> False]

We can incorporate all of this into a Module.

newtonSphere[fIn_, var_, resolution_, bail_: 50, r_: 0.01] := Module[
{f, n, limitInfo, color, colors, roots, preColors, step, limitData, rect},

f = Function[var, fIn];
n = Function[var, Evaluate[Simplify[var - f[var]/f'[var]]]];
limitInfo = With[{bailLoc = bail, rLoc = r, fLoc = f, nLoc = n},
Compile[{{z0, _Complex}},
Module[{z, cnt},
cnt = 0; z = z0;
While[Abs[fLoc[z]] > rLoc && cnt < bailLoc,
z = nLoc[z];
cnt = cnt + 1
];
{z, cnt}],
RuntimeAttributes -> {Listable},
Parallelization -> True,
RuntimeOptions -> "Speed"
]];
roots = z /. NSolve[f[z] == 0, z];
preColors = List @@@ Table[ColorData[61, k], {k, 1, Length[roots]}];
preColors = Append[preColors, {0.0, 0.0, 0.0}];
color = With[{bailLoc = bail, rootsLoc = roots, preColorsLoc = preColors},
Compile[{{z, _Complex}, {cnt, _Complex}},
Module[{arg, time, i},
arg = Arg[z];
time = Abs[cnt];
i = 1;
Scan[If[Abs[z - #] < 0.1, Return[i], i++] &, rootsLoc];
preColorsLoc[[i]]*(cnt/bailLoc)^(0.2)
]]
];
step = Pi/resolution;
limitData = limitInfo[
Table[Cot[phi/2] Exp[I*theta], {phi, step, Pi - step,
step}, {theta, -Pi, Pi, step}]];
colors = Apply[color, limitData, {2}];
rect = Image[colors, ImageSize -> 4/step];
ParametricPlot3D[{Cos[theta] Sin[phi], Sin[theta] Sin[phi],
Cos[phi]} ,
{theta, -Pi, Pi}, {phi, 0, Pi}, Mesh -> None, PlotPoints -> 100,
Boxed -> False, PlotStyle -> Texture[Show[rect]],
Lighting -> "Neutral", Axes -> False]
];

Now, if I had to guess, I'd say that example image in the original post was generated by a small perturbation of $z^8-z^2$.

newtonSphere[(2 z/3)^8 - (2 z/3)^2 + 1/10, z, 500]

Here are a few more examples.

pic1 = newtonSphere[z^2 - 1, z, 401];
SeedRandom[1];
pic2 = newtonSphere[Sum[RandomInteger[{-3, 5}] z^k, {k, 0, 8}], z, 400];
pic3 = newtonSphere[z^10 - z^5 - 1, z, 400, 200];
pic4 = newtonSphere[z^5 - z - 0.99, z, 400];
GraphicsGrid[{
{pic1, pic2},
{pic3, pic4}
}]

In the top row, we see that the result for quadratic polynomials is typically rather boring while that for a random degree 8 polynomial can be quite cool. On the bottom right, we see a black region. The color function is setup to default to black when none of the roots are detected. This can certainly happen; in fact the Newton iteration function for this example has an attractive orbit of period 6 leading to the quadratic like Julia set seen in the image. Sometimes black can occur simply because we didn't iterate enough, which is why I used the optional fourth argument for the image in the bottom left.

-
Now imagine those hanging from a fractal christmas tree. (+1) –  Jens Dec 24 '12 at 17:40

Here is my modest attempt, based on the formulae for stereographic projection in this Wikipedia entry (where the north pole corresponds to the point at infinity) and using a technique similar to the one in this answer:

newtonRaphson = Compile[{{n, _Integer}, {c, _Complex}},
Arg[FixedPoint[(# - (#^n - 1)/(n #^(n - 1))) &, c, 30]]]

tex = Image[DensityPlot[
newtonRaphson[3, Cot[ϕ/2] Exp[I θ]], {θ, -π, π}, {ϕ, 0, π},
AspectRatio -> Automatic,
ColorFunction -> (Which[# < .3, Red, # > .7, Yellow, True, Blue] &),
Frame -> False, ImagePadding -> None, PlotPoints -> 400,
PlotRange -> All, PlotRangePadding -> None],
ImageResolution -> 256];

(* yes, I know that I could have used SphericalPlot3D[]... *)
ParametricPlot3D[{Sin[ϕ] Cos[θ], Sin[ϕ] Sin[θ], Cos[ϕ]}, {θ, -π, π}, {ϕ, 0, π},
Axes -> None, Boxed -> False, Lighting -> "Neutral", Mesh -> None,
PlotStyle -> Texture[tex], TextureCoordinateFunction -> ({#4, #5} &)]

-
Thanks for your response. There are two problems. 1. How can one zoom in on a particluar place on this sphere without lowering the quality to observe the fractal behaviour of the method? 2. The "spcae size" of the output image? In fact, how one can save as the output fractal image with low "disk size" without lowering the quality in EPS format? For example, for $n=8$, its size is more than 2MB! –  Fazlollah Soleymani Nov 22 '12 at 21:16
You'll have to play with PlotPoints and ImageResolution on your own, of course... –  Guess who it is. Nov 22 '12 at 23:13
Thanks, the problem is that when we reduce PlotPoints or ImageResolution, the quality gets down dramatically. I am looking for a fast way to obtain high quality pics, with small space size, just like the one given in the question. Rasterize@... is a good choice, but it disables the feature of rotating the 3D pic, and also gets the quality lower. –  Fazlollah Soleymani Nov 23 '12 at 8:20
Well, I'd say this is one of those "no such thing as a free lunch" things. If you want to be able to zoom, you certainly need high resolution, which will demand more of your computer's resources... if you want small file sizes, you'll have to sacrifice quality somewhat. Scylla and Charybdis, you know... –  Guess who it is. Nov 23 '12 at 9:49
img2 = ImageCrop[Image[Graphics[{Raster[cols]}, PlotRangePadding -> 0,
ImagePadding -> 0, ImageMargins -> 0]], {343, 343}];

SphericalPlot3D[1 , {u, 0, Pi}, {v, 0, 2 Pi}, Mesh -> None,
TextureCoordinateFunction -> ({#1, #2} &),
PlotStyle -> Directive[Specularity[White, 10], Texture[img2]],
Lighting -> "Neutral", Axes -> False, ImageSize -> 500]

where img2 is cropped version of the 2D image in OP's question.

Check Texture for more examples.

-
Thanks for your reply, but there are big white circles at the middle of each basin. They should not be here. Please rotate your image, and then you will see the incomplete fractal image. Can you solve this drawback? –  Fazlollah Soleymani Nov 22 '12 at 20:59
img2 is the cropped version of your 2D image: img2=ImageCrop[ Image[Graphics[{Raster[cols]}, PlotRangePadding -> 0, ImagePadding -> 0, ImageMargins -> 0]], {343, 343}]. –  kglr Nov 22 '12 at 21:50
Now that I think about it, one could have directly produced an image instead of going through the Raster[] route: Image[cols]. –  Guess who it is. Nov 24 '12 at 14:12

I've decided to write a simplification+extension of Mark's routine as a separate answer. In particular, I wanted a routine that yields Riemann sphere fractals not only for Newton-Raphson, but also its higher-order generalizations (e.g. Halley's method).

