# Math Help - Rhombus Problem

1. ## Rhombus Problem

Given Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD

I know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12 but I don't think the diagonals of a rhombus are congruent.

2. ## Re: Rhombus Problem

Originally Posted by Cake
Given Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD

I know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12 but I don't think the diagonals of a rhombus are congruent.

$x^2+y^2=10^2=100$

$(10-x)^2+y^2=12^2=144$

$(10+x)^2+y^2=BD^2$

solving the first two equations we get $x=\frac{14}{5} and y=\frac{48}{5}$

plugging into the 3rd equation we get

$\left(10+\frac{14}{5} \right)^2+\left(\frac{48}{5} \right)^2=BD^2$

$\left(\frac{64}{5}\right)^2+\left(\frac{48}{5} \right)^2=\frac{6400}{25}=BD^2\Rightarrow BD=\frac{80}{5}=16$

3. ## Re: Rhombus Problem

Originally Posted by romsek

$x^2+y^2=10^2=100$

$(10-x)^2+y^2=12^2=144$

$(10+x)^2+y^2=BD^2$

solving the first two equations we get $x=\frac{14}{5} and y=\frac{48}{5}$

plugging into the 3rd equation we get

$\left(10+\frac{14}{5} \right)^2+\left(\frac{48}{5} \right)^2=BD^2$

$\left(\frac{64}{5}\right)^2+\left(\frac{48}{5} \right)^2=\frac{6400}{25}=BD^2\Rightarrow BD=\frac{80}{5}=16$
Holy cow! That's some math work! Thank you, is there any shorter way than this?

4. ## Re: Rhombus Problem

Originally Posted by Cake
Holy cow! That's some math work! Thank you, is there any shorter way than this?
Yes. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. Note that there are four congruent triangles formed by them.1/2 of AC =6. 1/2 of BD=8 5-4- 3 right triangle

5. ## Re: Rhombus Problem

Originally Posted by bjhopper
Yes. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. Note that there are four congruent triangles formed by them.1/2 of AC =6. 1/2 of BD=8 5-4- 3 right triangle
Right, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 6 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right?

6. ## Re: Rhombus Problem

Originally Posted by Cake
Right, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 6 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right?
hrm I should have seen this.

No what you do is notice that

$\left(\frac{BD}{2}\right)^2+\left(\frac{12}{2} \right)^2=10^2$

$\left(\frac{BD}{2}\right)^2=100-36=64 \Rightarrow \frac{BD}{2}=8 \Rightarrow BD=16$

7. ## Re: Rhombus Problem

Hello, Cake!

$\text{Given rhombus }ABCD,\;AB = 10\text{ and }AC = 12.\;\text{ Find }AD\text{ and }BD.$

I know that AD = 10 because the sides of a rhombus are all congruent.
I cannot find what BD equals though.
I thought it was 12, but I don't think the diagonals of a rhombus are congruent.
If they were, you'd have a square.

The diagonals of a rhombus are perpendicular and bisect each other.

Hence: . $AO = OC = 6.$

Code:
             A       10        B
o---------------o
/ *           * /
/   *6      *   /
/     *   *     /
10 /       o       /
/     *  O*     /
/   *       *6  /
/ *           * /
o---------------o
D                 C
In right triangle $AOB$, we find that $OB = 8.$

Therefore: . $BD = 16.$

8. ## Re: Rhombus Problem

Originally Posted by Soroban
Hello, Cake!

The diagonals of a rhombus are perpendicular and bisect each other.

Hence: . $AO = OC = 6.$

Code:
             A       10        B
o---------------o
/ *           * /
/   *6      *   /
/     *   *     /
10 /       o       /
/     *  O*     /
/   *       *6  /
/ *           * /
o---------------o
D                 C
In right triangle $AOB$, we find that $OB = 8.$

Therefore: . $BD = 16.$

Makes sense ^_____^
Thank you guys!


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# Solve this functional equation:

Functional equations such as this one appear only once every several years on exams, so I feel it's hard to have a sure-fire way to approach the problem, unlike, say, solving a series convergence problem, multiple variable integration, or proving some results using basic Fourier series.

So, when I do see a solution offered for one of these problems and study the solution for a substantial amount of time, I still cannot remember how to solve these types of problems, when I come across another one.

But the question is:

Find all the real-valued continuous functions $f$ on $\mathbb R$ which satisfy $$f(x)f(y)=f(x_1)f(y_1)$$

for all $x$, $y$, $x_1$, $y_1$ such that $x^2+y^2=x_1^2+y_1^2$.

Ideally, besides offering a solution, I would love to hear about your intuition on how to solve these functional equations.

Thanks,

• +1. I wanna know how the experts solve functional equations too.... – Jack's wasted life Jun 26 '15 at 22:56

You can linearize the problem by introducing $$g(u):=\ln\left(f(\sqrt u)\right).$$

Then with $u=x^2,v=y^2$, $$u+v=u'+v'\implies g(u)+g(v)=g(u')+g(v').$$

Setting $u'=0,v'=u+v$,

$$g(u)+g(v)=g(0)+g(u+v)$$

shows that the function must be affine,

$$g(u)=au+b,$$ and $$f(x)=e^{ax^2+b}=F_0\left(\frac{F_1}{F_0}\right)^{x^2}.$$

The intuitions/tricks behind this:

• it is often advantageous to linearize to benefit of what we know from linear algebra and make the equations look more familiar;
• products can be linearized by means of logarithms;
• non-linear functions can be linearized by means of a change of variable with the function inverse;
• when you have a property involving several variables, try to exploit it by assigning particular values to some of them.
• Hi @YvesDaoust - I really like this approach. I just have one follow-up question: how do you get the line g(u) = au+b, hence showing the function, g, is affine? – User001 Jun 29 '15 at 2:03
• From your previous line, it follows that g(u) = g(0) + g(u+v) - g(v)... – User001 Jun 29 '15 at 2:06
• @LebronJames: let $h(u)=g(u)-g(0)$, then $h(u)+h(v)=h(u+v)$. Assuming $h$ continuous, this is enough to say that $h$ is linear. – Yves Daoust Jun 29 '15 at 6:19

All solutions are the functions $f(x) = \alpha e^{\beta x^2}$, $\alpha,\beta \in \mathbb{R}$. Any of this functions satisfies the OP query: $$f(x)f(y) = \alpha e^{\beta x^2} \alpha e^{\beta y^2} = \alpha^2 e^{\beta(x^2 + y^2)} = \alpha^2 e^{\beta(x_1^2 + y_1^2)} = \alpha e^{\beta x_1^2} \alpha e^{\beta y_1^2} = f(x_1)f(y_1) \, .$$

Here is why these are the only ones. Observe that the hypothesis implies that there is a function $\psi$ such that $$f(x)f(y) = \psi(x^2 + y^2) \, .$$ If $f(0) = 0$ then $0.f(y) = \psi(y^2) = 0$ hence $f(x).f(y) \equiv 0$. So $f$ must be identically zero. Assume that $f(0) = \alpha \neq 0$. Then $\tilde{f}(x) := \frac{f(x)}{\alpha}$ also satisfies the OP hypotesis. So we can assume w.l.o.g. that $\alpha = 1$. From this we get that $f(x) = \psi(x^2)$ and $\psi(r)$ is a continuous function for $r \geq 0$. Moreover the function $\psi$ satisfies $$f(x) f(y) = \psi(x^2) \psi(y^2) = \psi(x^2 + y^2) \, .$$ By taking $x=y$ we see that $\psi \geq 0$. Actually, $\psi(x) > 0$. Indeed, if $\psi(x_0^2) = 0$ then $\psi(x_0^2 + r) = 0$ for $r \geq 0$. W.l.o.g. we can assume $x_0>0$. Observe that also $f(x_0)=0$. So there are values $y_0$ such that $f(y_0) = 0$ and $0 \leq y_0 < x_0$. But then also $\psi(y_0^2) = 0$. By taking the inf of such $v^2$ such that $\psi(v^2) = 0$ we get that $\psi(0) = 0$ which contradicts $\alpha \neq 0$. Finally, we can take logarithms. Namely, we define the function $\lambda(x) := \log(\psi(x))$, for $x \geq 0$. Then $$\lambda(x) + \lambda(y) = \lambda(x+y)$$ for all $x,y \geq 0$. Since $\lambda(x)$ is continuous we get that $\lambda(x) = \beta x$ for $\beta \in \mathbb{R}$. Then $\psi(x^2) = e^{\beta x^2}$. So $f(x) = e^{\beta x^2}$. But we had assumed that $f(0)=1$. So the general solution is as I claimed : $f(x) = \alpha e^{\beta x^2}$.

