Sum of random decreasing numbers between 0 and 1: does it converge??

Let's define a sequence of numbers between 0 and 1. The first term, $r_1$ will be chosen uniformly randomly from $(0, 1)$, but now we iterate this process choosing $r_2$ from $(0, r_1)$, and so on, so $r_3\in(0, r_2)$, $r_4\in(0, r_3)$... The set of all possible sequences generated this way contains the sequence of the reciprocals of all natural numbers, which sum diverges; but it also contains all geometric sequences in which all terms are less than 1, and they all have convergent sums. The question is: does $\sum_{n=1}^{\infty} r_n$ converge in general? (I think this is called almost sure convergence?) If so, what is the distribution of the limits of all convergent series from this family?

• Not sure if my question itself makes sense: But how about a modified question where r1 is chosen uniformly randomly from (-1,+1) and we iterate by choosing r2 from (-1 x abs(r1),+1 x abs(r1)) Would that series converge too? Feb 5 '17 at 17:24
• Your second series almost surely converges absolutely (i. e. $|r_1| +$|r_2|$+ \cdots$ converges) if and only if your first series almost surely converges. Combining that with Byron Schmuland's answer, the modified series almost surely converges absolutely, and so a fortiori almost surely converges. Feb 5 '17 at 18:11

4 Answers

Let $(u_i)$ be a sequence of i.i.d. uniform(0,1) random variables. Then the sum you are interested in can be expressed as $$S_n=u_1+u_1u_2+u_1u_2u_3+\cdots +u_1u_2u_3\cdots u_n.$$ The sequence $(S_n)$ is non-decreasing and certainly converges, possibly to $+\infty$.

On the other hand, taking expectations gives $$E(S_n)={1\over 2}+{1\over 2^2}+{1\over 2^3}+\cdots +{1\over 2^n},$$ so $\lim_n E(S_n)=1.$ Now by Fatou's lemma, $$E(S_\infty)\leq \liminf_n E(S_n)=1,$$ so that $S_\infty$ has finite expectation and so is finite almost surely.

• @ByronSchmuland $S_\infty$ will have the same distribution as $X(1+S_\infty)$ where $X$ is uniformly distributed from 0 to 1. Since that is a recursive expression it does not directly tell you the distribution, but it can be used to sanity check any distribution you came up with. Feb 5 '17 at 18:35
• ...And the distribution of $S_\infty$ is uniquely determined by its Laplace transform $$L(s)=E(e^{-sS_\infty})$$ which is the unique solution of the differential equation $$\left(sL(s)\right)'=e^{-s}L(s)\qquad L(0)=1$$ solved by $$L(s)=\exp\left(-\int_0^s\frac{1-e^{-t}}tdt\right)=\exp\left(-\int_0^se^{-t}\ln\left(\frac{s}t\right)dt\right)$$
– Did
Feb 5 '17 at 19:45
• @polfosol if your previously drawn number is $a$, a uniformly distributed number between $a$ and $0$ is $a \cdot u$, where $u$ is uniformly distributed between $0$ and $1$. The rest follows by recurrence ($u_1=u$, $u_2 = u_1 \cdot u$...) Feb 6 '17 at 16:21
• Can someone tell me what i.i.d. uniformity means? Feb 7 '17 at 13:28
• @zakoda i.i.d. means "independent identical distribution". Uniform is specifically what kind of distribution (a uniform one where the probability density function is constant over the whole range).
– Kyle
Feb 7 '17 at 16:18

The probability $f(x)$ that the result is $\in(x,x+dx)$ is given by $$f(x) = \exp(-\gamma)\rho(x)$$ where $\rho$ is the Dickman function as @Hurkyl pointed out below. This follows from the the delay differential equation for $f$, $$f^\prime(x) = -\frac{f(x-1)}{x}$$ with the conditions $$f(x) = f(1) \;\rm{for}\; 0\le x \le1 \;\rm{and}$$ $$\int\limits_0^\infty f(x) = 1.$$ Derivation follows

From the other answers, it looks like the probability is flat for the results less than 1. Let us prove this first.

Define $P(x,y)$ to be the probability that the final result lies in $(x,x+dx)$ if the first random number is chosen from the range $[0,y]$. What we want to find is $f(x) = P(x,1)$.

Note that if the random range is changed to $[0,ay]$ the probability distribution gets stretched horizontally by $a$ (which means it has to compress vertically by $a$ as well). Hence $$P(x,y) = aP(ax,ay).$$

We will use this to find $f(x)$ for $x<1$.

Note that if the first number chosen is greater than x we can never get a sum less than or equal to x. Hence $f(x)$ is equal to the probability that the first number chosen is less than or equal to $x$ multiplied by the probability for the random range $[0,x]$. That is, $$f(x) = P(x,1) = p(r_1<x)P(x,x)$$

But $p(r_1<x)$ is just $x$ and $P(x,x) = \frac{1}{x}P(1,1)$ as found above. Hence $$f(x) = f(1).$$

The probability that the result is $x$ is constant for $x<1$.

Using this, we can now iteratively build up the probabilities for $x>1$ in terms of $f(1)$.

First, note that when $x>1$ we have $$f(x) = P(x,1) = \int\limits_0^1 P(x-z,z) dz$$ We apply the compression again to obtain $$f(x) = \int\limits_0^1 \frac{1}{z} f(\frac{x}{z}-1) dz$$ Setting $\frac{x}{z}-1=t$, we get $$f(x) = \int\limits_{x-1}^\infty \frac{f(t)}{t+1} dt$$ This gives us the differential equation $$\frac{df(x)}{dx} = -\frac{f(x-1)}{x}$$ Since we know that $f(x)$ is a constant for $x<1$, this is enough to solve the differential equation numerically for $x>1$, modulo the constant (which can be retrieved by integration in the end). Unfortunately, the solution is essentially piecewise from $n$ to $n+1$ and it is impossible to find a single function that works everywhere.

For example when $x\in[1,2]$, $$f(x) = f(1) \left[1-\log(x)\right]$$

But the expression gets really ugly even for $x \in[2,3]$, requiring the logarithmic integral function $\rm{Li}$.

Finally, as a sanity check, let us compare the random simulation results with $f(x)$ found using numerical integration. The probabilities have been normalised so that $f(0) = 1$.

The match is near perfect. In particular, note how the analytical formula matches the numerical one exactly in the range $[1,2]$.

Though we don't have a general analytic expression for $f(x)$, the differential equation can be used to show that the expectation value of $x$ is 1.

Finally, note that the delay differential equation above is the same as that of the Dickman function $\rho(x)$ and hence $f(x) = c \rho(x)$. Its properties have been studied. For example the Laplace transform of the Dickman function is given by $$\mathcal L \rho(s) = \exp\left[\gamma-\rm{Ein}(s)\right].$$ This gives $$\int_0^\infty \rho(x) dx = \exp(\gamma).$$ Since we want $\int_0^\infty f(x) dx = 1,$ we obtain $$f(1) = \exp(-\gamma) \rho(1) = \exp(-\gamma) \approx 0.56145\ldots$$ That is, $$f(x) = \exp(-\gamma) \rho(x).$$ This completes the description of $f$.

• Can you explain your reasoning to multiply the probability that the first pick is less than $x$ by the probability that you land between $x$ and $x+dx$ starting at $x$? I don't understand that. Wouldn't the probability depend on what the first pick is, necessitating an integral like the one you use later on?
– user142299
Feb 6 '17 at 0:32
• $$P(x,x) = 1/x \int_0^x P(x-z,z) dz$$ $$P(x(<1),1)=1/1\int_0^xP(x-z,z)dz$$ $$P(x(<1),1) =xP(x,x)=P(1,1)$$ Feb 6 '17 at 6:16
• This is (proportional to) the Dickman function!
– user14972
Feb 6 '17 at 7:54
• @Hurkyl Wow, that's cool. Just added it to the answer. Feb 6 '17 at 8:21

Just to confirm the simulation by @curious_cat, here is mine:

It's a histogram, but I drew it as a line chart because the bin sizes were quite small ($0.05$ in length, with 10 million trials of 5 iterations).

Note: vertical axis is frequency, horizontal axis is sum after 5 iterations. I found a mean of approximately $0.95$.