I decided to use Kalantari's "basic iteration" family for the purpose. An $n$-th order member of the family looks like this:

$$x_{k+1}=x_k-f(x_k)\frac{\mathcal D_{n-1}(x_k)}{\mathcal D_n(x_k)}$$

where

$$\mathcal D_0(x_k)=1,\qquad\mathcal D_n(x_k)=\begin{vmatrix}f^\prime(x_k)&\tfrac{f^{\prime\prime}(x_k)}{2!}&\cdots&\tfrac{f^{(n-2)}(x_k)}{(n-2)!}&\tfrac{f^{(n-1)}(x_k)}{(n-1)!}\\f(x_k)&f^\prime(x_k)&\ddots&\vdots&\tfrac{f^{(n-2)}(x_k)}{(n-2)!}\\&f(x_k)&\ddots&\ddots&\vdots\\&&\ddots&\ddots&\vdots\\&&&f(x_k)&f^\prime(x_k)\end{vmatrix}$$

As noted in that paper, the basic family generalizes the Newton-Raphson iteration; $n=1$ corresponds to Newton-Raphson, while $n=2$ gives Halley's method. (Relatedly, see also Kalantari's work on polynomiography.)

Here's a routine for $\mathcal D_n(x)$:

iterdet[f_, x_, 0] := 1;
iterdet[f_, x_, n_Integer?Positive] := Det[ToeplitzMatrix[PadRight[{D[f, x], f}, n],
Table[SeriesCoefficient[Function[x, f]@\[FormalX], {\[FormalX], x, k}], {k, n}]]]

Here is the routine for generating the Riemann sphere fractals:

Options[rootFractalSphere] = {ColorFunction -> Automatic, ImageResolution -> 400,
MaxIterations -> 50, Order -> 1, Tolerance -> 0.01};

rootFractalSphere[fIn_, var_, opts : OptionsPattern[]] /; PolynomialQ[fIn, var] :=
Module[{γ = 0.2, bail, cf, colList, f, h, itFun, ord, roots, tex, tol},

f = Function[var, fIn];
ord = OptionValue[Order];
itFun = Function[var, var - Simplify[f[var] iterdet[f[var], var, ord - 1]/
iterdet[f[var], var, ord]] // Evaluate];

roots = var /. NSolve[f[var], var];
cf = OptionValue[ColorFunction];
If[cf === Automatic, cf = ColorData[61]];
colList = Append[Table[List @@ ColorConvert[cf[k], RGBColor], {k, Length[roots]}],
{0., 0., 0.}];

bail = OptionValue[MaxIterations]; tol = OptionValue[Tolerance];
makeColor = Compile[{{z0, _Complex}},
Module[{cnt = 0, i = 1, z},
z = FixedPoint[(++cnt; itFun[#]) &, z0, bail,
SameTest -> (Abs[f[#2]] < tol &)];
Scan[If[Abs[z - #] < 10 tol, Return[i], i++] &, roots];
Abs[colList[[i]] (cnt/bail)^γ]],
CompilationOptions -> {"InlineExternalDefinitions" -> True},
RuntimeAttributes -> {Listable}, RuntimeOptions -> "Speed"];

h = π/OptionValue[ImageResolution];
tex = DeveloperToPackedArray[makeColor[
Table[Cot[φ/2] Exp[I θ], {φ, h, π - h, h}, {θ, -π, π, h}]]];

ParametricPlot3D[{Cos[θ] Sin[φ], Sin[θ] Sin[φ], Cos[φ]}, {θ, -π, π}, {φ, 0, π},
Axes -> False, Boxed -> False, Lighting -> "Neutral", Mesh -> None,
PlotPoints -> 75, PlotStyle -> Texture[tex],
Evaluate[Sequence @@ FilterRules[{opts}, Options[Graphics3D]]]]]

Other notes:

• The compiled functions limitInfo[] and color[] have been merged into the single function makeColor[]. This function was not localized on purpose to allow its use even after executing rootFractalSphere[].

• Texture[] can directly accept an array of RGB triplets, so there is no need to use Image[] if these triplets are being generated directly by makeColor[].

Now, for some examples. The first two are Newton-Raphson fractals:

rootFractalSphere[z^3 - 1, z]

rootFractalSphere[(2 z/3)^8 - (2 z/3)^2 + 1/10, z]

Here is a fractal generated by Halley's method:

rootFractalSphere[(2 z/3)^8 - (2 z/3)^2 + 1/10, z, Order -> 2]

Finally, a fractal from a third order iteration:

rootFractalSphere[z^10 - z^5 - 1, z, ColorFunction -> ColorData[54],
MaxIterations -> 200, Order -> 3]
`

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Thanks. An excellent answer. Just one note. There might be some diverging points (black areas) in the fractal picture for some test problems. On the other hand, we used a color correspond to a point in a sphere or a rectangular domain. So, the domain of working (I mean the mesh of points) are finite. So is it possible to count the number of diverging points? I mean it would be nice to have the percentage of diverging points for each fractal picture. Is it possible to cunt the number of diverging points in your implementation? –  Fazlollah Soleymani Apr 6 '13 at 12:25