• why are we able to assert that $f(x)f(y) = \psi(x^2 + y^2)$? – Matematleta Jun 27 '15 at 1:30
• We know that $f(x)f(y) = f(\sqrt{x^2+y^2})f(0)$ for all $x,y$. So, define $\psi(r) = f(\sqrt{r})f(0)$ for $r \ge 0$. – JimmyK4542 Jun 27 '15 at 7:11
• @Chilango. $f(x)f(y)$ has the same value for all $x,y$ such that $x^2 + y^2$ have a fix value. Then $f(x)f(y)$ is a function of $x^2 + y^2$. – Holonomia Jun 27 '15 at 8:05
• Thanks so much @Holonomia. – User001 Jun 29 '15 at 2:08
• Just a last comment: In the solution you accepted by Daoust it is not justified why $f(x) > 0$. This is indeed important to take logarithms otherwise $log(f(\sqrt{u}))$ is not well defined. – Holonomia Jun 29 '15 at 5:33

The aim of the following is to address the intuition side of the question - I doubt that I have more experience than anybody else, and the following is certainly not rigorous - still... One way to the answer - at least in this case! - is to use calculus, i.e., assume everything in sight is differentiable, and perhaps try "to sweep up the loose ends" afterwards.

As pointed out by Holonomia, the function $g(x,y) = f(x)f(y)$ is constant on circles. This means that the gradient of the differentiable $g$ is parallel to the vector $(2x,2y)$, as the latter is normal to $x^2 +y^2 = c$. So $${\rm grad}\ g = \lambda\cdot (2x, 2y),$$ where where $\lambda = \lambda(x,y)$ is a scalar function. Comparing the components, one gets $$f'(x) f(y) = \lambda 2 x,$$ and $$f'(y) f(x) = \lambda 2 y.$$

Doing the algebra (formally), one obtains $${f'(x) \over 2x f(x)} = {f'(y) \over 2y f(y)}.$$ Therefore, both sides of the equality are constant, i.e., $${f'(x) \over 2x f(x)} = \beta,$$ with $\beta$ some constant. Cross-multiplying by $2x$ and integrating, one ends up with $$f(x) = \alpha e^{\beta x^2},$$ for some constant $\alpha$ - i.e., Holonomia's answer.

• Edit * Some "sweeping up," by request, to show that any $f$ satisfying the conditions of the problem (continuity, functional equation) is differentiable at $x=c$, for every $c$. Fix $c$, set $a = |c|$, and consider $$f(x) \int_{a+10}^{a+20} f(y)\, dy = \int_{a+10}^{a+20} f(x) f(y) \,dy = \int_{a+10}^{a+20} \psi( x^2 +y^2) \, dy,$$ where $\psi$ is Holonomia's $\psi$. The integral multiplying $f(x)$ is not zero if $f$ is not identically zero (using a Holonomia-style argument and $f(\sqrt 2 x)f(0) = f(x)^2$, for instance, to conclude that the continuous $f$ is nowhere $0$ if not identically $0$). With the change of variables $y= \sqrt{r^2-x^2}$, the integral on the right becomes $$\int_{\sqrt {(a+10)^2 +x^2}}^{\sqrt {(a+20)^2 +x^2}} \psi ( r^2) {r\over \sqrt{r^2-x^2} }\, dr,$$ which is differentiable at (in a neighborhood of) $x=c$, because $r^2-x^2 \ge (a+10)^2 >0$, $\psi$ is continuous, and the limits of the integral are differentiable. Thus $f(x)$ is differentiable.
• Hi @peterag, how do we then back out of the differentiability assumption? Really cool intuition... – User001 Jun 29 '15 at 1:44
• @LebronJames - actually, I just rolled back the edit to address your comment, as I messed something up - Tomorrow... – peter a g Jun 29 '15 at 3:47
• Ok, got it. Thanks @peterag! :-) – User001 Jun 29 '15 at 3:52
• @LebronJames see the 'edit'. However, as advertised in the first line of this answer, the point really was not 'rigor' - although the first version of this when you posed the question actually had a <problematic> version of this 'edit', but it seemed pointless given Holonomia's answer. Be that as it may, the argument in the edit is based on the standard proof that a measurable character on the reals is continuous (and differentiable). – peter a g Jun 29 '15 at 21:24