• The shape of that graph does not look like I would have expected it to. It looks flat from 0 to 1 (I assume the variance seen there is entirely due to only having sampled a finite number of times). I would have expected it to start decreasing right away. And it looks like the derivative is discontinuous at 1 and nowhere else. I would have expected the derivative to be continuous. And given that the distribution can be expressed recursively I can't quite see how a single discontinuity in the derivative wouldn't cause an infinite number of discontinuities. Feb 5 '17 at 18:44
• That doesn't mean I think you are wrong. Just that I am surprised. I will now try to simulate it myself to see if I can reproduce your result. Feb 5 '17 at 18:45
• I have proved why it has to be constant for $x<1$, please check my answer. Feb 5 '17 at 19:15
• I have done a simulation myself which produced a graph similar to yours except from an anomaly at 0 due to using discrete random numbers in the simulation. So it looks like the shape of the graph is pretty much correct. Feb 5 '17 at 19:36
• For an infinite sum, the mean should be $1$ since it satisfies $\mu=\frac12 +\frac12\mu$ where $\frac12$ is the mean of a uniform random variable on $[0,1]$. For five terms in the sum it should be $\frac{2^5-1}{2^5}=\frac{31}{32}=0.96875$ Feb 9 '17 at 14:13

I just ran a quick simulation and I get a mean sum of one (standard deviation of 0.7)

Caveat: Not sure I coded it all right! Especially since I didn't test convergence.

• How did you run a quick simulation of an infinite sequence?
– JiK
Feb 6 '17 at 12:24
• @JiK Right. I didn't. Feb 6 '17 at 12:39
• Chuck Norris could have run a quick simulation of an infinite sequence. After all, he counted to infinity. Twice. Feb 8 '17 at 5:49
• For what it's worth, I did the simulation in excel and let it cook for half a day. My histogram was essentially the same as posted above: flat from 0 to 1, then an exponential like decay. This leads to another question involving histograms, which I shall submit to the community... Recently the Israel prime minister dismissed his security staff. He was meeting Chuck Norris, and knew the Norris could personally handle any security problem. Feb 8 '17 at 19:12


# [SOLVED]Polynomial roots

#### wishmaster

##### Active member
I have to find all solutions for X when:
x4-2x3-25x2+50x

I have done it so,but im not sure if this is ok:

x(x3-2x2-25x+50)

= x(x2(x-2)-25(x+2)
= x(x2-25)(x-2)
=x(x-5)(x+5)(x-2)

Now i see that root/zeroes are +5,-5 and 2. I know that this polynomial has another zero that is 0,but how do i know that? Because x is in the front? Or did i make a mistake and should such problems deal another way? Thank you for all replies!

#### MarkFL

Staff member
I have to find all solutions for X when:
x4-2x3-25x2+50x

I have done it so,but im not sure if this is ok:

x(x3-2x2-25x+50)

= x(x2(x-2)-25(x-2))
= x(x2-25)(x-2)
=x(x-5)(x+5)(x-2)

Now i see that root/zeroes are +5,-5 and 2. I know that this polynomial has another zero that is 0,but how do i know that? Because x is in the front? Or did i make a mistake and should such problems deal another way? Thank you for all replies!
You had a minor typo in your working which did not affect the outcome. I have highlighted it in red, but is is obvious from your subsequent steps that you intended to write this. The four roots of the polynomial are found by equating each of the four factors to zero, including the factor $x$ in front, and solving for $x$. So you are correct that the four roots are (in ascending order):

$$\displaystyle x=-5,\,0,\,2,\,5$$

Great job!

#### wishmaster

##### Active member
You had a minor typo in your working which did not affect the outcome. I have highlighted it in red, but is is obvious from your subsequent steps that you intended to write this. The four roots of the polynomial are found by equating each of the four factors to zero, including the factor $x$ in front, and solving for $x$. So you are correct that the four roots are (in ascending order):

$$\displaystyle x=-5,\,0,\,2,\,5$$

Great job!
Thank you! Sorry,that was a type mismatch.
one question,when x stays alone in front of other terms,then this root is always zero?

#### MarkFL

Staff member
Thank you! Sorry,that was a type mismatch.
one question,when x stays alone in front of other terms,then this root is always zero?
Yes, equating that factor to zero, we get:

$$\displaystyle x=0$$

It is already solved for $x$, and it tells us that $x=0$ is a root.

#### Petrus

##### Well-known member
Thank you! Sorry,that was a type mismatch.
one question,when x stays alone in front of other terms,then this root is always zero?
Another way to se it if we want to divide by x Then we get 2 case
case 1 $$\displaystyle x=0$$
put $$\displaystyle x=0$$ to the equation and we see that $$\displaystyle 0=0$$ hence 0 is a root
case 2 $$\displaystyle x \neq 0$$
we know can divide by x (we get third degree polynom) and now it's just to use rational root Theorem and long polynom division and Then it becomes second grade polynom which is easy to solve

Regards,
$$\displaystyle |\pi\rangle$$

#### topsquark

##### Well-known member
MHB Math Helper
x4-2x3-25x2+50x

=x(x-5)(x+5)(x-2)
To be specific we have that if ab = 0 then either a = 0 or b = 0. In this case that means when
x(x - 5)(x + 5)(x - 2) = 0

we get solutions when:
x = 0, or x - 5 = 0, or x + 5 = 0, or x - 2 = 0

The solutions to these four equations are your roots.

-Dan


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# Find the smallest possible checksum

What is the smallest possible checksum $$C_{min}$$ of the sum of two three-digit numbers $$N_1$$ and $$N_2$$ that are formed from the six given digits 2, 3, 4, 5, 7 and 9 (each digit shall be used only once). The checksum of a number is defined as the sum of its digits. What are the sums that have this smallest possible checksum $$C_{min}$$?

Here my thoughts: I believe that the smallest checksum is 3. I obtained this by simple reasoning: With $$3 + 7 = 10$$ there is 0 in the sum and a carry-over "1". Then, with $$4 + 5 = 9$$ and the carry-over "1" we obtain another "0" in the sum, creating another carry-over "1". Finally, $$2 + 9 = 11$$, and with the carry-over it is "12". Thus, the checksum is $$1 + 2 + 0 + 0 = 3.$$ But is it right?

Examples:

$$243 + 957 = 1200,$$ checksum = 3

$$423 + 597 = 1020,$$ checksum = 3

If the answer is at all correct, how can I prove that it is? And then, how would I be able to find all the sums of numbers? My approach: Permute the given checksum, and check if I can find N1 and N2 that have this result.

There must be a more systematic or rigorous approach, I would believe. Perhaps someone can guide me on this. Thank you.

• You don't give a rigorous definition of the checksum that you're using. Do you mean "find the smallest digit sum of $a + b$ when $a, b$ are two three-digit numbers formed from the six given digits $2, 3, 4, 5, 7, 9$?" – orlp Oct 2 '18 at 12:20
• Sorry, I mean the sum over the digits. Will edit my post.If the numbers N1 and N2 are 539 and 247, their sum is 786 and the checksum of the sum is 7 + 8 + 6 = 21 – Parzifal Oct 2 '18 at 12:23
• I'm not posting this as an answer as it's a mathematics site, but there is a systematic way: check each case with a computer. Since $6! = 720$ this happens in the blink of an eye and indeed confirms $3$ is the smallest possible digit sum. – orlp Oct 2 '18 at 12:26
• Are you familiar with the notion of casting out nines? – Barry Cipra Oct 2 '18 at 12:26
• No - what do you mean by casting out nines? – Parzifal Oct 2 '18 at 12:29

Let's start with a brief explanation of "casting out nines" as it's needed for the problem at hand. If $$ABCD$$ is the (at most) four-digit sum of a pair of three-digit numbers, $$abc+def$$, then, since $$1000\equiv100\equiv10\equiv1$$ mod $$9$$, we have

\begin{align} A+B+C+D &\equiv(1000A+100B+10C+D)\\ &=(100a+10b+c)+(100d+10e+f)\\ &\equiv a+b+c+d+e+f\mod 9 \end{align}

so, if $$\{a,b,c,d,e,f\}=\{2,3,4,5,7,9\}$$, we have

$$A+B+C+D\equiv2+3+4+5+7+9=(2+7)+3+(4+5)+9\equiv3\mod 9$$

which means that the digit sum for any pair $$abc+def$$ belongs to $$\{3,12,21,\ldots\}$$. The OP has already found one example with digit sum $$3$$, namely $$243+957=1200$$, so we can conclude that the smallest possible digit sum is $$3$$. The remaining question, how many different pairs give digit sum $$A+B+C+D=3$$?

