# Math Help - dice probability question

1. ## dice probability question

This is a revision question for a phase test that i could sure use some help with!

A dice game consitsts of throwing 2 six sided dice and noting the score on them. If the score is greater than 9 or at least one dice shows a 6 then you score 3 points, otherwise you lose 1 point. Calculate the probability of gaining 3 points on a single throw. What is the average (or expected) number of points you will gain per throw.

i make the probability of scoreing 3 points as 12 over 36, but i dont know how to do the rest.

any suggestions?

Cheers

2. Originally Posted by marciano1976
This is a revision question for a phase test that i could sure use some help with!

A dice game consitsts of throwing 2 six sided dice and noting the score on them. If the score is greater than 9 or at least one dice shows a 6 then you score 3 points, otherwise you lose 1 point. Calculate the probability of gaining 3 points on a single throw. What is the average (or expected) number of points you will gain per throw.

i make the probability of scoreing 3 points as 12 over 36, but i dont know how to do the rest.

any suggestions?

Cheers
Colour the die red and blue, then the combinations that win are (r6,b4), (r6,b5), (r6,b6), (r5,b5), (r5,b6), (r4,b6), (r6,b1), (r6,b2), (r6,b3), (r1,b6), (r2,b6), (r3,b6). That is 12 winning configurations out of a total of 36.

p(3)=12/36=1/3

p(1)=1-p(3)=2/3

So the expected number of points is:

(-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3

RonL

3. Originally Posted by CaptainBlack
So the expected number of points is:
(1)p(1)+(3)p(3)=(2/3)+(3/3)=5/3
Surely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3

4. Hello, marciano1976!

A dice game consists of throwing two six-sided dice and noting the score on them.
If the score is greater than 9 or at least one dice shows a 6 then you score 3 points,
otherwise you lose 1 point.

(a)Calculate the probability of gaining 3 points on a single throw.

(b) What is the average (or expected) number of points you will gain per throw?

There are 36 possible outcomes . . .

$\begin{array}{cccccc}(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & {\color{blue}(1,6)} \\
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & {\color{blue}(2,6)} \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & {\color{blue}(3,6)} \\
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & {\color{blue}(4,6)}
\\ (5,1) & (5,2) & (5,3) & (5,4) & {\color{blue}(5,5)} & {\color{blue}(5,6)}
\end{array}$

$\begin{array}{cccccc}{\color{blue}(6,1)} & {\color{blue}(6,2)} & {\color{blue}(6,3)} & {\color{blue}(6,4)} & {\color{blue}(6,5)} & {\color{blue}(6,6)} \end{array}$

There are 12 in which the sum is greater than 9 or there is at least one 6.

Therefore: . $(a)\;\;P(\text{gain 3 points}) \:=\:\frac{12}{36} \:=\:\frac{1}{3}$

Hence: . $\begin{array}{ccc}P(\text{win 3 points}) &=& \frac{1}{3} \\ P(\text{lose 1 point}) & = & \frac{2}{3} \end{array}$

And: . $(b)\;\;\text{Average} \;=\;(+3)\left(\frac{1}{3}\right) + (-1)\left(\frac{2}{3}\right) \:=\:\frac{1}{3}$

You can expect to win an average of $\frac{1}{3}$ of a point per throw.

. . . as Plato already pointed out.
.

5. Originally Posted by Plato
Surely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3
Of course it should, I misread the question (again)

RonL


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# 6^(1/2)/(1/2^(1/2)+1/3^(1/2))

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$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$

A. $$\frac{6}{5}$$

B. $$\frac{5}{6}$$

C. 5

D. $$2\sqrt{3} +3\sqrt{2}$$

E. $$6\sqrt{3}-6\sqrt{2}$$

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The trick behind this question is to use the answer choices to guide your mathematics. (In other words, don't do math just because you can... do math because it gets you closer to your target!)