standard form of a quadratic function examples
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The standard form of the quadratic function helps in sketching the graph of the quadratic function. Two resistors are in parallel, like in this diagram: The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other. Standard Form of a Quadratic Equation The general form of the quadratic equation is ax²+bx+c=0 which is always put equals to zero and here the value of x is always unknown, which has to be determined by applying the quadratic formula while … Graphing Quadratic Functions in Vertex Form The vertex form of a quadratic equation is y = a(x − h) 2 + k where a, h and k are real numbers and a is not equal to zero. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.. The standard form of a quadratic function presents the function in the form $f\left(x\right)=a{\left(x-h\right)}^{2}+k$ where $\left(h,\text{ }k\right)$ is the vertex. The standard form of quadratic equations looks like the one below:. The constants ‘a’, ‘b’ and ‘c’ are called the coefficients. Quadratic functions follow the standard form: f(x) = ax 2 + bx + c. If ax 2 is not present, the function will be linear and not quadratic. Because (0, 8) is point on the parabola 2 units to the left of the axis of symmetry, x  =  2, (4, 8) will be a point on the parabola 2 units to the right of the axis of symmetry. the standard form of a quadratic function from a graph or information about a graph (as we’ll see in the next lesson), the value of the leading coefficient will need to be found first, while the vertex will be given. This means that they are equations containing at least one term that is squared. shows the profit, a company earns for selling items at different prices. Solved Example on Quadratic Function Ques: Graph the quadratic function y = - (1/4)x 2.Indicate whether the parabola opens up or down. Graphing a Quadratic Function in Standard Form. 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Therefore, the standard form of a quadratic equation can be written as: ax 2 + bx + c = 0 ; where x is an unknown variable, and a, b, c are constants with ‘a’ ≠ 0 (if a = 0, then it becomes a linear equation). What are the values of the two resistors? Let us solve this one by Completing the Square. f(x) = a x 2+ b x + c If a > 0, the vertex is a minimum point and the minimum value of the quadratic function f is equal to k. This minimum value occurs at x = h. If a < 0, the vertex is a maximum point and the maximum value of the quadratic function f is equal to k. This maximum value occurs at x = h. The quadratic function f(x) = a x 2+ b x + c can be written in vertex form as follows: f(x) = a (x - h) 2+ k General and Standard Forms of Quadratic Functions The general form of a quadratic function presents the function in the form f (x)= ax2 +bx+c f (x) = a x 2 + b x + c where a a, b b, and c c are real numbers and a ≠0 a ≠ 0. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. 1. y = x^{2} , y = 3x^{2} - 2x , y = 8x^{2} - 16x - 15 , y = 16x^{2} + 32x - 9 , y = 6x^{2} + 12x - 7 , y = \left ( x - 2 \right )^{2} . To find the roots of such equation, we use the formula, (root1,root2) = (-b ± √b 2-4ac)/2. Here, “a” is the coefficient of which is generally called as leading coefficient,“b” is the coefficient of “x” and the “c” is called as the constant term. Example 1. Find the roots of the equation as; (x + 2) … Using Vertex Form to Derive Standard Form. Quadratic functions in standard form: $$y=ax^2+bx+c$$ where $$x=-\frac{b}{2a}$$ is the value of $$x$$ in the vertex of the function. (Note: t is time in seconds). Examples of Quadratic Equations in Standard Form. In "Standard Form" it looks like: −5t 2 + 14t + 3 = 0. Quadratic equations pop up in many real world situations! The standard form of a quadratic equation: The standard form of a quadratic equation is given by It contains three terms with a decreasing power of “x”. Now we use our algebra skills to solve for "x". So, the selling price of $35 per item gives the maximum profit of$6,250. Substitute the value of h into the equation for x to find k, the y-coordinate of the vertex. But we want to know the maximum profit, don't we? y = a(x 2 - 2xh + h 2) + k. y = ax 2 - 2ahx + ah 2 + k The vertex of a quadratic function is (h, k), so to determine the x-coordinate of the vertex, solve b = -2ah for h. Because h is the x-coordinate of the vertex, we can use this value to find the y-value, k, of the vertex. Find the vertex of the quadratic function. Quadratic functions make a parabolic U-shape on a graph. Graph vertical compressions and stretches of quadratic functions. Answer: Boat's Speed = 10.39 km/h (to 2 decimal places), And so the upstream journey = 15 / (10.39â2) = 1.79 hours = 1 hour 47min, And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min. The quadratic equations refer to equations of the second degree. Write the vertex form of a quadratic function. Quadratic Equations are useful in many other areas: For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation. x2 − x − 6 < 0. Show Step-by-step Solutions Try the free Mathway calculator and problem solver below to practice various math topics. If a is negative, the parabola is flipped upside down. Write the vertex form of a quadratic function. And many questions involving time, distance and speed need quadratic equations. The vertex form of a quadratic equation is y = a (x − h) 2 + k where a, h and k are real numbers and a is not equal to zero. x = â0.39 makes no sense for this real world question, but x = 10.39 is just perfect! Area of steel after cutting out the 11 Ã 6 middle: The desired area of 28 is shown as a horizontal line. The standard form of quadratic equations looks like the one below:. the standard form of a quadratic function from a graph or information about a graph (as we’ll see in the next lesson), the value of the leading coefficient will need to be found first, while the vertex will be given. f (x)= a(x−h)2 +k f ( x) = a ( x − h) 2 + k. The Standard Form of a Quadratic Equation looks like this: 1. a, b and c are known values. Factoring Quadratic Functions. The "basic" parabola, y = x 2 , … Write the equation of a transformed quadratic function using the vertex form. It looks even better when we multiply all terms by −1: 5t 2 − 14t − 3 = 0. Examples of quadratic inequalities are: x 2 – 6x – 16 ≤ 0, 2x 2 – 11x + 12 > 0, x 2 + 4 > 0, x 2 – 3x + 2 ≤ 0 etc.. The squaring function f(x)=x2is a quadratic function whose graph follows. Quadratic functions are symmetric about a vertical axis of symmetry. Example : Graph the quadratic function : f(x) = x 2 - 4x + 8. ax² + bx + c = 0. Examples of Quadratic Equations in Standard Form. The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm2, The inside of the frame has to be 11 cm by 6 cm. Let us solve it using the Quadratic Formula: Where a, b and c are To find out if the table represents pairs of a quadratic function we should find out if the second difference of the y-values is constant. The following video shows how to use the method of Completing the Square to convert a quadratic function from standard form to vertex form. The x-axis shows the selling price and the y-axis shows the profit. 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Q)

# At 3.40, the hour hand and the minute hand of a clock form an angle of

( A ) 120
( B ) 135
( C ) 130
( D ) 125

Comment
A)

Comment
A)
130 is correct answer or not
Yes, that is correct

Comment
A)
Solution :
Angle traced by hour hand in 12 hrs $=360^{\circ}$
Angle traced by 1t in $\large\frac{11}{3}$$hrs = \bigg[\large\frac{360}{12} \times \frac{11}{3}\bigg] \qquad= 110^{\circ} Angle traced by minute hand in 60 mins =360^{\circ} Angle traced by it in 40 min =\bigg[\large\frac{360}{60}$$ \times 40 \bigg]$
$\quad= 240^{\circ}$
Required angle $=[240 -110]^{\circ}$
$\qquad= 130^{\circ}$


# Show why the value converges to $\pi$

$a_0=1$

$a_{n+1}=a_n+\sin{(a_n)}$

Explain why the following occurs:

$a_0=1$

$a_1=1+\sin{(1)}\approx 1.841470985$

$a_2=1+\sin{(1)}+\sin{(1+\sin{(1)})}\approx 2.805061709$

$a_3=1+\sin{(1)}+\sin{(1+\sin{(1)})}+\sin{(1+\sin{(1)}+\sin{(1+\sin{(1)})})}\approx 3.135276333$

$a_4\approx 3.141592612$

$a_5\approx 3.141592654\approx\pi$

Note by Jack Han
4 years, 11 months ago

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Let's see here...

The interesting thing I was talking about is the fact that the given series always converges into a value $a_n$ such that $sin(a_n)=0$ for different values of $a_0$. To put it in more exact terms, it always converges to a value of $a_n$ such that $cos(a_n)=-1\Rightarrow a_n=m\pi$, where $m$ is an odd integer. ($m$ can also be even, but that is a degenerate case where all terms are same)

I'm going to use something here that I actually learned from gradient descent. If you don't know what it is, you can google it. But, the mathematics used below is an extremely tame form and is easy to understand with little knowledge of calculus.

Consider the function $f\left( x \right) =\cos { (x) } \\ \Rightarrow \frac { df\left( x \right) }{ dx } =-\sin { (x) }$

Now see what happens when we take some arbitrary value of $x$ (say $x=1$)and then do the following repeatedly:

$x:=x-\frac { df\left( x \right) }{ dx }=x+\sin { (x) }$ (":=" is the assignment operator )

In the above figure, we can see two points marked. One is red, which represents the first value of $x$($=1$). The other is brown, and is after one iteration of above step.

We can see that when we do $x:=x+sin(x)$, what is actually happening is that $x$ is surfing along the slope of the curve $cos(x)$. We move the value of $x$ down the tangent. Change $x$ little by little, so that finally, after many iterations it moves closer and closer to the minima, i.e $x=\pi$.

I know this is not a definitive proof of what happens... I'm sure you will realize the importance of this once you understand what is happening.

In general, series defined as $a_n=a_{n-1}-\alpha\frac { df\left(a_{n-1} \right) }{ da_{n-1} }$

Will converge to the nearest value of $a$ (nearest to $a_0$) such that $f(a)$ is minimum, provided the value of $\alpha$ is not too large.

- 4 years, 11 months ago

Yes, that was awesome, that is basically the newton Rhapsody method of estimation of roots, doing the following iteration for any curve will eventually lead us to the nearest root, that is great, actually i think this is pretty much the solution +1

- 4 years, 11 months ago

Good work! Newton's method for estimating roots.

Pretty much seals the deal. Great solution +1.

- 4 years, 11 months ago

Good work.

- 4 years, 11 months ago

- 4 years, 11 months ago

As usual, since the series converges.. this means that when $n\to\infty$ ,

$a_{n+1}=a_{n}$. But $a_{n+1}=a_{n}+sin(a_{n})$

$\Rightarrow sin(a_{n})=0$

Now how do we know that $a_{n}=\pi$? We know this since $a_0=1$ and the series is constantly increasing. Therefore, it converges onto the first value of $x>1$ such that $sin(x)=0$.

- 4 years, 11 months ago

Bro , But It is not always true $\lim _{ n\rightarrow \infty }{ ({ a }_{ n }) } =\lim _{ n\rightarrow \infty }{ { (a }_{ n+1 }) }$ .