Note that

$$abc+def=dbc+aef=aec+dbf=dec+abf$$

so it suffices to count the number of solutions with $$a\lt d$$, $$b\lt e$$ and $$c\lt f$$ and then multiply by $$4$$ (or by $$8$$, if you want to distinguish $$(abc,def)$$ from $$(def,abc)$$). Now the only realistic possibilities for $$ABCD$$ are $$1200$$, $$1020$$, $$1002$$, $$1110$$, $$1101$$, and $$1011$$. (That is, $$500\lt abc+def\lt2000$$, so we must have $$A=1$$ since $$A=0$$ would imply $$B\ge5$$.) Let's consider these possibilities according to what's in the ones place.

If $$D=0$$ (i.e., if $$ABCD=1200$$, $$1020$$, or $$1110$$), we can only have $$c=3$$ and $$f=7$$, which will carry a $$1$$ into the tens place. That makes $$C=1$$ impossible, but it gives $$1+2+9=11$$ and $$1+4+5=10$$ as possibilities for $$C=2$$ and $$C=0$$. Indeed, we get x two solutions (with $$a\lt d$$, etc.), namely $$423+597=1020$$ and $$243+957=1200$$.

If $$D=1$$ (i.e., if $$ABCD=1101$$ or $$1011$$), we can have $$c+f=2+9=11$$ or $$c+f=4+7=11$$. In either case the carried $$1$$ from $$c+f$$ means we need $$b+d=9$$ or $$10$$ in order to get $$C=0$$ or $$1$$. For $$c+f=2+9$$, both values of $$C$$ are attainable, each in only one way: $$342+759=1101$$ and $$432+579=1011$$. For $$c+f=4+7$$, neither value of $$C$$ is attainable, since the only digits that sum to $$9$$ are $$2+7$$ and $$4+5$$ and, as already remarked, the only digits that sum to $$10$$ is $$3+7$$. Thus we get just two solutions with $$D=1$$, namely $$342+759=1101$$ and $$432+579=1011$$.

Finally, if $$D=2$$ (i.e., $$ABCD=1002$$), we can only have $$c+f=3+9$$ or $$c+f=5+7$$. In either case we'll need $$b+e=9$$, which is possible only as $$2+7$$ or $$4+5$$. If $$c+f=3+9$$, either of these is possible, while neither is possible if $$c+f=5+7$$. So again we get just two solutions, namely $$243+759=1002$$ and $$423+579=1002$$.

Altogether we get six solutions with $$a\lt d$$, $$b\lt e$$ and $$c\lt f$$:

\begin{align} 423+579&=1020\\ 243+957&=1200\\ 342+759&=1101\\ 432+579&=1011\\ 243+759&=1002\\ 423+579&=1002 \end{align}

The total count, without the restrictions $$a\lt d$$ and $$b\lt e$$, is thus $$24$$; removing the restriction $$c\lt f$$ brings the total number of solutions to $$48$$.

• Thank you for the comprehensive and very well explained answer. Insightful. Much appreciate the time you took! – Parzifal Oct 4 '18 at 11:04

Hint1:

$$\sum_{k=0}^n 10^ka_k \equiv \sum_{k=0}^n a_k \mod{9}$$

Hint2:

Note, that the digital root for your numbers is always the same, regardless of your choice.

• Thanks, I checked your link but I do not understand how your hint leads to the solution. Can you elaborate a bit, please? – Parzifal Oct 2 '18 at 12:53
• @Parzifal I've added one more hint. – Jaroslaw Matlak Oct 2 '18 at 13:08
• So indeed, the digital root always is 12. This means that any combination of the digits will yield this root. And I always get to this root as the minimum digital sum by re-applying the casting out of nines ... the argumentation runs in this direction? – Parzifal Oct 2 '18 at 13:36
• * Digital root is one-digit number, in this case $3$. * The control sum is equal to the digital root (mod 9). Therefore the control sum can't be smaller than the digital root. You've found the number with control sum equal to the digital root, so... – Jaroslaw Matlak Oct 2 '18 at 18:27
• Thank you - your comments and hints were very helpful to me. First time that I came across such concepts. – Parzifal Oct 2 '18 at 19:22

Building on the helpful hints of @Jaroslaw Matlak, the digital root of the sum will equal the digital root of the sum of the digital roots for the summed numbers. Consider your first example:

$$243 + 957 = 1200$$

The digital root of $$243$$ is

$$2+4+3 = 9$$

and the digital root of $$957$$ is

$$9 + 5 + 7 = 21,$$ continuing to sum until we have one digit, $$2+1 = 3$$.

So the sum of the digital roots for $$243$$ and $$957$$ is $$9+3 = 12$$ giving $$1+2 = 3$$. This is equal to the digital root of $$1200$$.

$$1+2+0+0 = 3$$.

You should see that the actual permutations of the numbers $$2,3,4,5,7,9$$ don't matter in the calculation of the digital root. The left hand side digital root will always be $$2+3+4+5+7+9=30$$, $$3+0=3$$, the same as the digital root of the sum.

The other answers have shown that the digit sum will converge to $$3$$ if you repeat it enough. It appears you only want to take the digit sum once, so you can get $$3,12,21$$. Once you find an example with $$3$$ you are done. Finding such an example can be done with clever searching. There is only one pair of digits, $$3,7$$, available that sum to $$10$$, so we put those in the ones place to get a $$0$$. There is another pair, $$4,5$$, that sum to $$9$$ so if they are added with a carry in you will get another $$0$$. The sum of two three digit numbers cannot have more than four digits, so if you get two zeros the digit sum has to be $$3$$ or $$12$$, but since the carry into the thousands is $$1$$ you can only get $$3$$. Good work.

• Milikan: That is how I actually searched but I lacked the proof. I learned a lot from you all. And about finding all possible sums - is my suggested approach ok? Thanks – Parzifal Oct 2 '18 at 14:33
• What you would usually permute is the digits in the sum and see what checksums come out. If you try all the possibilities (there are only $360$) you will get them all. We have $3$. It is easy to get $12$ from $243+957$ and $21$ from $342+957$. The sum of all the digits is $30$ and you can't avoid a carry, so I believe $30$ is impossible but I haven't proven it. – Ross Millikan Oct 2 '18 at 15:03
• How aout $439+572$? – Barry Cipra Oct 2 '18 at 15:11
• @Barry Cipra: Good one (CS=1101). But can I rule out 1110 and 1011 apart from trying out all possibilities? – Parzifal Oct 2 '18 at 15:46
• @Ross Milikan: Thank you for your time explaining, I appreciate this very much. I´ll try my luck (but as Barry Cipra points out, it´s not an "easy" task). – Parzifal Oct 2 '18 at 15:49