The Distributive Law is used in reverse to factorise a quadratic trinomial, as illustrated below.. We can use the box method to factorise a quadratic trinomial. For example: Here b = –2, and c = –15. But a "trinomial" is any three-term polynomial, which may not be a quadratic (that is, a degree-two) polynomial. Free online science printable worksheets for year 11, solve quadratic java, sample algebra test solving addition equations, College algebra tutorial. Trinomial. Examples are 7a2 + 18a - 2, 4m2, 2x5 + 17x3 - 9x + 93, 5a-12, and 1273. To figure out which it is, just carry out the O + I from FOIL. A quadratic trinomial is a polynomial with three terms and the degree of the trinomial must be 2. Think FOIL. Example 1; Example 2; Example 3; Example 4; Example 5; Example 1 Example. Quadratic equation of leading coefficient 1. Let’s consider two cases:  (1) Leading coefficient is one, a = 1, and (2) leading coefficient is NOT 1, a ≠ 1. Factor a Quadratic Trinomial. So (3x5)2 = 9x10. x2 + 20x + ___ x2 - 4x + ___ x2 + 5x + ___ 100 4 25/4 … Example 6: A quadratic relation has an equation in factored form. Yes, … Vocabulary. This math video tutorial shows you how to factor trinomials the easy fast way. By the end of this section, you will be able to: Factor trinomials of the form ; Factor trinomials of the form ; Before you get started, take this readiness quiz. Do you see how all three terms are present? The x-intercepts of the parabola are − 4 and 1. Again, think about FOIL and where each term in the trinomial came from. For example: $$x^2 + y^2 + xy$$ and $$x^2 + 2x + 3xy$$. The general form of a quadratic equation is. Expand the equation (2x – 3) 2 = 25 to get; 4x 2 – 12x + 9 – 25 = 0 4x 2 – 12x – 16 = 0. Example … You need to think about where each of the terms in the trinomial came from. Simplify: ⓐ ⓑ If you missed this problem, review . The general form of a quadratic trinomial is ax 2 + bx + c, where a is the leading coefficient (number in front of the variable with highest degree) and c is the constant (number with no variable). Log base change on the TI-89, cube root graph, adding/subtracting positive and negative numbers, java applet factoring mathematics algebra, trigonometry questions, algebra1 prentice hall. Following is an explanation of polynomials, binomials, trinomials, and degrees of a polynomial. Remember, when a term with an exponent is squared, the exponent is multiplied by 2, the base is squared. In a quadratic equation, leading coefficient is nothing but the coefficient of x 2. For example, x + 2. The answer would be 5 and 6. Step 1:  Identify if the trinomial is in quadratic form. We require two numbers that multiply to – 18 and add to 7. ax 2 + bx + c. 3x 2 + 7x – 6. ac = 3 × – 6 … Example 3. On this page we will learn what a trinomial in quadratic form is, and what a trinomial in quadratic form is not. In the next section, we will address the technique used to factor $$ax^2+bx+c$$ when $$a \neq 1$$. A quadratic trinomial is a trinomial of which the highest power of any variable is two. Learn how to factor quadratic expressions as the product of two linear binomials. A special type of trinomial can be factored in a manner similar to quadratics since it can be viewed as a quadratic in a new variable (x n below). A binomial is a … In general g(x) = ax 2 + bx + c, a ≠ 0 is a quadratic polynomial. (2x + ? Solution . Quality resources and hosting are expensive, Creative Commons Attribution 4.0 International License. Now you’ll need to “undo” this multiplication—to start with the product and end up with the factors. Below are 4 examples of how to use algebra tiles to factor, starting with a trinomial where A=1 (and the B and C values are both positive), all the way to a trinomial with A>1 (and negative B and/or C values). Jenn, Founder Calcworkshop ®, 15+ Years Experience (Licensed & Certified Teacher) Finding the degree of a polynomial is nothing more than locating the largest … To factorise a quadratic trinomial. $$\text{Examples of Quadratic Trinomials}$$ For example, let us apply the AC test in factoring 3x 2 + 11x + 10. trinomial, as illustrated below. Factoring Trinomials Formula, factoring trinomials calculator, factoring trinomials a 1,factoring trinomials examples, factoring trinomials solver. Summary:  A quadratic form trinomial is of the form axk + bxm + c, where 2m = k.  It is possible that these expressions are factorable using techniques and methods appropriate for quadratic equations. It is due to the presence of three, unlike terms, namely, 3x, 6x 2 and 2x 3. That would be a – 5 and a + 3. For example, 2x 2 − 7x + 5. If you experience difficulties when using this Website, tell us through the feedback form or by phoning the contact telephone number. In the given trinomial, the product of A and C is 30. We begin by showing how to factor trinomials having the form $$ax^2 + bx + c$$, where the leading coefficient is a = 1; that is, trinomials having the form $$x^2+bx+c$$. Quadratic equation of leading coefficient not equal to 1. Below are 4 examples of how to use algebra tiles to factor, starting with a trinomial where A=1 (and the B and … This is a quadratic form polynomial because the second term’s variable, x3, squared is the first term’s variable, x6. Solving Trinomial Equations Using The Quadratic Formula, Algebra free worked examples for children in 3rd, 4th, 5th, 6th, 7th & 8th grades, worked algebra problems, solutions to algebra questions for children, algebra topics with worked exercises on , inequalities, intergers, logs, polynomials, angles, linear equations, quadratic equation, monomials & more This page will focus on quadratic trinomials. Which of the following is a quadratic? All my letters are being represented by numbers. The last term is plus. A special type of trinomial can be factored in a manner similar to quadratics since it can be viewed as a quadratic in a new variable (x n below). NCERT Solutions For Class 12 Physics; NCERT Solutions For Class 12 Chemistry ; NCERT Solutions For Class 12 Biology; NCERT Solutions For Class 12 Maths; NCERT Solutions … And the middle term's coefficient is also plus. Let’s look first at trinomials with only the middle term negative. For example, the box for is: \begin{array}{|c|c|c} \hline x^2 & 3x & x \\ \hline 2x & 6 & 2 \\ \hline x & 3 \\ \end{array} Therefore Factoring quadratic trinomial and how to factor by grouping. Generally we have two types of quadratic equation. What happens when there are negative terms? This will help you see how the factoring works. The solution a 1 = 2 and a 2 = 1 of the above system gives the trinomial factoring: (x 2 + 3x+ 2) = (x + a 1)(x + a 2) … Factorise 3x 2 + 7x – 6. This part will focus on factoring a quadratic when a, the x 2-coefficient, is 1. Solution. x is being squared. Likewise, 11pq + 4x 2 –10 is a trinomial. The product of two linear factors yields a quadratic trinomial; and the D is a perfect square because it is the square of 5. The tricky part here is figuring out the factors of 8 and 30 that can be arranged to have a difference of 43. b) Write the equation in vertex form. So either -5 × 1 or 5 × -1. Now hopefully, we have got the basic difference between Monomial, Binomial and Trinomial. If you're behind a web filter, please make sure that … Solution: Find the product of the first and the last constants. Examples, solutions, videos, worksheets, ... Scroll down the page for more examples and solutions of factoring trinomials. Solve the following quadratic equation (2x – 3) 2 = 25. NCERT Solutions. Following is an example of trinomial: x 3 + x 2 + 5x 2x 4 -x 3 + 5 A trinomial meaning in math is, it is a type of polynomial that contains only three terms. For example, a univariate (single-variable) quadratic function has the form = + +, ≠in the single variable x.The graph of a univariate quadratic function is a parabola whose axis of symmetry is parallel to the y-axis, as shown at right.. (y+a) (y+b) = y (y+b) + a (y+b) = y 2 + by + ay + ab = y 2 + y (a+b) + ab … Expand the equation (2x – 3) 2 = 25 to get; 4x 2 – 12x + 9 – 25 = 0 4x 2 – 12x – 16 = 0. Perfect Square Trinomial – Explanation & Examples A quadratic equation is a polynomial of second degree usually in the form of f(x) = ax 2 + bx + c where a, b, c, ∈ R and a ≠ 0. Donate Login … How to factor a quadratic trinomial: 5 examples and their solutions. A trinomial is a sum of three terms, while a multinomial is more than three. A trinomial is a polynomial or algebraic expression, which has a maximum of three non-zero terms. Problem 1. This part will focus on factoring a quadratic when a, the x 2 -coefficient, is 1. a x 2 + b x + c = 0 → (x + r) (x + s) Let's solve the following equation by factoring the trinomial: Quadratic is another name for a polynomial of the 2nd degree. A few examples of trinomial expressions are: – 8a 4 +2x+7; 4x 2 + 9x + 7; Monomial: Binomial: Trinomial: One Term: Two terms: Three terms: Example: x, 3y, 29, x/2: Example: x 2 +x, x 3-2x, y+2: Example: x 2 +2x+20: Properties . Use the tabs below to navigate through the notes, video, and practice problems. How to factor a quadratic trinomial: 5 examples and their solutions. In this post, I want to focus on that last topic -- using algebra tiles to factor quadratic trinomials. Remember: To get a negative sum and a positive product, the numbers must both be