- 4 years, 11 months ago

Why not? Do you have a counter-example?

- 4 years, 11 months ago

Which value converges to pi ??

- 4 years, 11 months ago

If you call L the value of the limit you obtain sin(L)=0. Now L can be pi or zero but zero is impossible because of the initial condition. More precisely you can say that the value of the sequence is LOW bounded

- 4 years, 11 months ago

This is turning out to be very interesting...

I want to know if we can find a general form for a function $f(x)$ such that the series $a_1,a_2,...$ defined by:

$a_{n+1}=a_{n-1}+f(a_{n-1})$

Converges for a given value of $a_0$.

Further, is it true that all of the values of such $a_n$ as $n\to \infty$ satisfy $f(a_n)=0$?

- 4 years, 11 months ago

I think the answer to this is going to be extremely interesting.... I have a feeling... Is anyone else thinking what I'm thinking?

- 4 years, 11 months ago

What are you thinking?

- 4 years, 11 months ago


# In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digits must not be adjacent?

In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digit must not be adjacent?

I tried this by first taking total permutation as $\dfrac{8!}{2^4}$
Now $n_1$ as $22$ or $33$ or $44$ or $55$ occurs differently
$N_1 = \left(^7C_1\times \dfrac{7!}{8}\right)$
And $n_2 = \left(^4C_1 \times 4!\right)$
Using the inclusion-exclusion principle I got:
$\dfrac{8!}{16}-\left(^7C_1\times\dfrac{7!}{8}\right)+\left(^4C_1\times4!\right)$
This question is from combinatorics and helpful for RMO

• I made some edits to help the layout and appearance - see this linked article for more help on formatting - but I am not clear how you derived your formulas. Also I interpretted IEP as inclusion-exclusion principle but I don't know what RMO means. Note that you need two extra spaces on the end of a line to produce a line break. Aug 20 '17 at 2:26

You need a few more inclusion-exclusion steps to complete this approach.

Without constraints, you do indeed have $\dfrac {8!}{2^4} = 2520$ arrangements.

Then there are $\dfrac {7!}{2^3} = 630$ cases where a $22$ is found in the arrangement, and similarly for the other digits.

Then there are $\dfrac {6!}{2^2} = 180$ cases where both a $22$ and a $33$ are found, and similarly for other pairs, etc.

So by inclusion-exclusion, we have to subtract the paired cases then add back the double-paired cases, then subtract off triple-paired again and finally add in the cases where all digits appear in pairs.

$$\frac {8!}{2^4} - \binom 41\frac {7!}{2^3} + \binom 42\frac {6!}{2^2} - \binom 43\frac {5!}{2} + \binom 44\frac {4!}{1} \\[3ex] =2520 -4\cdot 630 +6\cdot 180-4\cdot60 + 24 = 864$$

[Sharp eyes might notice that $\frac {8!}{2^4} = \binom 41\frac {7!}{2^3}$, shortening the calculation process.]

Here is a variation based upon generating functions of Smirnov words. These are words with no equal consecutive characters. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.)

We encode the digits \begin{align*} 2,3,4,5 \qquad\text{as}\qquad a,b,c,d \end{align*} and look for Smirnov words of length $8$ built from $a,b,c,d$ with each letter occurring exactly twice.

A generating function for the number of Smirnov words over a four letter alphabet $V=\{a,b,c,d\}$ is given by \begin{align*} \left(1-\frac{4z}{1+z}\right)^{-1} \end{align*}

We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series $A(z)$. The number of all Smirnov words of length $8$ over a four letter alphabet is therefore \begin{align*} [z^8]\left(1-\frac{4z}{1+z}\right)^{-1} \end{align*}

Since we want to count the number of words of length $8$ with each character in $V$ occurring twice, we keep track of each character. We obtain with some help of Wolfram Alpha \begin{align*} [a^2b^2c^2d^2]\left(1-\frac{a}{1+a}-\frac{b}{1+b}-\frac{c}{1+c}-\frac{d}{1+d}\right)^{-1}=\color{blue}{864} \end{align*}

• (+1) This approach is so elegant and considerably malleable! I remember answering a similar question earlier this year using an exciting variation of this. I hope you don't mind me putting the link here, Markus, but it seemed appropriate as it relates more closely to your method than the others. Aug 21 '17 at 2:07

Here is another approach:

Assume for the moment that the first appearance of the four digits is in increasing order. The places of their first appearance can be distributed in six ways, see the following figure. The places for the prospective second appearances have been marked by empty boxes, next to which is written the number of choices we have when filling them in. The last column shows the product of these numbers in each row. $$\matrix{ 2&3&4&5&\square_3&\square_3&\square_2&\square_1&&18\cr 2&3&4&\square_2&5&\square_2&\square_2&\square_1&&8\cr 2&3&4&\square_2&\square_2&5&\square_1&\square_1&&4\cr 2&3&\square_1&4&5&\square_2&\square_2&\square_1&&4\cr 2&3&\square_1&4&\square_1&5&\square_1&\square_1&&1\cr 2&3&\square_1&\square_1&4&5&\square_1&\square_1&&1\cr}$$ Summing the last column gives $36$. This has to be multiplied by $4!$ in order to compensate for the chosen order $2345$. It follows that there are $864$ admissible arrangements of the eight digits.

• One of the charms of mathematics is that there is always another way.... Aug 20 '17 at 13:52


# With 4 rooks on a $4\times4$ chessboard such that no rook can attack another, what is the probability there are no rooks on the diagonal?

Four rooks are randomly placed on a $4 \times 4$ chessboard. Suppose no rook can attack another. Under this condition, what is the probability that the leading diagonal of the chessboard has no rooks at all?

Since no rook can attack another, we know that each row and each column contains exactly one rook each. Let $A_i$ be the event that row $i$ has its rook on the diagonal. Then $P\{A_i\} = \frac{1}{4}$ for each $i = 1,\dots,4$.

We want to find the probability that the diagonal of the chessboard has no rooks at all, or equivalently that none of the rows have their rook on the diagonal. Therefore we have

\begin{align} P\{A^c_1 \cap A^c_2 \cap A^c_3 \cap A^c_4\} & = P\{(A_1 \cup A_1 \cup A_3 \cup A_4)^c\} \\ & = 1 - P\{A_1 \cup A_1 \cup A_3 \cup A_4\} \\ & = 1 - (P\{A_1\} + P\{A_2\} + P\{A_3\} + P\{A_4\} - P\{A_1 \cap A_2\} - P\{A_1 \cap A_3\} - P\{A_1 \cap A_4\} - P\{A_2 \cap A_3\} - P\{A_2 \cap A_4\} - P\{A_3 \cap A_4\} + P\{A_1 \cap A_2 \cap A_3\} + P\{A_1 \cap A_2 \cap A_4\} + P\{A_1 \cap A_3 \cap A_4\} + P\{A_2 \cap A_3 \cap A_4\} - P\{A_1 \cap A_2 \cap A_3 \cap A_4\}) \\ & = 1 - (4 \cdot \frac{1}{4} - 6 \cdot \frac{1}{16} + 4 \cdot \frac{1}{64} - \frac{1}{256}) \\ & = 1 - \frac{175}{256} \\ & = \frac{81}{256} \end{align}

using De Morgan's Law and the inclusion-exclusion principle.

However, it seems that this is incorrect since if we consider the number of ways that we can place the rooks such that no rook can attack each other we have $\frac{(4!)^2}{4!} = 4! = 24$ [as per this answer for a similar problem] and so the answer should have denominator of 24. Having said that I don't see where my answer is wrong, so would someone be able to show me the correct solution?