# A Coin Is Tossed Three Times What Is The Probability Of Getting 3 Heads

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And what I want to think about in this video is the probability of getting exactly three heads. Three coins are tossed up in the air, one at a time. A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. When you register, you support the site and your question history is saved. In possibility theory, a The probability of happening an event can easily be found using. Determine a probability assignment for the simple events of the sample space S ={H,T} that reflects this bias. A fair coin is tossed 3 times, the probability of gitting 3 heads is? Let us understand that this is the probability of getting exactly 3 tails out of the 5 chances too. 5 we get this probability by assuming that the coin is fair, or heads and tails are equally likely. What is the probablity that 3 heads will occur?. So the difference is we multiply the probabilities in the independent events. If a coin is tossed 12 times, the maximum probability of getting heads is 12. Spun coins can exhibit "huge bias" (some spun. Sollicitatievragen voor Senior Solutions Developer in Maadi Cornish. Note that the last three cases for the coins chosen expand into eight diﬀerent cases when we care whether coin 2 or coin 3 is chosen. The reason the probability of flipping 3 heads in a row is due to conditional probability. Example 1: A fair coin is tossed 5 times. Algebra -> Probability-and-statistics-> SOLUTION: A fair coin is tossed 5 times. When the Seahawks won the coin toss and started overtime with the ball, was there any doubt about what was going to happen?. Willie Wong served as mayor of Mesa from 1992 to 1996. Youngs sends the kick long and Vermuelen is met by May. Best Answer: Probability of getting a head: 0. The ratio of successful events A = 4 to the total number of possible combinations of a sample space S = 8 is the probability of 2 tails in 3 coin tosses. The chances of getting a girl should be the same whether or not the rst child was a girl (after all, the coin doesn't know whether it came down heads or tails last time). Later in this section we shall see a quicker way. A math-ematical model for this experiment is called Bernoulli Trials (see Chapter 3). The final test of a criminal investigation is in the presentation of the evi¬dence in court. Probability of getting First Tail = 1/2 Probability of getting Second tail (Such that first tail has occurred, this incidentally is also the probability when first was head and second is tail) = 1/2 * 1/2 = 1/4 Probability of getting Third. The probability of getting a head on any one toss of this coin is 3/4. We never know the exact probability this way, but we can get a pretty good estimate. The Bucs got 67 yards and a touchdown on 18 carries from Jones, with two catches for 15 yards on two targets, tying the most touches Jones has ever had in a game. I want a free account. Trump's legal team to mount a defense. 5 probability of a particular two. was there for 30mins before parents noticed. The probability of getting a head on any one toss of this coin is 3/4. ( ) 1 1 7! 7 63 4 7,3 2 2 3! 4! C ⋅ = = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅5 4 3 2 1 3 2 1⋅ ⋅ ⋅ ⋅ ⋅ ⋅4 3 2 1 7 35 128 1 2 = 2. But, 12 coin tosses leads to #2^12#, i. If the coin is spun, rather than tossed, it can have a much-larger-than-50% chance of ending with the heavier side down. Gmm to three decimal places. a coin flip; a coin toss PK ¹^nK\¨ ¥†À%áˆ''02_Tier 1 PBIS & PSC Coaching Guide. It describes the probability distribution of a process that has two possible outcomes. What is the probability of obtaining exactly 3 heads. i think exactly like that, but then i would realize if we do that, what are the odds of gettin 3heads and 2 tails in no order???? wouldnt be 100%????? cuz its only a two sided coin and odds of gettin heads or tails is equal. What is the probability that (a) At least one of the dice shows an even number? P(at least one is even) = 1 - P(both are odd). What is the probability of getting (i) all heads, (ii) two heads, (iii) at least one head, (iv) at least two heads? Sol. self Question Solution - A fair coin is tossed three times in succession. asked by Dee on April 23, 2014; Math. com loan 1 Let an experiment consist of tossing a fair coin three times. Date: 04/21/2003 at 17:12:44 From: Maggie Subject: Probability In a box there are nine fair coins and one two-headed coin. One of two coins is selected at random and tossed three times. Tossing a Coin A balanced coin is tossed three times. One is a two-headed coin ( having head on both faces ), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. They are: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT. find the probability of getting a result of an odd number of heads in 1/2 of the cases. A coin flip: A fair coin is tossed three times. 5 Probability of getting a tail = 0. In a short trial, heads may easily come up on every flip. There are three ways that can happen. The outcomes of the three tosses are recorded. There are many other kinds of situations, however, where the probability of an event is not independent but dependent — that is, where the. sum of three times a. Probability of Coin Tosses. One Head : 160 times c. 3 10 Illowsky et al. enter your value ans - 5/16. There were five heads and three tails in the eight tosses. The probability of rolling a six on one die is 1/6. Subjective probability is more connected to the original meaning of the word probable for example, because simply calculating the probability of any horse winning There are three types of probability problems that occur in mathematics. Poker night 2 steam achievement manager. Solution: n=20, p=0. 5 Number of heads$\leq$Number of tails. When three coins are tossed at random, what is the probability of getting : (i) 3 heads ? (ii) 2 heads Best Answer for Silver Coins, Round, Sloven A fair coin is tossed 5 times. Let E = event of getting exactly 3 heads. The Questions and Answers of A coin is tossed three times. The answer would be a cumulative probability. If$X\$ is the number of heads. A consecutive streak or a run can happen in random. what's the probability of getting three heads? You can put this solution on YOUR website! a fair coin is tossed 3 times, 1. A coin has a probability of 0. So let's start again with a fair coin. The chances of getting a girl should be the same whether or not the rst child was a girl (after all, the coin doesn't know whether it came down heads or tails last time). Note that this is the same as summing the probability of the 3 mutually exclusive events Given that we have at least one head the probability that there are at least 2 heads. Number of times two heads appeared = 55. Consider, you toss a coin once, the chance of occurring a head is 1 and chance of occurring a tail is 1. Wonder if Papa will left the coin decide if Naz gets a real Guinea Pig this time? Heads or tails will decide. What is the probability that exactly 3 heads occur?. Complimentary Events. Unfortunately, I do not believe I was successful in explaining to Kent why my figures were correct. what's the probability of getting three heads? You can put this solution on YOUR website! a fair coin is tossed 3 times, 1. An Easy GRE Probability Question. 5 Probability of getting a tail = 0. Httt tttt ttht thtt. Question: 520. The probability that an electronic device produced by a company does not function properly. Answer to You toss a coin three times. 5 Probability of getting a tail: 0. They are: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT. Example: Suppose 20 biased coins are flipped and each coin has a probability of 75% of coming up heads. Annie will toss a fair coin three times. We use the experiement of tossing a coin three times to create the probability distribution This is a basic introduction to a probability distribution table. Three coins are tossed once. What is the probability of getting (i) all heads, (ii) two heads, (iii) at least one head, (iv) at least two heads? Sol. 1 Let an experiment consist of tossing a fair coin three times. it can occur first, second or third. If you toss a coin 10 times, how many heads and how many tails would you expect to get? How about the results of your partners tosses? How close was each set of results to what was Probability and Segregation. What is the probability of getting neither an ace nor a king card? a. A coin is tossed n times. The probability that an electronic device produced by a company does not function properly. Find the probability of getting (i) 3 heads (ii) 2 heads (iii) at least 2 heads (iv) at most 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) at most two tails. Let be the number of heads on the first two tosses, the number of heads on the last two tosses. There are three tosses. the level of inventory at the end of a given month, or the number of production runs on. Coin flipping, coin tossing, or heads or tails is the practice of throwing a coin in the air and checking which side is showing when it lands This penny has been flipped times. So the probability is ----- b) What is the probability of obtaining tails on each of the first 3 tosses That only happens 2 times. As it gets cumbersome to write the repeated multiplication, we can use exponents to simplify work. The probability of getting heads three times in 5 tries is 10/32. There are 15 girls and 15 boys in 8th period. a coin flip; a coin toss PK ¹^nK\¨ ¥†À%áˆ''02_Tier 1 PBIS & PSC Coaching Guide. When tossed, the first shows heads with probability 4/10, the second shows heads with probability 7/10. The results are similarly formatted. (a) What is the Posted 4 years ago. Probability of Getting 2 Heads in 3 Coin Tosses. 2) b) Use simulations to find an empirical probability for the probability of getting exactly 5 heads in 10 tosses of an unfair coin in which the probability of heads is 0. Find the probability that a common year (not a leap year) contains If one is randomly chosen, then what is the probability of getting non-faulty pen? 11. : Let 'S' be the sample - space. A coin is biased in such a way that a head is twice as likely to come up as a tails. Probability of Coin Tosses. If three fair coins are tossed randomly 175 times and it is found that three heads appeared 21 times, two heads appeared 56 times, one head appeared 63 times and zero head appeared 35 times. Ex: Consider tossing a fair coin. Based on the experimental probability, how many times should Kuan expect to get 3 heads in the next 55 tosses?. Fifty-three years later, family members cling to the hope that her murder will be solved. For instance, we can toss the coin many times and determine the proportion of the tosses that the coin comes up head. Users may refer the below solved example work with steps to learn how to find what is the probability of getting at-least 2 heads, if a coin is tossed three times or 3 coins tossed together. It indicates a way to close an interaction, or dismiss a notification. It's got this cool variegated blue glaze, with swirls of grays and blues with bits of browns and whites. What is the probability of flipping 3 coins and getting all heads flipping three coins and the possibility of getting all Probability Math Help For College. Download with Google Download with Facebook or download with email. The frequency with which the coin lands heads is three out of four, and it can never be tossed again. Every time we toss a fair coin, the probability of seeing heads is 1/2 regardless of what previous tosses have revealed. If the coin is fair, then by symmetry the probability of getting at least 2 heads is 50%. The question is: Suppose a fair coin is thrown three times. Probability of getting First Tail = 1/2 Probability of getting Second tail (Such that first tail has occurred, this incidentally is also the probability when first was head and second is tail) = 1/2 * 1/2 = 1/4 Probability of getting Third. 5% chance of all three tosses landing on heads. It is well known that the probability that a head turns up in a single toss of a coin is 0. what is the probability of getting exactly 8 heads in a row? c. Besides adding together the. quince medios de un numero es igual a cuarenta y cinco?. What is the probability that all 3 tosses will result in heads? A)1/2. Hhtt hhth ttth thht. ( ) 1 1 7! 7 63 4 7,3 2 2 3! 4! C ⋅ = = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅5 4 3 2 1 3 2 1⋅ ⋅ ⋅ ⋅ ⋅ ⋅4 3 2 1 7 35 128 1 2 = 2. Let x = probability of winning after no heads (or a tail). If a coin is tossed three times what is the pribability of getting at most one tail and what is the probability of getting at most 2 head And plz also explain why you are taking hhh in case of at most one tail - Math - Probability. The probability of success, i. 7E-20 A fair coin is tossed 20 times. The reason the probability of flipping 3 heads in a row is due to conditional probability. Determine the probability of getting 2 heads in two successive tosses of a balanced coin. Perhaps you are expected to use the normal approximation to the binomial. One after another what is the probability that the first ball is red and second ball is yellow?. Solution 2: How could you use the entries in Pascal’s Triangle to solve this problem?. Pregunta de entrevista para Senior Solutions Developer en Maadi Cornish. As the number of tosses of a coin increases, the number of heads will be getting closer and closer to the number of tails. I need to write a python program that will flip a coin 100 times and then tell how many times tails and heads were flipped. Httt tttt ttht thtt. quince medios de un numero es igual a cuarenta y cinco?. EDIT - It's not 1/2^6, it's (1/2)^6. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry. An array of Asian-related grocery stores, restaurants and shops organically sprouted in Mesa. ) What is the probability of getting heads on only one of your flips? B. What is the probability that you will get heads all three times? of getting a heads. In general when a coin is tossed n times , the total number of possible outcomes = 2^n). Find simple formulae in terms of n and k for. Opening scene of Rosencrantz & Guildenstern Are Dead contemplating probability. What is the probability that exactly two tosses result in heads? Express your answer as a common fraction. If you look at a chart you will see that the possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. getting a 6, one each trial is 1/6. What is the probability of rolling a six on all four rolls?' and find homework help for other Math questions at eNotes A coin is tossed. One of two coins is selected at random and tossed three times. 7870 and the probability of What is the probability that three of the six possible outcomes do not show up and each of the other three possible. I respond to three objections, which claim that time-restricted laws lessen the coincidence of observed regularities without making it Those claims will be defended within three broad domains: quantum mechanics, classical statistical mechanics, and macroscopic chance events such as coin tosses. If you toss a coin, it will come up a head or a tail. find the probability of getting a head at least once Follow. Then S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT } (i) Let 'E1' = Event of getting all heads, Then E1 = { HHH }. I believe that I have the trick coin, which has a probability of landing heads 40% of the time, p= 0. When two coins are tossed at random, what is the probability of getting a. 5 of coming up heads. Avasaram Component Library We are thrilled. Thus, the required probability is 2 ÷ 8 = 0. They are: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT. Let E be an event of getting heads in tossing the coin and S be the sample space of maximum possibilities of getting heads. According to this table, the theoretical probability of getting 6 heads in 10 tosses is 20. Probabilities can also be thought of in terms of relative frequencies. Curry gathers the kick and carries the ball back. Since these all have a probability of 12. Given N number of coins, the task is to find probability of getting at least K number of heads after tossing all the N Time Complexity: O(n) Auxiliary space: O(n). Answer : Option A Explanation : Total number of balls = 4 + 5 + 6 = 15 Let S be the sample space. The probability of getting a head on any one toss of this coin is 3/4. A pair of dice are rolled. Solution: Total number of trials = 250. Question: A Coin Flip: A Fair Coin Is Tossed Three Times. thats why our thought process is wrong. If you toss a coin 3 times, you're going to get at least two heads or at least two tails, but you can't get _both_ 2 heads and 2 tails. This is a problem that takes some time and a few steps to solve. What is the theoretical probability of both coins landing on heads?. The total number of outcomes = 2 ^6 =64 (it is because each toss has two possibilities Head or Tail. #color(blue){3" Heads"}#. If you need the probability of getting EXACTLY 2 tails then the probability is (3/8) or.