You should notice immediately that the answer choices are much simpler. Three of the five answer choices contain no fractions at all. None of the answer choices contain $$\sqrt{6}$$. Use this to your advantage to help you think about the question. One quick way to get rid of both the \sqrt{6}[/m] and the "fractions inside of fractions" issue is to do what I call "Multiply by 1". In this case, multiply the original expression by $$\sqrt{6}/\sqrt{6}$$. Since $$\sqrt{6}/\sqrt{6}=1$$, we don't actually change the value; we just change the shape of the expression. Thus,

$$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}*(\frac{\sqrt{6}}{\sqrt{6}})=\frac{6}{\sqrt{3}+\sqrt{2}}$$

Now the expression that remains is starting to look more like the answer choices, but we still need to get rid of the compound denominator $$(\sqrt{3}+\sqrt{2})$$. Once again, look to the answer choices for clues. Both answer choices A and E contain a $$6$$. But it should be clear that answer choice A cannot be an option. $$(\sqrt{3}+\sqrt{2})$$ does not equal 5. So, let's try to turn what we have into answer choice E.

Answer choice E contains a factor of $$(\sqrt{3}-\sqrt{2})$$, which should immediately get us thinking about the possibility of difference of squares. Multiplying our reduced expression by "1" again gives us:

$$(\frac{6}{\sqrt{3}+\sqrt{2}})*(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}})$$

Using the concept of difference of squares, we can quickly see that the denominator of this fraction simplifies down completely. Watch this "disappearing math" trick:

$$(\sqrt{3}+\sqrt{2})*(\sqrt{3}-\sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3-2 = 1$$

What began as a "Mathugly" arithmetic issue quickly simplifies down to $$6\sqrt{3}-6\sqrt{2}$$.

The answer is E.

Addendum: As with many quantitative questions on the GMAT, there is "more than one way to skin a CAT" for this question (pun intended.) We can also use the answer choices not just as targets that guide our math, but as leverage for approximating. Watch this. If we recognize that $$\sqrt{3}\approx{1.7}$$ and $$\sqrt{2}\approx{1.4}$$, then the we can approximate our answer very quickly:
$$\frac{6}{\sqrt{3}+\sqrt{2}}\approx{\frac{6}{1.4+1.7}}\approx{\frac{6}{3.1}}\approx{2}$$

Now, we look at the answer choices to see which one is roughly equal to 2.

(A) $$\frac{6}{5}$$ <-- quickly eliminated. Too small.

(B) $$\frac{5}{6}$$ <-- quickly eliminated. Too small.

(C) $$5$$ <-- quickly eliminated. Too big.

(D) $$2\sqrt{3} + 3\sqrt{2}$$ <-- quickly eliminated. Too big.

(E) $$6\sqrt{3} - 6\sqrt{2}$$ <-- The right answer. $$6(\sqrt{3} - \sqrt{2})\approx{6(1.7-1.4)}\approx{2}$$

Anyway you look at it, the answer is still E.
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Originally posted by AaronPond on 08 Jan 2018, 17:04.
Last edited by AaronPond on 09 Jan 2018, 07:13, edited 1 time in total.
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AbdurRakib wrote:
$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$

A. $$\frac{6}{5}$$

B. $$\frac{5}{6}$$

C. 5

D. $$2\sqrt{3} +3\sqrt{2}$$

E. $$6\sqrt{3}-6\sqrt{2}$$

$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=\frac{\sqrt{6}}{(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}\sqrt{3}})}=\frac{\sqrt{6}}{(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{6}})}=\frac{\sqrt{6}*\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{6}{\sqrt{3}+\sqrt{2}}$$.

Multiply by $$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$:

$$\frac{6}{\sqrt{3}+\sqrt{2}}*\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{6*(\sqrt{3}-\sqrt{2})}{3-2}=6*\sqrt{3}-6\sqrt{2}$$.