negative. ax 2 + bx + c = 0. (Lesson 13: Exponents.) For example, the polynomial (x 2 + 3x + 2) is an example of this type of trinomial with n = 1. )(x + ?) Australian Business Number 53 056 217 611, Copyright instructions for educational institutions. So the book's section or chapter title is, at best, a bit off-target. Start from finding the factors of +2. This form is factored as: + + = (+) (+), where + = ⋅ =. Let’s see another example, here where a is not one. 15 Factor Quadratic Trinomials with Leading Coefficient 1 Learning Objectives. $$(x − 5)$$ and $$(x + 3)$$ are factors of $$x^2 − 2x … A quadratic trinomial is factorable if the product of A and C have M and N as two factors such that when added would result to B. c) Sketch a graph of the relation and label all features. Quadratic Polynomial. Since (x2)2 = x4, and the second term is x4, then n = 2. Non-Example: These trinomials are not examples of quadratic form. Step 3: Apply the appropriate factoring technique. Example 1. Binomials. For more practice on this technique, please visit this page. A binomial is a sum of two terms. Factorise by grouping the four terms into pairs. Website and our Privacy and Other Policies. Previously, we went over how to factor out a quadratic trinomial with a leading coefficient of 1. They take a lot of the guesswork out of factoring, especially for trinomials that are not easily factored with other methods. 6 or D = 25. If a polynomial P(x) is … It is the correct pair … NCERT Solutions For Class 12. Let’s look at this quadratic form trinomial and a quadratic with the same coefficients side by side. Example 1: Factor the trinomial x^2+7x+10 x2 + 7x + 10 as a product of two binomials. Obviously, this is an “easy” case because the coefficient of the squared term x x is just 1. If a is one, then we just need to find what two numbers have the product c and the sum of b. Factorising an expression is to write it as a product of its factors. Well, it depends which term is negative. In this quadratic, 3x 2 + 2x − 1, the constants are 3, 2, −1. Factoring quadratic trinomials using the AC Method. So, n = 3. For example, 2x²+7x+3=(2x+1)(x+3). Factoring Trinomials (Quadratics) : Method With Examples Consider the product of the two linear expressions (y+a) and (y+b). Don't worry about the difference, though; the book's title means … It means that the highest power of the variable cannot be greater than 2. 6, the independent term, is the product of 2 and 3. If you’re a teacher and would like to use the materials found on this page, click the teacher button below. The argument appears in the middle term. Then, find the two factors of 30 that will produce a sum of 11. A polynomial formed by the sum of only three terms (three monomials) with different degrees is known as a trinomial. Let’s begin with an example. It consists of only three variables. write the expression in the form ax 2 + bx + c; find two numbers that both multiply to ac and add to b; split the middle term bx into two like terms using those two numbers as coefficients. Let’s look at an example of multiplying binomials to refresh your memory. There are 4 methods: common factor, difference of two squares, trinomial/quadratic expression and completing the square. A polynomial is an algebraic expression with a finite number of terms. A quadratic trinomial is a trinomial in which the highest exponent or power is two, or the second power. Please read the Terms and Conditions of Use of this Quadratic trinomials with a leading coefficient of one. Factoring quadratic trinomial and how to factor by grouping. a, b, c are called constants. factors, | Home Page | Order Maths Software | About the Series | Maths Software Tutorials | The x-intercepts of the parabola are − 4 and 1. Let's take an example. FACTORING 2. Example are: 2x 2 + y + z, r + 10p + 7q 2, a + b + c, 2x 2 y 2 + 9 + z, are all trinomials having three variables. A polynomial having its highest degree 2 is known as a quadratic polynomial. Tie together everything you learned about quadratic factorization in order to factor various quadratic expressions of any form. This video contains plenty of examples and practice ... Factoring Perfect Square Trinomials Factoring Perfect Square Trinomials door The Organic Chemistry Tutor 4 jaar geleden 11 minuten en 3 seconden 267.240 weergaven This algebra video tutorial focuses on , factoring , perfect square , trinomials , . In general, the trinomial of the ax 2 + bx + c is a perfect square if the discriminant is zero; that is, if b 2 -4ac = 0, because in this case it will only have one root and can be expressed in the form a (xd) 2 = (√a (xd)) 2 , where d is the root already mentioned. COMPLETING THE SQUARE 3. Consider the expansion of (x + 2)(x + 3).We notice that: 5, the coefficient of x, is the sum of 2 and 3.; 6, the independent term, is the product of 2 and 3.; Note: The product of two linear factors yields a quadratic trinomial; and the factors of a quadratic trinomial are linear factors.. Now consider the expansion of … There is one last factoring method you’ll need for this unit: Factoring quadratic form polynomials. So, n = 5. Exercise 2.1. factors of a quadratic trinomial are linear factors. There are three main ways of solving quadratic equations: 1. Example 3. \((x − 5)(x + 3) = x^2 − 2x − 15$$ Here, we have multiplied two linear factors to obtain a quadratic expression by using the distributive law. Factoring Polynomials - Standard Trinomials (Part 1) Factoring Polynomials of the form ax … Solution. Consider making your next Amazon purchase using our Affiliate Link. Factoring quadratic is an approach to find the roots of a quadratic equation. A polynomial having its highest degree 3 is known as a Cubic polynomial. Year 10 Interactive Maths - Second Edition. Example: x 2 - 12x + 27. a = 1 b = -12 c = 27. Solving quadratic equations by factoring is all about writing the quadratic function as a product of two binomials functions of one degree each. Divide each term by 4 to get; x 2 – 3x – 4 = 0 (x – 4) (x + 1) = 0 x = 4 or x = -1. Some of the important properties of polynomials along with some important polynomial theorems are as follows: Property 1: Division Algorithm. a + b. Multiply: If you missed this problem, review . b) Write the equation in vertex form. FACTORING QUADRATIC TRINOMIALS Example 4 : X2 + 11X - 26 Step 2 Factor the first term which is x2 (x )(x ) Step 4 Check the middle term (x + 13)(x - 2) 13x multiply 13 and x + -2x multiply -2 and x 11x Add the 2 terms. The numbers that multiply to – 50 and add to + 5 are – 5 and + 10. If you need a refresher on factoring quadratic equations, please visit this page. Each factor is a difference of squares! quadratic trinomial, independent term, coefficient, linear factor Simplify: ⓐ ⓑ If you missed this problem, review . Here is the form of a quadratic trinomial with argument x: ax 2 + bx + c. The argument is whatever is being squared. Let’s factor a quadratic form trinomial where a = 1. The term ‘a’ is referred to as the leading coefficient, while ‘c’ is referred to as the absolute term of f (x). These terms are in the form “axn” where “a” is a real number, “x” means to multiply, and “n” is a non-negative integer. Polynomials. Choose the correct … Australian Business Number 53 056 217 611. In Equation (i), the product of coefficient of y 2 and the constant term = ab and the coefficient of y = a+b = sum of the factors of … The … Solving Quadratic Equations by Factoring with a Leading Coefficient of 1 - Procedure (i) In a quadratic … If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Factorising an expression is to write it as a product of its factors. Solution: Check: Key Terms. It does not mean that a quadratic trinomial always turns into a quadratic equation when we equate it to zero. x is being squared. It is called "Factoring" because we find the factors (a factor is something we multiply by) Example: Multiplying (x+4) and (x−1) together (called Expanding ) gets x 2 + 3x − 4: So (x+4) and (x−1) are factors of x 2 + 3x − 4. There are 4 methods: common factor, difference of two squares, trinomial/quadratic expression and completing the square. Types of Quadratic Trinomials . 10 Surefire Video Examples! An example of a quadratic trinomial is 2x^2 + 6x + 4. Factors of Quadratic Trinomials of the Type x2 + bx + c. The Distributive Law is used in reverse to factorise a quadratic trinomial, as illustrated below. For example- 3x + 6x 2 – 2x 3 is a trinomial. To "Factor" (or "Factorise" in the UK) a Quadratic is to: find what to multiply to get the Quadratic . Show Step-by-step Solutions. 5, the coefficient of x, is the sum of 2 and 3. 6, the independent term, is the product of 2 and 3. In other words, if you have a trinomial with a constant term, and the larger exponent is double of the first exponent, the trinomial is in quadratic form. Tie together everything you learned about quadratic factorization in order to factor various quadratic expressions of any form. In the examples so far, all terms in the trinomial were positive. The product’s factor pair that when added yields the middle constant, –8 is –14 and 6. The middle term's coefficient is plus. Just as before, the first … However, this quadratic form polynomial is not completely factored. 