# Set a fixed colour scale with DensityPlot

I am trying to plot some Zernike polynomials, but I want all of them to be plotted by using the same colour scale from -1 to +1 (i.e. the range of these polynomials). I am using the following code:

DensityPlot[
ZernikeR[n, m, Norm[{x, y}]] Cos[m ArcTan[x, y]], {x, -1, 1}, {y, -1,1}
, PlotRange -> {{-1, 1}, {-1, 1}, {-1, 1}}
, ColorFunction -> (Hue[2 (1 - #1)/3] &)
, PlotPoints -> 200
, PlotLegends -> Automatic
, RegionFunction -> Function[{x, y, z}, -1 < x^2 + y^2 < 1]
]

I get what I want with some polynomials, e.g. when m=0 and n=2:

and not with others, e.g. when m=0 and n=4:

where the colour range goes down to -0.5 and not to -1. I want all polynomials to be coloured with a scale from -1 to 1, from blue to red. I haven't found a way to fix this. What am I missing? Thanks.

• Look up ColorFunctionScaling and ColorFunction. – Szabolcs Jul 16 '18 at 11:08
• ColorFunctionScaling -> False will prevent scaling but 2 (1 - #1)/3 &@Interval[{-1, 1}] gives: Interval[{0, 4/3}] which can be missleading when Hue is applied ( {0,1} base domain). What exactly do you want to achieve, how your manual rescaling is related to the problem and are you only concerned about colors or the range in the bar legend aswell? – Kuba Jul 16 '18 at 11:49
• I'd like the color scale to always vary between blue set for -1 and red set for +1, independently from the min and max values of the function in that range. – MicheleG Jul 16 '18 at 11:58
• A related question. – J. M. is away Sep 26 '18 at 7:50

(This answer will be very similar to halirutan's answer (+1). The difference is that I preserved the color function from the OP, and that I made a fixed bar legend.)

We can do it like in this answer, which is to say, we can turn off the color function scaling and scale the values ourselves in a way that is the same for all plots. We can make our own bar legend that matches this scaling. We only need to change three options:

ColorFunction -> (Hue[2 (1 - Rescale[#1, {-1, 1}])/3] &),
ColorFunctionScaling -> False,
PlotLegends -> BarLegend[{Hue[2 (1 - Rescale[#1, {-1, 1}])/3] &, {-1, 1}}]

Now we get plots like this:

• Hi C.E., this is exactly the solution I just found thanks to the link you posted before. Thanks to all of you guys for you help. – MicheleG Jul 16 '18 at 12:13

You have currently two mistakes. The first one is your color-function itself. It needs to give Hue[0] for a value of -1 and Hue[1] for a value of 1. The transformation is

col[v_] := Hue[Rescale[v, {-1, 1}]]

Secondly, you need to turn off that Mathematica rescales the values for each plot which can be done with the ColorFunctionScaling -> False option.

plot[m_, n_] :=
DensityPlot[
ZernikeR[n, m, Norm[{x, y}]] Cos[m ArcTan[x, y]], {x, -1,
1}, {y, -1, 1}, PlotRange -> {{-1, 1}, {-1, 1}, {-1, 1}},
ColorFunction -> col, PlotPoints -> 200, PlotLegends -> Automatic,
ColorFunctionScaling -> False,
RegionFunction -> Function[{x, y, z}, -1 < x^2 + y^2 < 1]]

And then, plot[0,2] and plot[0,4] will use the same scaling for the colors:

If you don't want a circular color scheme like Hue is (begins and ends in red), then use one of the other schemes.


# On Cauchy-Schwarz inequality

What is Schwarz inequality in $$\mathbb R^2$$ or $$\mathbb R^3$$? Give another proof of it in these cases.