• No rooks on either diagonal, or just one specified diagonal? Oct 31 '15 at 20:47
• Look at those 24 rook configurations. Is $P(A_i)=1/4$ ? Oct 31 '15 at 20:51
• Just the leading diagonal, i.e. a4, b3, c2, d1 @BrianTung Oct 31 '15 at 20:51
• OK, thanks. I've given the answer to both interpretations, just in case. Oct 31 '15 at 20:59

Each non-attacking placement of the rooks defines a permutation $c_1c_2c_3c_4$ of $\{1,2,3,4\}$: $c_k$ is the number of the column containing the rook in row $k$. There are $4!=24$ such permutations, all equally likely. Those that have no rook on the main diagonal are derangements, and there are $9$ of them, so the desired probability is $\frac9{24}$.

If you know the formula for the number of derangements of a set of $n$ objects, you can use it, but $4$ is small enough that it’s almost as easy just to list them:

\begin{align*} &2143,2341,2413\\ &3142,3412,3421\\ &4123,4312,4321 \end{align*}

Your answer assumes independence between rook placements. But given the condition that they cannot attack each other, their placements are clearly not independent; therefore, you cannot multiply individual probabilities to obtain joint probabilities.

You are correct in observing that the total number of possible non-attacking arrangements is $4! = 24$. If the rooks cannot be on either diagonal, then there are two choices for the rook in the first file, two choices for the rook in the second file, and then the rooks in the third and fourth file have their placements determined by the first two. There are therefore $2 \times 2 = 4$ placements that avoid both diagonals.

If you only need to avoid one diagonal (say, the black diagonal), we merely need the number of derangements of four objects. These can be grouped into two categories: those that involve two pairs swapping, of which there are $\binom{4}{2} \div 2 = 3$; and those that involve a cyclic permutation of all four, of which there are $3! = 6$; for a total of $9$ derangements.

The number of allowable placements should be the number of derangements of $4$ items, which is $9$. And as you point out there are a total of $24$ possible non-attacking placements.

You are misapplying Inclusion-Exclusion. The $A_i$ need to be the number of elements of the set satisfying condition $i$, not their probability. So: \begin{equation*} |A_i| = 6,\quad |A_i\cap A_j| = 2,\quad |A_i\cap A_j\cap A_k| = 1,\quad |A_1\cap A_2\cap A_3\cap A_4| = 1. \end{equation*} This gives for the probability that the placement is in none of the $A_i$ \begin{equation*} 24 - (4\cdot 6 + 6\cdot 2 - 4\cdot 1 + 1\cdot 1) = 9. \end{equation*} So the probability is $\frac{9}{24} = \frac{3}{8}$. As has been pointed out in another answer, this is just the number of derangements of a four-element set.