1. How To Factorize Quadratic Expressions? = 2x2 + …  The last term, – 5, comes from the L, the last terms of the polynomials. Now here is a quadratic whose argument is x 3: 3x 6 + 2x 3 − 1. x 6 is the square of x 3. Example 12. To see the answer, pass your mouse over the colored area. Hence, the given trinomial is factorable. Study Materials. 1) x2 − 7x − 18 2) p2 − 5p − 14 3) m2 − 9m + 8 4) x2 − 16 x + 63 5) 7x2 − 31 x − 20 6) 7k2 + 9k 7) 7x2 − 45 x − 28 8) 2b2 + 17 b + 21 9) 5p2 − p − 18 10) 28 n4 + 16 n3 − 80 n2-1-©4 f2x0 R1D2c TKNuit 8aY ASXoqfyt GwfacrYed fL KL vC6. This is true, of course, when we solve a quadratic equation by completing the square too. Start from finding the factors of +2. Factor a Quadratic Trinomial. Let’s see another example, here where a is not one. If sum of the terms is the middle term in the given quadratic trinomial then the factors are correct. The polynomial root is a number where the polynomial becomes zero; in other words, a number that, by replacing it with x in the polynomial … X2 + 14x + ____ Find the constant term by squaring half the coefficient of the linear term. Solution. It’s really all about the exponents, you’ll see. Algebra tiles are a perfect way to introduce and practice this concept. Think of a pair of numbers whose sum is the coefficient of the middle term, +3, and whose product is the last term, +2. | Year 7 Maths Software | Year 8 Maths Software | Year 9 Maths Software | Year 10 Maths Software | Facebook Tweet Pin Shares 156 // Last Updated: January 20, 2020 - Watch Video // This lesson is all about Quadratic Polynomials in standard form. Don't worry about the difference, though; the book's title means the same thing as what this lesson explains.) (14/2)2 X2 + 14x + 49 Perfect Square Trinomials Create perfect square trinomials. x is called the argument. If you know how to factor a quadratic expression, then you can factor a trinomial in quadratic form without issue. If the quadratic function is set equal to zero, then the result is a quadratic equation.The solutions to the univariate equation are called the roots of the univariate function.. a, b, c are called constants. Example 7: Factor the trinomial 4x^2-8x-21 as a product of two binomials. This video provides a formula that will help to do so. Algebra - More on Factoring Trinomials Algebra - … This is a quadratic form polynomial because the second term’s variable, x3, squared is the first term’s variable, x6 . Answer: (x + 13)(x - 2) Step 1 Write 2 parenthesis. The are many methods of factorizing quadratic equations. x is called the argument. An example of a quadratic polynomial is given in the image. The argument appears in the middle term. The above trinomial examples are the examples with one variable only, let's take a few more trinomial examples with multiple variables. Solution. To factorise a quadratic trinomial, find two numbers whose sum is equal to the coefficient of x, and whose product is equal to the independent term. … Once the … Factoring Quadratic Expressions Date_____ Period____ Factor each completely. Factoring quadratic is an approach to find the roots of a quadratic equation. I. Example 5: Consider the quadratic relation y = 3 x 2 − 6 x − 24. a) Write the equation in factored form. | Homework Software | Tutor Software | Maths Software Platform | Trial Maths Software | The degree of a quadratic trinomial must be '2'. In other words, there must be an exponent of '2' and that exponent must be the greatest exponent. Some examples are: x 2 + 3x - 3 = 0 4x 2 + 9 = 0 (Where b = 0) x 2 + 5x = 0 (where c = 0) One way to solve a quadratic equation is by factoring the trinomial. Simplify: ⓐ ⓑ If you … This form is factored as: + + = (+) (+), where + = ⋅ =. 0 Comment. This is a quadratic form trinomial because the last term is constant (not multiplied by x), and (x5)2 = x10. THE QUADRATIC FORMULA FACTORING -Every quadratic equation has two values of the unknown variable usually known as the roots of the equation (α, β). Trinomials – An expressions with three unlike terms, is called as trinomials hence the name “Tri”nomial. Just to be sure, let us check: (x+4)(x−1) = x(x−1) + 4(x−1) = x 2 − x + 4x − 4 = x 2 + 3x − 4 . Divide each term by 4 to get; x 2 – 3x – 4 = 0 (x – 4) (x + 1) = 0 x = 4 or x = -1. Solve Quadratic Equations of the Form x 2 + bx + c = 0 by Completing the Square. Worked out Examples; 1.Solving quadratic equations by factoring: i) What is factoring the quadratic equation? To see the answer, pass your mouse over the colored area. For … Here is a look at the tiles in this post: In my set of algebra tiles, the same-size tiles are double-sided with + on one side and - on the other. If you're seeing this message, it means we're having trouble loading external resources on our website. The expressions $$x^2 + 2x + 3$$, $$5x^4 - 4x^2 +1$$ and $$7y - \sqrt{3} - y^2$$ are trinomial examples. NCERT Exemplar Class 10 Maths Chapter 2 Polynomials. Non-Example: These trinomials are not examples of quadratic form. And not all quadratics have three terms. But a "trinomial" is any three-term polynomial, which may not be a quadratic (that is, a degree-two) polynomial. Here’s an example: The first term, 2x2, comes from the product of the first terms of the binomials that multiply together to make this trinomial. Contents. One way to solve a quadratic equation is by factoring the trinomial. And not all quadratics have three terms. How to factor quadratic equations with no guessing and no trial and error? If sum of the terms is the middle term in the given quadratic trinomial then the factors are correct. Worked out Examples; 1.Solving quadratic equations by factoring: i) What is factoring the quadratic equation? The are many methods of factorizing quadratic equations. If you're seeing this message, it means we're having trouble loading external resources on our website. This is a quadratic form trinomial, it fits our form:  Here n = 2. A quadratic form polynomial is a polynomial of the following form: Before getting into all of the ugly notation, let’s briefly review how to factor quadratic equations. The Distributive Law is used in reverse to factorise a quadratic You get the same prices, service and shipping at no extra cost, but a small portion of your purchase price will go to help maintaining this site! Guess and check uses the factors of a and c as clues to the factorization of the quadratic. For example, w^2 + 7w + 8. Show Step-by-step Solutions. Quadratic trinomials. Solution (Detail) Think of a pair of numbers whose product is the last term, +2, and whose sum is the coefficient of the middle term, +3. For example, 2x²+7x+3=(2x+1)(x+3). Courses. A trinomial is a polynomial with 3 terms.. In this quadratic, 3x 2 + 2x − 1, the constants are 3, 2, −1. Some examples of quadratic trinomials are as... See full answer below. It might be factorable. Example 5: Consider the quadratic relation y = 3 x 2 − 6 x − 24. a) Write the equation in factored form. So the book's section or chapter title is, at best, a bit off-target. Here are examples of quadratic equations lacking the constant term or "c": x² - 7x = 0; 2x² + 8x = 0-x² - 9x = 0; x² + 2x = 0-6x² - 3x = 0-5x² + x = 0-12x² + 13x = 0; 11x² - 27x = 0; Here are examples of quadratic equation in factored form: (x + 2)(x - 3) = 0 [upon computing becomes x² -1x - 6 = 0] (x + 1)(x + 6) = 0 [upon computing becomes x² + 7x + 6 = 0] (x - 6)(x + 1) = 0 [upon computing becomes x² - 5x - … QUADRATIC EQUATION A quadratic equation is a polynomial of degree 2 or trinomial usually in the form of ax 2 + bx + c = 0. Example 6: A quadratic relation has an equation in factored form. Factor by making the leading term positive. FACTORING QUADRATIC TRINOMIALS Example 4 : X2 + 11X - 26 Step 2 Factor the first term which is x2 (x )(x ) Step 4 Check the middle term (x + 13)(x - 2) 13x multiply 13 and x + -2x multiply -2 and x 11x Add the 2 terms. Solve the following quadratic equation (2x – 3) 2 = 25. Since factoring can be thought of as un-distributing, let’s see where one of these quadratic form trinomials comes from. That is (4)(–21) = –84. In this article, our emphasis will be based on how to factor quadratic equations, in which the coefficient of x … To get a -5, the factors are opposite signs. For example, f (x) = 2x 2 - 3x + 15, g(y) = 3/2 y 2 - 4y + 11 are quadratic polynomials. Example Factor x 2 + 3x + 2. Cubic Polynomial. There are a lot of methods to factor these quadratic equations, but guess and check is perhaps the simplest and quickest once master, though mastery does take more practice than alternative methods. quadratic trinomial, linear Equation (i) is Simple Quadratic Polynomial expressed as Product of Two linear Factors and Equation (ii) is General Quadratic Polynomial expressed as Product of Two linear Factors Observing the two Formulas, leads us to the method of Factorization of Quadratic Expressions. Fast way the last term, – 5, the numbers that multiply –. Are − quadratic trinomial examples and 1 clues to the factorization of the 2nd.! Factored as: + + = ( + ) ( x ) = –84 3 ) 2 +! Square too of 2 and 2x 3 + bx + c, a bit.. Of These quadratic form without issue tutorial shows you how to factor a! Are three main ways of solving quadratic equations by factoring: i ) what is factoring the trinomial from! 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# Minimal polynomial of $x$ over $\mathbb{Q}(x+y+z, x^2+y^2+z^2, x^3+y^3+z^3)$