Here is my attempt in $$\mathbb R^2$$. Let $$x=(x_1,x_2)$$ and $$y=(y_1,y_2)$$ both in $$\mathbb R^2$$. The Cauchy-Schwarz inequality is

$$\lvert\langle x,y \rangle \rvert \leq \lVert x\rVert \lVert y\rVert.$$ Then our claim is

$$(x_1y_1+x_2y_2)^2 \leq (x_1^2+x_2^2)(y_1^2+y_2^2).$$

Proof for this inequality: since $$(x_1y_2-x_2y_1)^2 \geq 0$$, add $$(x_1y_1+x_2y_2)^2$$ to both sides: \begin{align*} (x_1y_1+x_2y_2)^2 & ≤ (x_1y_2-x_2y_1)^2+(x_1y_1+x_2y_2)^2 \\ & =(x_1y_2)^2+(x_2y_1)^2 -2(x_1y_2)(x_2y_1) \\ & \qquad +(x_1y_1)^2+(x_2y_2)^2+2(x_1y_1)(x_2y_2) \\ & =x_1^2(y_1^2+y_2^2) +x_2^2(y_1^2+y_2^2) \\ & =(x_1^2+x_2^2)(y_1^2+y_2^2), \end{align*} namely $$(x_1y_1+x_2y_2)^2\leq (x_1^2+x_2^2)(y_1^2+y_2^2).$$ This proves the claim.

My question is: is this answer complete to the given question? As there is a choice for $$\mathbb R^2$$ or $$\mathbb R^3$$. If any step is missing please identify it.

• I guess you should do a similar thing in the case of $\mathbb R^3$. The steps look good to me. – Gibbs May 28 '20 at 12:10
• Note that I have modified the format of the entire question. Please, use MathJax next times. See here. – Gibbs May 28 '20 at 12:22

you wrote: $$\quad "$$ our claim is
$$(x_1y_1+x_2y_2)^2 \leq (x_1+x_2)(y_1+y_2)."$$
$$(x_1y_1+x_2y_2)^2 \leq (x_1^2+x_2^2)(y_1^2+y_2^2).$$


# Transforming numbers of irreducible polynomials

Let $M(n)(q)$, where $q$ is a prime power and $n$ a natural number, stand for the number of irreducible monic polynomials of degree $n$ in the polynomial ring $\mathbf{F}_{q}[X]$ over the finite field $\mathbf{F}_{q}$. Since \begin{equation*} M(n) = \frac{1}{n} \sum_{d \mid n} \mu(d) q^{n/d} \end{equation*} we have that $M(1)=q$, $M(2) = \frac{1}{2}(q^2-q)$, etc. Also, let $P(n)$ be the set of partitions $[n_1^{e_1},\ldots, n_s^{e_s}]$ of $n = n_1e_1 + \cdots + n_se_s$. Consider the sum of polynomials in $\mathbf{Q}[q]$ \begin{equation*} \sum_{[n_1^{e_1},\ldots, n_s^{e_s}] \in P(n)} (-1)^{\sum e_i} \prod_{i=1}^s \binom{M(n_i)}{e_i} \end{equation*} ranging over all partitions of $n$. This sum equals $-q$ if $n=1$ and it equals $0$ for $n=2,\ldots,50$. I suspect it equals $0$ for all $n>1$. Is this true?

• I think you mean $M(n)(q)$ to be the number of irreducible polynomials of degree $n$ in $\mathbb F_q[X]$. Oct 22 '16 at 10:30
• The sum is actually $-q$ for $n=1$. Oct 22 '16 at 10:46
• Thanks for the comments - my original question has been corrected accordingly. Oct 22 '16 at 11:14

Yes, it is true. Your expression is the coefficient of $x^n$ in the following product: $$\prod_{P\text{ monic irreducible}} (1-x^{\deg P}) = \prod_{n} (1-x^n)^{M(n)}.$$ The Zeta function of $\mathbb{F}_q[X]$ is $$\sum_{f \text{ monic}} x^{\deg f} = \sum_{n \ge 0} q^n x^n = \frac{1}{1-qx}.$$ The Euler product identity tells us that $$\frac{1}{1-qx} = \prod_{P\text{ monic irreducible}} (1-x^{\deg P})^{-1} =\prod_{n} (1-x^n)^{-M(n)}.$$ Taking its reciprocal, we find that $$1-qx = \prod_{P\text{ monic irreducible}} (1-x^{\deg P}).$$ Now it is just a matter of comparing coefficients on both sides.
Interpretation: Let $\mu: \mathbb{F}_q[X] \to \mathbb{C}$ be the polynomial Möbius function, defined by $$\mu(f) = \begin{cases} 0 & f \text{ not squarefree} \\ (-1)^k & f=p_1 \cdots p_k (p_i \text{ distinct irreducibles}) \end{cases}.$$ The term $\prod_{i=1}^{s} \binom{M(n_i)}{e_i}$ counts the number of monic, squarefree polynomials of degree $n$ whose factorization consists of $e_i$ irreducibles of degree $n_i$. The polynomial Möbius function $\mu(\bullet)$ assumes the value $(-1)^{\sum e_i}$ for each such polynomial. In other words, your sum may be rewritten as $$\sum_{f \text{ monic, squarefree of degree }n} \mu(f).$$ Since $\mu(f)=0$ for $f$ which is not squarefree, your claim is the same as $$n>1 \implies \sum_{f \text{ monic of degree }n} \mu(f) = 0.$$ This is classical, and due to L. Carlitz, whose proof was exactly as above. It should be compared with the following equivalent formulation of the Riemann Hypothesis: $$\sum_{n \le x } \mu(n) = O(x^{\frac{1}{2}+\varepsilon}),$$ where this time $\mu$ is the integer Möbius function.
It is interesting to ask whether $\sum_{f\text { monic of degree }n} \mu(f)$ may be evaluated without the use of formal power series and the Euler product identity. I will present such a way. Let $M_q$ denote the set of monics in $\mathbb{F}_q[X]$. Given two functions $\alpha, \beta : M_q \to \mathbb{C}$, one defines their "Dirichlet convolution" as the following function: $$(\alpha*\beta) (f) = \sum_{d_1 \cdot d_2 = f} \alpha(d_1) \beta(d_2).$$ Let $\zeta$ denote the constant function $1$ on $M_q$. Let $\delta_{1}$ denote the indicator function of the constant polynomial $1$. The "defining" property of $\mu$ is $$\mu * \zeta = \delta_1.$$ By summing the above over all monic polynomials of degree $n$ (>0) and changing the order of summation, we get that $$\sum_{\deg d \le n} \mu(d) q^{n-\deg d} = 0,$$ or equivalently $$\sum_{\deg d \le n} \frac{\mu(d)}{q^{\deg d}} =0.$$ Plugging $n=m,m+1$ in the above and subtracting, we get that $$m>0 \implies \sum_{\deg d = m+1} \mu(d) = 0.$$
Here is a reformulation of Ofir Gorodetsky's excellent answer to my question: Let $(a(n))_{n \geq 1}$ be a sequence of rational numbers. Define the transformed sequence $T(a)$ of $a$ to have $n$th element \begin{equation*} T(a)(n) = \sum_{n_1^{e_1} \cdots n_s^{e_s} \in P(n)} (-1)^{\sum e_i} \prod_{i=1}^s \binom{A(n_i)}{e_i}, \qquad n \geq 1 \end{equation*} Then \begin{equation*} 1+ \sum_{n \geq 1} T(a)(n)x^n = \prod_{n \geq 1} (1-x^n)^{a(n)} \end{equation*} We now apply this to the sequence $M(n)(q)$. From the classical formula \begin{equation*} \frac{1}{1-qx} = \prod_{n \geq 1} (1-x^n)^{-M(n)(q)} \end{equation*} we get that \begin{equation*} 1+ \sum_{n \geq 1} T(M(n)(q))x^n = \prod_{n \geq 1} (1-x^n)^{M(n)(q)} = 1-qx \end{equation*} This shows that $T(M)(1)=-q$ and $T(M)(n)=0$ for all $n>1$.