This video explains to graph graph horizontal and vertical stretches and compressions in the A point on the object gets further away from the vertical axis on the image. J. JonathanEyoon. x). 1. This problem has been solved! Embedded content, if any, are copyrights of their respective owners. Horizontal And Vertical Graph Stretches And Compressions (Part 1) The general formula is given as well as a few concrete examples. This graph has a vertical asymptote at $$x=–2$$ and has been vertically reflected. Retain the y-intercepts’ position. Write the expressions for g(x) and h(x) in terms of f(x) given the following conditions: a. Images/mathematical drawings are created with GeoGebra. 8. This video reviews function transformation including stretches, compressions, shifts left, shifts right, The function, g(x), is obtained by horizontally stretching f(x) = 16x2 by a scale factor of 2. When in its original state, it has a certain interior. Apply the transformations to graph g(x). Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. A horizontal stretch or shrink by a factor of 1/kmeans that the point (x, y) on the graph of f(x) is transformed to the point (x/k, y) on the graph of g(x). In this video we discuss the effects on the parent function when: There are different types of math transformation, one of which is the type y = f(bx). This is called a horizontal stretch. ... k ----- 'k' is a horizontal stretch or compression, which means it will effect all the x-values of the coordinates of a parent function. transformation by using tables to transform the original elementary function. So I don't want to change any scales or values or limits. Related Pages Translation Of 2 Units Left IV. Horizontally stretched by a scale factor of 1/3. See the answer. horizontal/vertical stretch? Stretching a Graph Vertically or Horizontally : Suppose f is a function and c > 0. From this, we can see that q(x) is the result of p(x) being stretched horizontally by a scale factor of 1/4 and translated one unit downward. The graph of $$y = f(0.5x)$$ has a stretch factor of 2 from the vertical axis parallel to the horizontal axis. But do not divide outside of the parenthesis, it remains close to the X. Use the graph of f(x) shown below to guide you. We welcome your feedback, comments and questions about this site or page. Horizontal and vertical translations, as well as reflections, are called rigid transformations because the shape of the basic graph is left unchanged, or rigid. Use the graph of f(x) shown below to guide you. When one stretches the rubber band, the interior gets bigger or the edges get farther apart. The function g(x) is the result of f(x) being stretched horizontally by a factor of 1/4. The new x-coordinate of the point will be, 1. You da real mvps! Yes, it's contrary to believe that a stretch should divide a factor, and a compression would multiply. Cosine of x would be the same as these, but shifted πb/2 to the left. If you're seeing this message, it means we're having trouble loading external resources on our website. We can also stretch and shrink the graph of a function. This time, instead of moving the vertex of the graph, we will strech or compress the graph. More Pre-Calculus Lessons. What are the transformations done on f(x) so that it results in g(x) = 3√(x/2)? This video provides two examples of how to express a horizontal stretch or compression using function notation.Site: http://mathispower4u.com The general formula is given as well as a few concrete examples. This shifted the graph down 1 unit. When f (x) is stretched horizontally to f (ax), multiply the x-coordinates by a. The image below shows the graph of f(x). :) https://www.patreon.com/patrickjmt !! 2f (x) is stretched in the y direction by a factor of 2, and f (x) is shrunk in the y direction by a factor of 2 (or stretched by a factor of ). 4. more examples, solutions and explanations. 5. I just didn’t know how to animate that with my program. This video talks about reflections around the X axis and Y axis. Viewed 28k times 15. The simplest way to consider this is that for every x you want to put into your equation, you must modify x before actually doing the substitution. Vertical Stretch and Vertical Compression y = af(x), a > 1, will stretch the graph f(x) vertically by a factor of a. y = af(x), 0 < a < 1, will stretch the graph f(x) vertically by a factor of a. Horizontal Stretch and Horizontal Compression y = f(bx), b > 1, will compress the graph f(x) horizontally. When using transformations to graph a function in the fewest steps, you can apply a and k together, and then c and d together. by horizontally stretching f(x) by a factor of 1/k. (Part 3). Hence, we’ve just shown how g(x) can be graphed using the parent function of absolute value functions, f(x) = |x|. Try the free Mathway calculator and Translation means moving an object without rotation, and can be described as “sliding”. 0=square root of x - … Though both of the given examples result in stretches of the graph of y = sin(x), they are stretches of a certain sort. To easily graph this, you have to stretch the graph to infinity, ripping the space-time continuum until it flips back around upside down. Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or compression. A point $\,(a,b)\,$ on the graph of $\,y=f(x)\,$ moves to a point $\,(k\,a,b)\,$ on the graph of [beautiful math coming... please be patient] $\,y=f(\frac{x}{k})\,$. Ask Question Asked 7 years ago. Try the given examples, or type in your own Teams. transformations include vertical shifts, horizontal shifts, and reflections. 7. Vertical stretch on a graph will pull the original graph outward by a given scale factor. In all seriousness, you flip your graph upside down. We carefully make a 90° angle around the third peg, so that one side is vertical and the other is horizontal. Lastly, let’s translate the graph one unit downward. Please submit your feedback or enquiries via our Feedback page. This type of The function, f(x), passes through the point (10, 8). Meaning, n(x) is the result of m(x) being vertically stretched by a scale factor of 3 and horizontally stretched by a scale factor of 1/4. problem and check your answer with the step-by-step explanations. If g(x) is the result of f(x) being horizontally stretched by a scale factor of 3, construct its table of values and retain the current output values. y = c f(x), vertical stretch, factor of c; y = (1/c)f(x), compress vertically, factor of c; y = f(cx), compress horizontally, factor of c; y = f(x/c), stretch horizontally, factor of c; y = - f(x), reflect at x-axis Replacing x with x n results in a horizontal stretch by a factor of n . and reflections across the x and y axes. Q&A for Work. Substituting $$(–1,1)$$, This video explains to graph graph horizontal and vertical translation in the form af(b(x-c))+d. This video discusses the horizontal stretching and compressing of graphs. problem solver below to practice various math topics. I want a simple x,y plot created with matplotlib stretched physically in x-direction. Scroll down the page for So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of $\frac{1}{4}$ in our function: $f\left(\frac{1}{4}x\right)$. Vertically stretched by a scale factor of 2. Thanks to all of you who support me on Patreon. b. Horizontal Stretching and Compression of Graphs This applet helps you explore the changes that occur to the graph of a function when its independent variable x is multiplied by a positive constant a (horizontal stretching or compression). Jul 2007 290 3. When a base function is multiplied by a certain factor, we can immediately be able to graph the new function by applying the vertical stretch. Horizontal Stretch and Shrink. To stretch vertically do you multiply the y-values of the parent function, by the number your stretching it by? Copyright © 2005, 2020 - OnlineMathLearning.com. It looks at how a and b affect the graph of f(x). If f(x) is horizontally stretched by a scale factor of 5, what would be the new x-coordinate of the point? This means that the input values must be four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching. It might be simpler to think of a stretch or a compression in terms of a rubber band. The graphs below summarize the key features of the resulting graphs of vertical stretches and compressions of logarithmic functions. The table of values for f(x) is shown below. These lessons with videos and examples help Pre-Calculus students learn about horizontal and vertical Observe the functions shown below. The function, g(x), is obtained by horizontally stretching f(x) = 16x, Horizontal Stretch – Properties, Graph, & Examples, Since the y-coordinates will remain the same, the, We can only horizontally stretch a graph by a factor of. Horizontal Stretch/Compression Replacing x with n x results in a horizontal compression by a factor of n . physically stretch plot in horizontal direction in python. When f(x) is stretched horizontally to f(ax). We know so far that the equation will have form: $$f(x)=−a\log(x+2)+k$$ It appears the graph passes through the points $$(–1,1)$$ and $$(2,–1)$$. We can only horizontally stretch a graph by a factor of 1/a when the input value is also increased by a. To perform a horizontal compression or stretch on a graph, instead of solving your equation for f(x), you solve it for f(c*x) for stretching or f(x/c) for compressing, where c is the stretch factor. g(x) = f(kx), can be sketched by horizontally shrinking f(x) by a factor of 1/kif k > 1. or. Let’s go ahead and express g(x) in terms of f(x). 2. Translation Of 2 Units Right O I And IV O II And III Oll And IV I And III. Take a look at the following graph. form af(b(x-c))+d. Lastly, let’s observe the translations done on p(x). In describing transformations of graphs, some textbooks use the formal term “translate”, while others use an informal term like “shift”.Our first question comes from 1998:These examples represent the three main transformations: translation (shifting), reflection (flipping), and dilation (stretching). Functions that are multiplied by a real number other than 1, depending on the real number, appear to be stretched vertically or stretched horizontally. How To: Given a logarithmic function Of the form $f\left(x\right)=a{\mathrm{log}}_{b}\left(x\right)$, $a>0$, graph the Stretch … Define functions g and h by g (x) = c f (x) and h (x) = f (cx). Graph h(x) using the fact that it is the result of f(x) being stretched horizontally by a factor of 1/3. This means that the translations on f(x) to obtain g(x) are: Let’s slowly apply these transformations on f(x) starting with horizontally stretching f(x). Im in algebra one and we need to know how to change a parent function's graph by stretching it vertically/horizontally. Expert Answer . This transformation type is formally called horizontal scaling (stretching/shrinking). Horizontal Stretch By A Factor Of 3 II. if 0 < k< 1. Parent Functions And Their Graphs Other important 6. The intention is to get a result were it is easier for me to detect features in the signal. You make horizontal changes by adding a number to or subtracting a number from the input variable x, or by multiplying x by some number. We do not know yet the vertical shift or the vertical stretch. Subtracting from x makes the function go right. Question: How Is The Graph Y =3(x - 2)2 Related To The Graph Of Y = 1. You start with y=square root of (x-1) it becomes 0<=x-1. The following table gives a summary of the Transformation Rules for Graphs. Apply the transformations to graph g(x). \$1 per month helps!! then 1 <=x. Show transcribed image text . Active 2 years ago. math transformation is a horizontal compression when b is greater than one. PLEASE give an easy way to stretch! Which of the following is the correct expression for g(x)? We can graph this math What is the relationship between f(x) and g(x)? Sal graphs y=-2.5*cos(1/3*x) by considering it as a vertical stretch and reflection, and a horizontal stretch, of y=cos(x). Now we stretch one part of the rubber band straight up from the left peg and around a third peg to make the sides of a right triangle as shown in Figure $$\PageIndex{2}$$. y = c f(x), vertical stretch, factor of c, y = (1/c)f(x), compress vertically, factor of c, y = f(cx), compress horizontally, factor of c, y = f(x/c), stretch horizontally, factor of c. Stretching a graph involves introducing a coefficient into the function, whether that coefficient fronts the equation as in y = 3 sin(x) or is acted upon by the trigonometric function, as in y = sin(3x). Describe the transformations done on the following functions shown below. 3. Horizontal Stretches/Compressions - multiply the x value directly. The resulting function will have the same range but may have a different domain. To stretch or shrink the graph in the y direction, multiply or divide the output by a constant. Let’s now stretch the resulting graph vertically by a scale factor of 2. All horizontal transformations, except reflection, work the opposite way you’d expect: Adding to x makes the function go left. It looks at how c and d affect the graph of f(x). What are the transformations done on f(x) so that it results to g(x) = 2|x/3| – 1? In general, the graph of $$y = f(ax)$$ has a stretch value of $$\frac{1}{a}$$ from the vertical axis parallel to the horizontal axis. if we say we stretched it by 1/4, that means it only increased by 1/4 of its original length as opposed to 4 times its original length . A horizontal stretch can be applied to a function by multiplying its input values by a scale factor, Let’s go ahead and take a look at how f(x) = x, Remember that when we horizontally stretch a function by, When we stretch a graph horizontally, we multiply the base function’s x-coordinate by the given scale factor’s denominator, Hence, we have (6, 4) → (2 ∙ 6, 4). You use the graph and solve it as you would for any function using small values first, then you have y=square root of x - 1, the domain 0<=x. When we horizontally stretch g(x) by a scale factor of 1/3, we obtain h(x). Transformations Of Trigonometric Graphs Then. graph stretches and compressions. the graph will be stretched horizontally so that its horizontal length on any finite interval will be 4 times what it was originally, stretching by a factor of 4 is the way we would describe that. For a horizontal stretch of 2, x 2 would become (x/2) 2. Vertical Stretch By A Factor Of 3 III. 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# If I flip $1$ of $3$ modified coins $3$ times, what's the probability that I will get tails?

We have $3$ modified coins: $M_1$ which has tails on the both sides, $M_2$ which has heads on the both sides and $M_3$ which is a fair coin. We extract a coin from the urn and we flip it $3$ times.

1. What is the probability that if I flip the coin $3$ times I will get all tails?
2. If I got all tails at all $3$ flips what is the probability that the extracted coin is $M_3$?

My attempt:

1. I have tried this way: There is a $\frac{1}{3}$ chance to get $M_1$ or $M_2$ or $M_3$. If we get $M_1$ the probability to get tails is $1$, for $M_2$ is $0$ and for $M_3$ is $\frac{1}{2}$. Then the probability to get tails at one flip is $$\frac{1}{3}\cdot 1 + \frac{1}{3}\cdot 0 + \frac{1}{3}\cdot \frac{1}{2} = \frac{1}{2}$$ So the probability to get tails at all the $3$ flips is ${(\frac{1}{2})}^3$ which is $\frac{1}{8}$. Is this right?