Let $$K$$ be the field of fractions of $$\mathbb{Q}[x,y,z]$$, $$\alpha=x+y+z,\quad \beta=x^2+y^2+z^2,\quad \gamma = x^3+y^3+z^3\in K$$ and $$M=\mathbb{Q}(\alpha, \beta, \gamma)$$.

1. Is $$x\in K$$ algebraic over $$M$$? What is the minimal polynomial?

2. What is the degree of transcendence of $$M$$ over $$\mathbb{Q}$$?

For part 1 I really have no idea. I tried some combinations but it didn't get me anywhere. Is there a systematic way of approaching this kind of problems?

For part 2 I'm guessing the degree is 3, but I'm not sure how to prove it.

• Please slow down on your questions, particularly given that that you're asking two back to back "I have no clue" questions. Mar 19, 2021 at 18:44
• Sorry, I have just been doing some questions today and I couldn't figure out these two. Since I couldn't find any similar questions online, so I decided to post them here Mar 19, 2021 at 18:46
• Okay, but please try to add more relevant context: what have you most recently covered? Do you fully understand the terminology, e.g., degree of transcendence? Try to include anything that might be relevant, even if you cannot yet connect it. We just want to see you as invested as we are in helping you. Mar 19, 2021 at 18:49
• If the degree is $3$ can you find a cubic with the roots $x, y, z$? Mar 19, 2021 at 18:51

Consider the polynomial $$g(t)=(t-x)(t-y)(t-z)$$ in $$\mathbb{Q}(x,y,z)[t]$$. Expanding it out yields \begin{align} g(t)&=t^3-(x+y+z)t^2 +\\ &=(yz+xz+xy)t-xyz. \end{align} Since $$x$$ is certainly a root of $$g$$, to show that $$x$$ is algebraic over $$\mathbb{Q}(\alpha,\beta,\gamma)$$ it suffices to show that each of the coefficients of $$g$$ lies in $$\mathbb{Q}(\alpha,\beta,\gamma)$$. To see this, note:

• $$(x+y+z)=\alpha\in\mathbb{Q}(\alpha,\beta,\gamma)$$.
• $$\alpha^2=\beta+2(yz+xz+xy)$$, so $$(yz+xz+xy)=(\alpha^2-\beta)\big/2\in\mathbb{Q}(\alpha,\beta,\gamma)$$.
• $$\alpha^3=3\alpha\beta-2\gamma+6xyz$$, so $$xyz=(\alpha^3-3\alpha\beta+2\gamma)\big/6\in\mathbb{Q}(\alpha,\beta,\gamma)$$.