Clearly all fractions are of that it can't be the terminating decimals because they're rational. How To Find Irrational Numbers Between Two Decimals DOWNLOAD IMAGE. 0.5 can be written as ½ or 5/10, and any terminating decimal is a rational number. The vast majority are irrational numbers, never-ending decimals that cannot be written as fractions. The only candidates here are the irrational roots. Rational numbers. In mathematics, a number is rational if you can write it as a ratio of two integers, in other words in a form a/b where a and b are integers, and b is not zero. 9 years ago. Include the decimal approximation of the ...” in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. Learn the difference between rational and irrational numbers, and watch a video about ratios and rates Rational Numbers. Take this example: √8= 2.828. In short, rational numbers are whole numbers, fractions, and decimals — the numbers we use in our daily lives.. We can actually split this into thirds. 10 How to Write Fractions Between Two Decimal Numbers. But there's at least one, so that gives you an idea that you can't really say that there are fewer irrational numbers than rational numbers. A set of real numbers is uncountable. Real numbers also include fraction and decimal numbers. Practice Problems 1. And these non recurring decimals can never be converted to fractions and they are called as irrational numbers. Yes. So the question states: what is a decimal number between each of the following pairs of rational numbers and then it gives the fractions -5/6, 1 and -17/20 and -4/5! (Examples: Being able to determine the value of the √2 on a number line lies between 1 and 2, more accurately, between 1.4 … So number of irrational numbers between the numbers should be infinite or something finite which we cannot tell ? In summary, this is a basic overview of the number classification system, as you move to advanced math, you will encounter complex numbers. Method of finding a number lying exactly midway between 2 given numbers, is add those 2 numbers & divide by 2. Rational and Irrational numbers both are real numbers but different with respect to their properties. To find the rational number between two decimals, we can simply look for the average of the two decimals. Step 3: Place the repeating digit(s) to the left of the decimal point. Irrational Numbers on a Number Line. In other words, irrational numbers require an infinite number of decimal digits to write—and these digits never form patterns that allow you to predict what the next one will be. Make use of this online rational or irrational number calculator to ensure the rationality and find its value. ⅔ is an example of rational numbers whereas √2 is an irrational number. But rational numbers are actually rare among all numbers. Irrational numbers tend to have endless non-repeating digits after the decimal point. Which means that the only way to find the next digit is to calculate it. Step 5: Using the two equations you found in step 3 and step 4, subtract the left sides of the two equations. Examples of Rational and Irrational Numbers For Rational. They can be written as a ratio of two integers. O.13 < 3/n < 0.14 13/100 < 3/n < 14/100 When is 13/100 < 3/n ? Irrational numbers' decimal representation is non terminating , non repeating.. From the below figure, we can see the irrational number is √2. Representation of irrational numbers on a number line. 1/3. An Irrational Number is a real number that cannot be written as a simple fraction.. Irrational means not Rational. Many people are surprised to know that a repeating decimal is a rational number. What is the Difference Between Rational and Irrational Numbers , Intermediate Algebra , Lesson 12 - Duration: 3:03. So irrational number is a number that is not rational that means it is a number that cannot be written in the form $$\frac{p}{q}$$. The decimal expansion of a rational number always either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over. one irrational number between 2 and 3 . Some decimals terminate which means the decimals do not recur, they just stop. It makes no sense!!! Solution: If a and b are two positive numbers such that ab is not a perfect square then : i ) A rational number between and . List Of Irrational Numbers 1 100. Step 4: Place the repeating digit(s) to the right of the decimal point. since 3^2 = 9 and 6^2 = 36, √41 > 6. apply the same kind of thinking to the numbers in B and they are all between 3 and 6. Wednesday, October 14, 2020 Find two rational numbers between decimals Ex: Find two rational numbers that have 3 in their numerator and are in between 0.13 and 0.14 with no calculator. Irrational Number between Two Rational Numbers. Rational Number is defined as the number which can be written in a ratio of two integers. Number System Notes. I can approximately locate irrational numbers on a number line ; I can estimate the value of expression involving irrational numbers using rational numbers. On the other hand, all the surds and non-repeating decimals are irrational numbers. The major difference between rational and irrational numbers is that all the perfect squares, terminating decimals and repeating decimals are rational numbers. Subtract the rounded numbers to obtain the estimated difference of 0.5; The actual difference of 0.988 - 0.53 is 0.458; Some uses of rounding are: Checking to see if you have enough money to buy what you want. The decimal expansion of an irrational number continues without repeating. I tried to cross multiply like my teacher said but that was for finding what lay in between fractions it worked for that I have no idea how to do this. Getting a rough idea of the correct answer to a problem Irrational Numbers. A rational number is the one which can be represented in the form of P/Q where P and Q are integers and Q ≠ 0. They include the counting numbers and all other numbers that can be written as fractions. Get an answer to your question “Part B: Find an irrational number that is between 9.5 and 9.7.Explain why it is irrational. Lv 6. The first number we have here is five, and so five is five to the right of zero, five is right over there. This amenability to being written down makes rational numbers the ones we know best. Learn Math Tutorials 497,805 views When placing irrational numbers on a number line, note that your placement will not be exact, but a very close estimation. New Proof Settles How To Approximate Numbers Like Pi Quanta Magazine. Rational numbers are contrasted with irrational numbers - numbers such as Pi, √ 2, √ 7, other roots, sines, cosines, and logarithms of numbers. An irrational number is a number which cannot be expressed in a ratio of two integers. How many irrational numbers can exist between two rational numbers ? So 1/3 is between zero and one. That's our five. Irrational Numbers. A rational number is of the form $$\frac{p}{q}$$, p = numerator, q= denominator, where p and q are integers and q ≠0.. Irrational Numbers. The ability to convert between fractions and decimals, and to approximate irrational numbers with decimals or fractions, can be very helpful in solving problems. 0 0. cryptogramcorner. Example: Find two irrational numbers between 2 and 3. Examine the repeating decimal to find the repeating digit(s). Key Differences Between Rational and Irrational Numbers. A rational number is a number that can be written as a ratio. As there are infinite numbers between two rational numbers and also there are infinite rational numbers between two rational numbers. DOWNLOAD IMAGE. Are there any decimals that do not stop or repeat? Real numbers include natural numbers, whole numbers, integers, rational numbers and irrational numbers. - [Voiceover] Plot the following numbers on the number line. But an irrational number cannot be written in the form of simple fractions. The number $\pi$ (the Greek letter pi, pronounced ‘pie’), which is very important in describing circles, has a decimal form that does not stop or repeat. Essentially, irrational numbers can be written as decimals but as a ratio of two integers. Terminating, recurring and irrational decimals. Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. The difference between rational and irrational numbers can be drawn clearly on the following grounds. Hence, almost all real numbers are irrational. A Rational Number can be written as a Ratio of two integers (ie a simple fraction). how cuanto mide una cama matrimonial haunted house in san diego that lasts 4 7 hours journey to the west full movie with english subtitles. To study irrational numbers one has to first understand what are rational numbers. Then we get 1/3. The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. Difference between rational and irrational numbers has been clearly explained in the picture given below. DOWNLOAD IMAGE. Difference between Rational and Irrational Numbers. m/n —> 3/n N has to be an integer, n cannot be 0. For instance, when placing √15 (which is 3.87), it is best to place the dot on the number line at a place in between 3 and 4 (closer to 4), and then write √15 above it. Suppose we have two rational numbers a and b, then the irrational numbers between those two will be, √ab. . Irrational numbers include √2, π, e, and θ. Rational Numbers. Irrational numbers are those which do not terminate and have no repeating pattern. 2 and 3 are rational numbers and is not a perfect square. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers. Example 15: Find a rational number and also an irrational number between the numbers a and b given below: a = 0.101001000100001…., b = 0.1001000100001… Solution: Since the decimal representations of a and b are non-terminating and non-repeating. Irrational Numbers: We have different types of numbers in our number system such as whole numbers, real numbers, rational numbers, etc. Rational numbers are whole numbers, fractions, and decimals - the numbers we use in our daily lives. ii) An irrational number between and . How to Write Irrational Numbers as Decimals Before studying the irrational numbers, let us define the rational numbers. Apart from these number systems we have Irrational Numbers. You can use the decimal module for arbitrary precision numbers: import decimal d2 = decimal.Decimal(2) # Add a context with an arbitrary precision of 100 dot100 = decimal.Context(prec=100) print d2.sqrt(dot100) If you need the same kind of ability coupled to speed, there are some other options: [gmpy], 2, cdecimal.

## how to find irrational numbers between decimals

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# What consitutes an exponential function?

I was recently having a discussion with someone, and we found that we could not agree on what an exponential function is, and thus we could not agree on what exponential growth is.

Wikipedia claims it is $e^x$, whereas I thought it was $k^x$, where k could be any unchanging number. For example, when I'm doing Computer Science classes, I would do everything using base 2. Is $2^x$ not an exponential function? The classical example of exponential growth is something that doubles every increment, which is perfectly fulfilled by $2^x$. I'd also thought $10^x$ was a common case of exponential growth, that is, increasing by an order of magnitude each time. Or am I wrong in this, and only things that follow the natural exponential are exponential equations, and thus examples of exponential growth?