2. The probability seems to be intuitively $\frac{1}{3}$, but I don't know how to formally prove it.

By Bayes' theorem \begin{align}P(M_3\mid TTT)=\frac{P(TTT\mid M_3)P(M_3)}{P(TTT)}=\frac{\left(\frac12\right)^3\cdot\frac13}{\frac38}=\frac19\end{align}

Note: The denominator was calculated using the Law of total probability as is common when applying the Bayes rule. You did this in part 1. but not correctly. To see this write \begin{align}P(TTT)&=P(TTT\mid M_1)P(M_1)+P(TTT\mid M_2)P(M_2)+P(TTT\mid M_3)P(M_3)\\[0.2cm]&=1\cdot\frac13+0\cdot\frac13+\left(\frac12\right)^3\frac13\\[0.2cm]&=\frac13\left(1+\frac18\right)=\frac38\end{align}

• Why $(\frac{1}{2})^3$. Where did this come frome? So for each coin you computed the probability that we will get tails and for that probability the probability to get 3 tails in a row? Apr 2, 2016 at 16:40
• $P(TTT\mid M_3)=\left(\frac12\right)^3$ You roll a fair coin three times and you want three times tails, so $\frac12\cdot\frac12\cdot\frac12$. Apr 2, 2016 at 16:43
• Yes, exactly as you say it. Apr 2, 2016 at 16:44

Paint the double-head coin yellow on one side and red on the other.
Paint the double-tail coin blue and green.
There are 24 possible outcomes. One of the outcomes is:

Run through the 24 outcomes, how many of them give you three tails?
Of those outcomes, how many were with the fair coin, how many were with the double-tail coin?

I will extend my comment: remember that $\text{probability of A}=\frac{\text{number of A cases}}{\text{all possible cases}}$.

Then, how many ways we can get (tail, tail, tail)? If we take the fair coin with this coin we only can take (tail, tail, tail) i.e. only exist one way we can take the desired result.

But if we took the double-tail coin we take (tail, tail, tail) any time i.e. the full 8 ways that a coin can show when it is tossed three times.

And when we get the double-head coin we cant take (tail, tail, tail).

Then the total amount of ways we can take (tail, tail, tail) is just $1+8$, and the cases for the fair coin is just $1$ so the probability that you want is $1/9$.

This is a visual way to see the problem but the formal way to solve it is the answer of @JimmyR i.e. using the basic definitions and theorems of probability theory.

There are some correct answers here. Many use Bayes' rule, which is correct and elegant but takes getting used to. Let me try instead to help you think through this particular example, to train your intuition.

In your answer to #1 you correctly compute that the probability of one $T$ is 1/2. But that doesn't mean the probability of $TTT$ is 1/8 unless you put the coin back and choose independently again for each of the next two tosses. The way the problem is stated, you use the same coin all three times. Then the right way to compute the weighted average is $$\frac{1}{3}⋅1+ \frac{1}{3}⋅0+ \frac{1}{3}⋅\frac{1}{8}= \frac{3}{8}.$$

For the second question, you know that you don't have the middle coin, so you need the probability of the first compared to the last. If you imagine that you can tell the two sides of the two-tailed coin apart, there are 8 ways to do three flips, all of which are all tails. For the fair coin, only one triple is all tails. So when you see all tails the probability that you had the all-tail coin is 8/9.


The lack of closed form solution for the arc length of an elliptic arc led to the development of the elliptic integrals. Measurement by arc length Definition of arc length and formula to calculate it from the radius and central angle of the arc. Improve your skills with free problems in 'Arc measure and arc length' and thousands of other practice lessons. The red arc measures 120. Measurement by central angle . length of arc AB = (5/18)(2r) = (5/18)(2(18)) = 10. Any diameter of a circle cuts it into two equal semicircles. Yes, Arc Length and Circumference isnt particularly exciting. Section 11.1 Circumference and Arc Length 595 Using Arc Lengths to Find Measures Find each indicated measure. The blue arc measures 240. $Ans = 2\pi a$ How to obtain the ans? The area of a semicircle is half the area area of the circle from which it is made. Related Book. 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The length of arc is equal to radius The length of an arc of a circle which subtends an angle radian at the center is equal to r where r is the radius of the circle. Calculating a circle's arc length, central angle, and circumference are not just tasks, but essential skills for geometry, trigonometry and beyond. CHAPTER 5A Central Angles, Arc Length, and Sector Area ... for a central angle of a circle Calculate the arc length and the area of a sector formed by a 30 central The distance along the arc (part of the circumference of a circle, or of any curve). Question: Find the arc length of the circle given by $x^2+y^2=a^2$. It depends on the radius of a circle and the central angle. Geometry Teachers Never Spend Time Trying to Find Materials for Your Lessons Again! Relate the length of an arc to the circumference of a whole circle and the central angle subtended by the arc. The circumference of a circle is an arc measuring 360o. Step 1 : Here, radius = 7cm central angle= 30 degrees. We dare you to prove us wrong. How to Find Arc Length. ... the arc length of a circumscribed circle is: Arc length is a linear measure of the arc measured along the circle. Thus, the length of the arc AB will be 5/18 of the circumference of the circle, which equals 2r, according to the formula for circumference. The relationship of arc length to a ... as the length x of an arc of the unit circle. Join Our Geometry Teacher Community Today! * An alternative definition is that it is an open arc. a. arc length of AB b. circumference of Z c. m RS A connected section of the circumference of a circle. Here's how to calculate the circumference, radius, diameter, arc length and degrees, sector areas, inscribed angles, and other shapes of the circle. and l stand for arc length. Our arc length calculator can calculate the length of an arc of a circle and the area of a sector. Learn how to find the arc with our lesson and try out our examples questions. Geometry calculator solving for circle arc length given radius and central angle An arc is any portion of the circumference of a circle.http://www.mathwords.com/a/arc_circle.htm Arc length is the $Ans = 2\pi a$ How to obtain the ans? The arc length formula is used to find the length of an arc of a circle. To find the arc length, we now need to find the circumference It can be understood, that the arc length is a fraction of the circumference of the circle. But it can, at least, be enjoyable. The length of the circumference is given by the formula: C = d, where d is the diameter of the circle. I would like to calculate the arc length of a circle segment, i.e. The length of an arc is a connected section of the circumference of a circle. These curves are called rectifiable and the number is defined as the arc length. Use the formula C = d to calculate the circumference of a circle when the diameter is given. Question: Find the arc length of the circle given by $x^2+y^2=a^2$. Arc length of a circle is the distance measured as the length. For a circle: Arc Length = r (when is in radians) Fun math practice! I have no ideas after doing the following thing. Formula is S = r. See note at end of page. How to Determine the Length of an Arc. I have no ideas after doing the following thing. This formula can also be given as: C = 2r, where r is the radius. Step 2 : Calculation of an arc length without its central angle is a tough problem since the arc length is based on the angle.
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Looking for more algorithms for quasi-random numbers
11-29-2019, 01:06 PM (This post was last modified: 11-30-2019 06:16 AM by Namir.)
Post: #1
 Namir Senior Member Posts: 690 Joined: Dec 2013
Looking for more algorithms for quasi-random numbers
Hi All Math Lovers,

In another thread of mine, ttw mentions quasi-random numbers. Quasi-random numbers (QRNs) present a better spread over a range of values than pseudo-random numbers (PRNs). On the other hand, QRNs will often fail randomness tests. They true purpose to to cover more uniformly a range of values in one of more dimensions.