This shows part $$1$$ of the problem, and in fact the analogous result holds for $$\mathbb{Q}(x_1,\dots,x_n)$$ for any $$n\in\mathbb{N}$$. For more on this I recommend reading about symmetric polynomials.

Now, to compute $$\operatorname{tr}.\operatorname{deg}_\mathbb{Q}\mathbb{Q}(\alpha,\beta,\gamma)$$, note that $$g(t)$$ witnesses that all of the generators of $$\mathbb{Q}(x,y,z)$$, and hence the field $$\mathbb{Q}(x,y,z)$$ itself, are algebraic over $$\mathbb{Q}(\alpha,\beta,\gamma)$$. What can you conclude about the respective transcendence degrees of $$\mathbb{Q}(\alpha,\beta,\gamma)$$ and $$\mathbb{Q}(x,y,z)$$ over $$\mathbb{Q}$$? Answer below, but try to figure it out yourself first!

Suppose $$S=\{m_1,\dots,m_k\}\subset\mathbb{Q}(\alpha,\beta,\gamma)$$ is a transcendence basis for $$\mathbb{Q}(\alpha,\beta,\gamma)$$ over $$\mathbb{Q}$$. By definition this means (i) there is no polynomial in $$\mathbb{Q}[t_1,\dots,t_k]$$ satisfied by $$S$$, and (ii) $$\mathbb{Q}(\alpha,\beta,\gamma)$$ is algebraic over $$\mathbb{Q}(S)$$. Now we claim that $$S$$ is in fact a transcendence basis for $$\mathbb{Q}(x,y,z)$$. Indeed, condition (i) holds immediately, and condition (ii) follows from the fact that an algebraic extension of an algebraic extension is algebraic; we have shown above that $$\mathbb{Q}(x,y,z)$$ is algebraic over $$\mathbb{Q}(\alpha,\beta,\gamma)$$, so it follows that $$\mathbb{Q}(x,y,z)$$ is algebraic over $$\mathbb{Q}(S)$$, as desired. In particular, we have $$\operatorname{tr}.\operatorname{deg}_\mathbb{Q}\mathbb{Q}(\alpha,\beta,\gamma)=\operatorname{tr}.\operatorname{deg}_\mathbb{Q}\mathbb{Q}(x,y,z)=3,$$ just as you suspected. Note that this argument works much more generally, and we have, for any triple of fields $$E\subseteq F\subseteq G$$, if $$G$$ is algebraic over $$F$$, then $$\operatorname{tr}.\operatorname{deg}_E F=\operatorname{tr}.\operatorname{deg}_E G.$$

• Nice answer and crystal clear! Thank you :) Mar 19, 2021 at 19:03
• @14159 my pleasure, happy it helped!! :) Mar 19, 2021 at 19:05


# Finding probability functions

• Nov 12th 2011, 10:27 PM
deezy
Finding probability functions
Four fair coins are tossed simultaneously. Find the probability of the random variable X = number of heads and compute the following probabilities:
a) obtaining no heads
b) precisely 1 head
c) at least 1 head
d) not more than 3 heads.

I'm not sure how to set up these probabilities.

For example problems, the book has $f(x) = \begin{pmatrix}n\\ x \end{pmatrix}p^xq^{n-x}$ for a binomial distribution and

$f(x) = \begin{pmatrix}n\\ x\end{pmatrix}(\frac{1}{2})^n$ for a symmetric case. There are other probability functions that they list as well.

My biggest problem is finding out the proper way to write these probabilities out, or what probability function can be used (if any).

For example, in a), I can logically see that it would be (1/2)*(1/2)*(1/2)*(1/2), but how can I write this as a probability function?
• Nov 12th 2011, 10:48 PM
CaptainBlack
Re: Finding probability functions
Quote:

Originally Posted by deezy
Four fair coins are tossed simultaneously. Find the probability of the random variable X = number of heads and compute the following probabilities:
a) obtaining no heads
b) precisely 1 head
c) at least 1 head
d) not more than 3 heads.

I'm not sure how to set up these probabilities.

For example problems, the book has $f(x) = \begin{pmatrix}n\\ x \end{pmatrix}p^xq^{n-x}$ for a binomial distribution and

$f(x) = \begin{pmatrix}n\\ x\end{pmatrix}(\frac{1}{2})^n$ for a symmetric case. There are other probability functions that they list as well.

My biggest problem is finding out the proper way to write these probabilities out, or what probability function can be used (if any).

For example, in a), I can logically see that it would be (1/2)*(1/2)*(1/2)*(1/2), but hibution B(ow can I write this as a probability function?

The number of heads has a binomial distribution $\text{B}(4,\ 0.5)$, so the probability of $r$ heads is:

$p(numb\_ heads=r)=b(n;4,\ 0.5)= \begin{pmatrix}4\\ r \end{pmatrix}(0.5)^r (0.5)^{4-r}=\frac{4!}{(4-r)! r!}(0.5)^4$

So if $r=0$ we have:

$p(numb\_ heads=0)=\frac{4!}{(4-0)! 0!}(0.5)^4=(0.5)^4$

CB
• Nov 13th 2011, 11:19 AM
deezy
Re: Finding probability functions
Not sure how to write c).

I've tried doing the probability of 1 head, 2 heads, 3 heads, and 4 heads separately and adding/multiplying them together.

Is it correct to do these probabilities separately?
• Nov 13th 2011, 11:53 AM
Plato
Re: Finding probability functions
Quote:

Originally Posted by deezy
Not sure how to write c).

At least one is the complement of none.
$1-P(\text{none}).$


1. ## Induction Proof Problem

I'm trying to prove using induction that 5^n-2^n is divisible by 3 for all values of n. I'm used to using induction to get k+1 within an expression that verifies a given formula, and have never used it to try and get to an answer divisible by 3.

So far I have 1 in the truth set because 5^1-2^1=5-2=3
And with a value k in the truth set
5^(k+1)-2^(k+1)=5(5^k)-2(2^k)
but don't know how i can further manipulate this or if i'm even on the right track.

If anyone could show me where to go from here i would be very grateful.

2. Originally Posted by uconn711
I'm trying to prove using induction that 5^n-2^n is divisible by 3 for all values of n. I'm used to using induction to get k+1 within an expression that verifies a given formula, and have never used it to try and get to an answer divisible by 3.

So far I have 1 in the truth set because 5^1-2^1=5-2=3
And with a value k in the truth set
5^(k+1)-2^(k+1)=5(5^k)-2(2^k)
but don't know how i can further manipulate this or if i'm even on the right track.
Start by writing down what it means to say that k is in the truth set: 5^k-2^k is a multiple of 3, or in other words 5^k-2^k=3p, for some integer p.

Now you can write that as 5^k = 2^k +3p. Substitute that expression for 5^k into the right-hand side of the equation 5^(k+1)-2^(k+1)=5(5^k)-2(2^k), and see where that leads you.