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Whether one thinks of exponential growth as $e^{ct}$ ($c$ positive) or $a^t$ ($a\t 1$), one is dealing with the same phenomenon, just take $a=e^c$. –  André Nicolas Jun 21 '12 at 0:45

$e^x$ is the exponential function, but $c\cdot k^x$ is an exponential function for any $k$ ($> 0, \ne1$) and $c$ ($\ne 0$).

The terminology is a bit confusing, but is so well settled that one just has to get used to it.

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Therefore, I can refer to anything that can be described as $c⋅k^x$ as exponential growth, correct? –  Canageek Jun 20 '12 at 23:24
You can do whatever you want. The question is whether people will understand you, and in this case I think most people will understand. –  Qiaochu Yuan Jun 20 '12 at 23:26
The discession was based around the incorrect use of exponential growth (i.e. Hollywood usage) so yeah, I think I'll stick with the proper definitions. Thank you, hope the question wasn't too basic. –  Canageek Jun 21 '12 at 0:49
Sometimes, to highlight how special $\exp\,x$ is, it is given the adjective natural, as in "natural exponential function". –  Ｊ. Ｍ. Jun 21 '12 at 3:20
$c\cdot k^x=c \cdot e^{x\cdot \ln k}$, so if constants in the exponent don't bother you then sure. –  Robert Mastragostino Jun 21 '12 at 3:36

If $x$ has units (e.g. time), then there's no way to distinguish between these possibilities; they're all equivalent up to change of units.

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Predictive Hacks

# Estimate Pi with Monte Carlo

One method to estimate the value of $$\pi$$ is by applying the Monte Carlo method. Let’s consider a circle inscribed in square. Then for a radious $$r$$ we have:

Area of the Circle = $$\pi r^2$$

Are of the Square = $$4 r^2$$

Hence, we can say that $$\frac{AreaCircle}{AreaSquare}=Prob=\frac{\pi}{4}$$. This implies that $$\pi = 4Prob$$

We know that the Equation of Circle with center (j,k) and radius (r) is:

$$(x_1-j)^2+(x_2-k)^2 = r^2$$ where in our case the center is (0,0) and the radius is 1. This means, that in order to calculate the Area of the Circle we need to consider all the points from the uniform distribution which have a distance from the center (0,0) less than or equal to 1, i.e:

$$\sqrt{ x_1^2+x_2^2} \leq 1$$

Notice that since we chose the center to be (0,0) and the radius to be equal to 1, it means that the co-ordinates of the square will be from -1 to 1. For that reason we simulate from the uniform distribution with min and max values the -1 and 1 respectively.

## Example Using R

Below, we represent how we can apply the Monte Carlo Method in R to estimate the pi.

# set the seed for reproducility
set.seed(5)

# number of simulations
n=1000000

# generate the x1 and x2 co-ordinates from uniform
# distribution taking values from -1 to 1
x1<-runif(n, min=-1, max=1)
x2<-runif(n, min=-1, max=1)

# Distrance of the points from the center (0,0)
z<-sqrt(x1^2+x2^2)

# the area within the circle is all the z
# which are smaller than the radius^2
4*sum((z<=1))/length(z)

[1] 3.14204

Let’s have a look also at the simulated data points.

InOut<-as.factor(ifelse(z<=1, "In", "Out"))

plot(x1,x2, col=InOut, main="Estimate PI with Monte Carlo")

As we can see, with 1M simulations we estimated the PI to be equal to 3.14204. If want to get a more precise estimation we can increase the number of simulations.

## Example Using Python

import pandas as pd
import numpy as np
import math
import seaborn as sns
%matplotlib inline

# number of simulations
n=1000000

x1=np.random.uniform(low=-1,high=1, size=n)
x2=np.random.uniform(low=-1,high=1, size=n)

z=np.sqrt((x1**2)+(x2**2))

print(4*sum(z<=1)/n)

3.143288

Let’s have a look also at the simulated data points.

inout=np.where(z<=1,'in','out')

sns.scatterplot(x1,x2,hue=inout,legend=False)

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Python

#### How to Process Requests in Flask

The most common payload of incoming data are the query strings, the form data and the JSON objects. Let’s provide

Miscellaneous


# Converting a probability density function into its Probability Distribution

I have the following probability density function

$$f(x) =\begin{cases} 4x & \mbox{for }0< x < 1/2 \\ 4-4x & \mbox{for }1/2 \leq x < 1 & \\ 0 & \mbox{otherwise}\end{cases}$$

I am tasked to now find its probability distribution function in the same piece wise format. I have the solution to this problem however I do not really understand the solution.

The solution is the integral from 0 to x of 4x(for the first interval of 0 < x<= 1/2)+ integral of 0 to 1/2 of 4x + integral of 1/2 to x of 4-4x(for the second interval).

I understand the first interval but I am stumped as to why we add the integral of 0 to 1/2 of 4x to the integral of 1/2 to x for the second interval.

Any explanation would be appreciated.

• If you want the cumulative probability, there is a probability of $\int_0^{1/2} 4x \, dx =\frac12$ of being $\frac12$ or less, which you have to add to the probability of being from $\frac12$ up to the value you are interested in – Henry Feb 17 '16 at 23:23
• Formatting tips here. – Em. Feb 17 '16 at 23:30

I guess it is asking for the CDF. It's hard to tell what you are writing since you didn't format.

Essentially, it should be

$$F_X(x) = P(X\leq x)=\begin{cases} 0& x<0\\ \int_0^x 4t\,dt& 0\leq x< \frac{1}{2}\\ \int_0^{1/2} 4t\,dt+\int_{1/2}^x4-4t\,dt&\frac{1}{2}\leq x <1\\ 1& x\geq 1\end{cases}$$

Loosely speaking, this is the case because the CDF is the accumulation of probability (area) under the curve up to $x$.

• So x is always defined as the point at which we are accumulating to then? As I am confused when to include actual integers as the end points vs the variable x. That is I am confused as what x is when talking about the area under the curves of these functions. – JmanxC Feb 17 '16 at 23:51
• This is an odd example to try to understand from. Take instead $X$ to follow a continuous unif(0,1). The cdf of $X$ is $$F_X(x)=P(X\leq x) = \int_0^xf_X(t)\,dt = \int_0^x 1\,dt = x$$ when $0\leq x < 1$. Does this help? Do you see that this is the area under the curve $f_X(x)$ up to $x$? – Em. Feb 17 '16 at 23:58
• So x is really the end point of a specific interval? So if it was a piece wise from 0 to lets say 5 for f1(x) and 5 to 10 for f2(x) you would add from 0 to x for the first interval then from 5 onward to x for the next interval? – JmanxC Feb 18 '16 at 0:02
• I believe that is correct. It would be easier to tell if you format. But, yes naively speaking the cdf $F_X(x) =P(X\leq x)$ is the "area to the left" under the pdf. To see this using a piecewise function, use your current exercise. Graph the pdf, and notice that if $0\leq x <\frac{1}{2}$, then the area to the left of $x$ is the first intergal I provided. If $\frac{1}{2}\leq x<1$, then the area to the left of $x$ is the sum of integrals I provided. – Em. Feb 18 '16 at 0:09
• Perfect thank you. And yes I realize I'm not formatting correctly but I think I get it. You've been a big a help. I am starting to get it in a 1d environment but I soon have to look at it in a 2d environment(joint pdf's). May have more questions on this later that you may see on here haha. – JmanxC Feb 18 '16 at 0:13

You have $$f(x) =\begin{cases} 4x & \mbox{for }0< x < 1/2 \\ 4-4x & \mbox{for }1/2 \leq x < 1 & \\ 0 & \mbox{otherwise}\end{cases}$$

Then you want to find

$$F(x) =\begin{cases} 0 & \mbox{for } x\leq 0 \\[1ex] \int_0^x 4s\operatorname d s & \mbox{for }0< x < 1/2 \\[1ex] \int_0^{1/2} s\operatorname d s + \int_{1/2}^x 4-4s\operatorname d s & \mbox{for }1/2 \leq x < 1 & \\[1ex] 1 & \mbox{for } 1\leq x\end{cases}$$

• You read my mind :p – Em. Feb 17 '16 at 23:29
• Hi Graham I have left a comment for @probablyme above(the last comment) which you may be able to answer as well as I know you have helped me with a few questions on this site. – JmanxC Feb 18 '16 at 2:04