This is part of ttw's response in my other thread, where he mentions QRNs:

Quote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequence. One uses the fractional part of multiples of the square roots of primes. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the irrationals and updates. (List mode). The sequence is also called the Kronecker or Weyl sequence at times.

The above text includes the algorithm of setting x(1)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). The array of P() represents prime numbers starting with 2. You can change x(1) to had a uniform random number as a seed (to generate different sequences every time you apply the algorithm) or simply set x(1) = sqrt(P(1)) = sqrt(2).

I am curious about other formulas to calculate sequences of quasi-random numbers. You are welcome to use your imagination. My first attempt was something like:

Code:
n = number of x to generate m = 100*n Calculate P() for primes in the range of 1 to m X(1) = rand or Frac(ln(P(3)) * sqrt(P(1)) j = 2 count = 0 for i=2 to n   X(i) = Frac(X(i) + ln(P(j+1)) * sqrt(P(j-1))   j = j + 1   if j > m then    count = count + 1     j = 2 + count   end end

The above code produces x() with a mean near 0.5 and standard deviation near 0.28. The auto correlations for the first 50 lags are in the orde rof 10^(-2) to 10^(-4).

I am curious about other formulas to calculate sequences of quasi-random numbers. You are welcome to use your imagination. You can even commit math heresy!!! As long as it works, you are fine (and forgiven) :-)

Namir
11-29-2019, 04:49 PM (This post was last modified: 11-29-2019 06:24 PM by SlideRule.)
Post: #2
 SlideRule Senior Member Posts: 1,013 Joined: Dec 2013
RE: Looking for more algorithms for quasi-random numbers
Perusal of Quasi-random sequences in art and integration, John D. Cook Consulting, illumes the phenomena with references to more descriptive books; Random Number Generation and Quasi-Monte Carlo Methods & Monte Carlo and Quasi-Monte Carlo Methods, on the same.

BEST!
SlideRule
11-30-2019, 01:36 AM
Post: #3
 mfleming Senior Member Posts: 498 Joined: Jul 2015
RE: Looking for more algorithms for quasi-random numbers
(11-29-2019 01:06 PM)Namir Wrote:  This is part of ttw's response in my other thread, where he mentions QRNs:

Quote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequence. One uses the fractional part of multiples of the square roots of primes. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the irrationals and updates. (List mode). The sequence is also called the Kronecker or Weyl sequence at times.

Using "quote" in place of "code" will autowrap large blocks of text!

Who decides?
11-30-2019, 01:29 PM
Post: #4
 Namir Senior Member Posts: 690 Joined: Dec 2013
RE: Looking for more algorithms for quasi-random numbers
(11-30-2019 01:36 AM)mfleming Wrote:
(11-29-2019 01:06 PM)Namir Wrote:  This is part of ttw's response in my other thread, where he mentions QRNs:

Quote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequence. One uses the fractional part of multiples of the square roots of primes. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the irrationals and updates. (List mode). The sequence is also called the Kronecker or Weyl sequence at times.

Using "quote" in place of "code" will autowrap large blocks of text!

I learned that the hard way :-)
11-30-2019, 07:52 PM
Post: #5
 Namir Senior Member Posts: 690 Joined: Dec 2013
RE: Looking for more algorithms for quasi-random numbers
The few leads I got from the nice folks on this web were able to lead me to methods that generate sequences of quasi-random numbers that are practically perfectly distributed. I got what I was looking for. Thanks!!!

Namir
12-01-2019, 05:52 AM
Post: #6
 ttw Member Posts: 186 Joined: Jun 2014
RE: Looking for more algorithms for quasi-random numbers
This is one of the sequences from my paper in "Computational investigations of low-discrepancy point sets II" from the 1994 Las Vegas Conference on Monte Carlo and Quasi Monte Carlo Methods. I have made a single ad-hoc change (described below) that improves distribution for small numbers of points.

The Halton Sequence Phi(N,P) (for odd primes, 2 is a special case not considered here) can be described as:
1. Generate the digits of N in base P (for P an odd prime). Call these digits a(1) to a(k) where k is the maximum number of digits needed. (There should be lots of subscripts but I'll treat each prime separately to reduce index management.)
N=Sum from j=1 to k of a(j)*P^(j-1), that is: a(k)a(k-1)...a(2)(a(1).
2. Reverse the digits: a(1),a(2)....a(k-1),a(k) is resulting string.
3. Treat this string as a fraction with a decimal point (p-ary point?) in front.
Example: P=3, N=5: 5(3)=12. Reverse Phi(5,3)=.21(3) or 2/3+1/9 = 7/9
The sequence is very well distributed for example one starts with 0, 1/3, 2/3, 1/9, 4/9, 7/9, 2/9, 5/9, 8/9, 1/27, 10,27...
This sequence is uniformly in the unit d-cube using d different primes. For example in 3 dimensions using primes 3 and 5 gives the points: (skipping 0 which sits on the corner of the cube).
(1/3, 1/5, 1/7)
(2/3, 2/5, 2/7)
(1/9, 3/5, 3/7)
(4/9, 4/5, 4/7)
(7/9,1/25,5/7)
(2/9, 6/25, 6/7)
(5/9, 11/25, 1/49)
(8/9. 16/25, 8/49)
etc.
The process is sometimes termed a Kakatumi-von Neumann odometer.

There is a problem that I noticed about 1967 or so when I started working on quasi-Monte Carlo. For large base, the Halton Sequence produces strongly correlated points until enough points are generated. (This happens with all quasi-random sequences but not as severely.) Take the first few points using bases 101 and 103.
(1/101, 1/103)
(2/101, 2/103)
...
(100/101, 100/103)
(1/10201, 101/103)
(102/10201, 102/103)
(203/10203, 1/10609)
etc.
In 1993-1995 period, I figured out to multiply each numerator by a number (I called a spin) to break this up. Then I used the fact the fractional parts of square roots of primes are independent to do the following seemingly strange rule.

Give a prime P, to find a multiplier S, do the following.
1. Compute the nearest integers to the fractional part of the Sqrt(P), call these H and L (high and low, one is above and one below the number).
2. Compute the continued fraction of H/P and L/P; each generates a string of partial quotients. The multiplier S is the one of these satisfying the following:
3. A. Chose the one for which the sum of the partial quotients is smallest.
B. If tied, chose the one with the smallest maximum partial quotient.
C. If tied, chose whichever H/P or L/P is closest to the fractional part of the square root.
D. Ad Hoc Alert: if P=41, use 16. (To avoid 17/41 being close to 12/29. The only such case in all primes less than 2^32)

4. To generate the modified Phi sequence Phi(N,P,S): generate the digits of N base P as above and reverse. Multiply each digit by S modulo P and sum as above.
Examples: 3 dimensions: P=3, 5, 7 have S= 2, 2,and 5 respectively.

(2/3, 2/5, 5/7)
(1/3, 4/5, 3/7)
(2/9, 1/5, 1/7)
(8/9, 3/4, 6/7)
etc.
For the pathological case 101 and 103, the multipliers are 6 and 16 respectively (not the best but that's another post sometime).
(6/101, 16/103)
(12/101, 32/103)
(18/101, 48/103)
Clearly more spread out than the original Halton Sequence.

I've got some more but Mordechay Levin's paper ArXiv 1806 shows that even the original Halton Sequence hits the theoretical lower bound for dimensions 2 and up so great changes cannot be had by tinkering. I do have a bit better, but it's even longer to compute and I haven't tested the new ideas thoroughly.

HP50g code:(Number, Prime, Spin, Top, Bottom)

<< 0 1 -> N P S T B << WHILE N REPEAT
N S * P MOD T P * + 'T' STO
P 'B' STO* N P IQUOT 'N' STO END
T B />> >>

Not necessarily the fastest but eas to work with. (If I've done this right.)
12-01-2019, 11:46 AM
Post: #7
 Csaba Tizedes Senior Member Posts: 412 Joined: May 2014
RE: Looking for more algorithms for quasi-random numbers
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