3. Hello, uconn711!

Use induction to prove: . $5^n-2^n$ is divisible by 3 for all values of $n.$

Verify $S(1)\!:\;\;5^1 - 2^1 \:=\:3$ . . . True!

Assume $S(k)$ is true: . $5^k - 2^k \:=\:3a$ .for some integer $a$.

Add $4\!\cdot\!5^k - 2^k$ to both sides;

. . $\underbrace{5^k + 4\!\cdot\!5^k} - \underbrace{2^k - 2^k} \:=\:3a + 4\!\cdot\!5^k - 2^k$
. . $(1 + 4)\!\cdot\!5^k - 2\!\cdot\!2^k \;=\;3a + \overbrace{3\!\cdot\!5^k + 5^k} - 2^k$

. . $5\!\cdot\!5^k - 2^{k+1} \;=\;3\left[a + 5^k\right] + \underbrace{5^k - 2^k}_{\text{This is }3a}$

Hence: . $5^{k+1} - 2^{k+1} \;=\;3\left[a + 5^k\right] + 3a \;=\;3\left[2a + 5^k\right]$ . . . a multiple of 3

Therefore: . $S(k+1)$ is true.
. . The inductive proof is complete.

4. Ah, I had to think through that a couple times, but that method actually makes alot of sense; thanks alot!

5. I like this approach.
If $5^N - 2^N = 3m$ then take a look at
$\begin{array}{rcl}
5^{N + 1} - 2^{N + 1} & = & 5^{N + 1} - 5\left( {2^N } \right) + 5\left( {2^N } \right) - 2^{N + 1} \\
& = & 5\left( {\underbrace {5^N - 2^N }_{3m}} \right) + 2^N \left( {\underbrace {5 - 2}_3} \right) \\
\end{array}

$


# How many triangular numbers have exactly $d$ divisors?

The triangular numbers $$T_n$$ are defined by $$T_n = \frac{n(n + 1)}{2}.$$

Given a positive integer $$d$$, how many triangular numbers have exactly $$d$$ divisors, and how often do such numbers occur?

For $$d = 4, 8$$ the answer seems to be "infinitely many, and often"; for $$d = 6$$, it seems to be "infinitely many, but rarely"; and for $$d \geq 3$$ prime the answer is "none" (I think I can prove this). Given $$d$$ such that there are infinitely many such triangular numbers, can we say anything about the asymptotic gaps between them?

Here is a plot of the number of divisors of $$T_n$$ as $$n$$ ranges from $$0$$ to $$50,000$$:

The OEIS contains some sequences related to this question, namely A292989 and A068443, but I can't learn enough from the comments there to settle this question for arbitrary $$d$$.

Edit: The "none" claim for prime $$d$$ only holds when $$d > 2$$, as @BarryCipra pointed out.

• An interesting question to ask is are there any number with odd number of divisors. One example is ${{1681\times 1682} \over 2} = 1413721$ which is a square number itself and has $9$ divisors. I think this might actually be the only example. – cr001 Feb 7 '19 at 16:07
• @cr001: The only numbers with an odd number of divisors are square numbers. The triangular numbers which are also square are given at oeis.org/A001110 – Michael Lugo Feb 7 '19 at 16:20
• @cr001 There is also $n = 8$ and $n = 49$, in addition to your numbers. I agree with you and suspect that each odd number of divisors has only finitely many examples. – Robert D-B Feb 7 '19 at 16:21
• See oeis.org/A063440 . The comments there give conditions for $\sigma_0(T_n) = 4$ and $\sigma_0(T_n) = 6$. The conditions for 4 seem "easier" than the conditions for 6, although I'm having trouble making this precise. – Michael Lugo Feb 7 '19 at 16:29
• Let $a(n) = \sigma_0(T_n)$. It seems more generally true that $a(n)$ is usually a multiple of 4, from the data at A063440. The comments at A063440 give $a(2k) = \sigma_0(k) \sigma_0(2k+1)$ and $a(2k+1) = \sigma_0(2k+1) \sigma_0(k+1)$. The function $\sigma_0$ takes even values except when its argument is square, so $a(n)$ is almost always a mulitple of 4. – Michael Lugo Feb 7 '19 at 16:36

This is a partial answer. Write $$\sigma_0(k)$$ for the number of divisors of $$k$$. Note that $$n$$ and $$n+1$$ are relatively prime. If $$\sigma_0(T_n)$$ is odd, then either $$n$$ is even and both $$\frac{n}{2}$$ and $$n+1$$ are squares or $$n$$ is odd and both $$n$$ and $$\frac{n+1}{2}$$ are squares (note that in particular this implies that in the first case $$n\equiv 0\mod{8}$$ and in the second $$n\equiv 1\mod{8}$$). Simplifying, one sees that odd values of $$\sigma_0(T_n)$$ arise from solutions to the Pell equation $$a^2-2b^2 = \pm 1.$$ So there are an infinite number of $$n$$ for which $$\sigma_0(T_n)$$ is odd. However, since $$\sigma_0$$ is multiplicative, $$\sigma_0(T_n)$$ cannot be prime unless $$n=2$$.
Next, note that $$\sigma_0(T_n)=4$$ means that either $$n$$ is even and $$\sigma_0\left(\frac{n}{2}\right) = \sigma_0(n+1) = 2$$, or $$n$$ is odd and $$\sigma_0(n) = \sigma_0\left(\frac{n+1}{2}\right) = 2$$. Thus $$\sigma_0(T_n)=4$$ if and only if either $$\frac{n}{2}$$ and $$n+1$$ are both prime or if $$n$$ and $$\frac{n+1}{2}$$ are both prime. The first of these is A005097; the second is A006254.
A similar analysis shows that $$\sigma_0(T_n)=6$$ requires that one of the two factors (i.e., either $$\frac{n}{2}$$ and $$n+1$$, or $$n$$ and $$\frac{n+1}{2}$$) be prime and the other be the square of a prime, so the values of $$n$$ below $$200$$ are $$n=7, 9, 17, 18, 25, 97, 121$$. These are presumably rarer than the values for $$\sigma_0(T_n)=4$$.
In response to the OP's comment below, for fixed odd $$d$$, both factors must be squares in order that $$\sigma_0$$ be odd for each. If $$T_n = 2\prod p_i^{2r_i}$$, then you are looking for a way to write $$\prod p_i^{2r_i} = \prod p_i^{2s_i}\prod p_i^{2t_i}$$ such that $$\prod(s_i+1)\prod(t_i+1) = d$$. This doesn't seem like a problem with a straightforward solution in general.
• I see that this implies that there are infinitely many odd $d$ for which we can find $n$ that works. Does this method tell us anything about fixed $d$? – Robert D-B Feb 7 '19 at 16:26
• After some searching, I suspect actually there might be infinitely many solution for at least $d=9$. Apparently given a pair of solutions $(x,y)=(41, 29)$ for example the next solution can be constructed using $(3x+4y, 2x+3y)$ and the iff condition for infinite $d=9$ is such pairs being both prime numbers infinitely many times. And this looks pretty much true to me. – cr001 Feb 7 '19 at 17:09