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AbdurRakib wrote:
$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$

A. $$\frac{6}{5}$$

B. $$\frac{5}{6}$$

C. 5

D. $$2\sqrt{3} +3\sqrt{2}$$

E. $$6\sqrt{3}-6\sqrt{2}$$

Nice question!

Here's another approach:

Let's first deal with the denominator: 1/√2 + 1/√3 = √3/√6 + √2/√6
= (√3 + √2)/√6

So....$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$ √6/[(√3 + √2)/√6]

= (√6)(√6)/(√3 + √2)

= 6/(√3 + √2)

Now we can use the fact that √3 ≈ 1.7 and √2 ≈ 1.4 to get....
≈ 6/(1.7 + 1.4)
≈ 6/(3.1)

Since 6/3 = 2, we know that 6/(3.1) = a number a bit smaller than 2

Now check the answer choices to see which one evaluates to be a number a bit smaller than 2

A. $$\frac{6}{5}$$ ELIMINATE

B. $$\frac{5}{6}$$ ELIMINATE

C. 5 ELIMINATE

D. $$2\sqrt{3} +3\sqrt{2}$$ ≈ 2(1.7) + 3(1.4) ≈ a number that's bigger than 6 ELIMINATE

E. $$6\sqrt{3}-6\sqrt{2}$$[/quote] = 6(√3 - √2) ≈ 6(1.7 - 1.4) ≈ 6(0.3) ≈ 1.8 PERFECT!

Cheers,
Brent
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08 Jan 2018, 16:36
$$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}=$$

(A) $$\frac{6}{5}$$

(B) $$\frac{5}{6}$$

(C) $$5$$

(D) $$2\sqrt{3} + 3\sqrt{2}$$

(E) $$6\sqrt{3} - 6\sqrt{2}$$
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08 Jan 2018, 20:33
AaronPond wrote:
$$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}=$$

(A) $$\frac{6}{5}$$

(B) $$\frac{5}{6}$$

(C) $$5$$

(D) $$2\sqrt{3} + 3\sqrt{2}$$

(E) $$6\sqrt{3} - 6\sqrt{2}$$

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Re: 6^(1/2)/(1/2^(1/2)+1/3^(1/2)) &nbs [#permalink] 08 Jan 2018, 20:33
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# Help with linear algebra linear transformations

Is there a linear transformation $T : R ^2 → R ^3$ such that $T(1, 1) = (1, 0, 2)$ and $T(2, 3) = (1, −1, 4)$. Justify your answer.

I'm not sure what exactly this question is asking for. How would you go about tackling this question. I know I'm supposed to show some form of working on this website but I really have no idea where to start

• Try to start with the assumption that there is such a transformation. Then try to explicitly construct it by finding a matrix $A$ such that $T(x) = Ax$ for all $x\in \Bbb R^2$. If you're able to, then one clearly exists. If you run into some contradiction, then one clearly doesn't exist. – user137731 Oct 4 '15 at 4:25
• As always, reviewing the terms in the problem (what a linear transformation is) can help. – Nate 8 Oct 4 '15 at 4:30

In general if V and W are vector spaces and $\{e_i\}_{i\in I}$ is a basis for $V$. And $f(e_i)=w_i\in W$ for all $i \in I$.

Then f can be extended to a linear mapping $T: V \rightarrow W$ such that $T(e_i)=f(e_i)$. (You can write down the proof in the same way as the above answers)

In your case, $V= \mathbb R ^2$ and $W=\mathbb R^3$ $e_1=(1,1), e_2=(2,3)$ is a basis for $\mathbb R^2$ and $f(e_1)=(1,0,2)$ and $f(e_2)=(1,-1,4)$.

Therefore, you can extend it to a linear mapping by the above discussion.

Note: 1) $(1,0,2)$ and $(1,-1,4)$ doesn't have any significance role to play.If $(a,b,c), (d,e,f)$ still you could extend it to a linear map.

2) The above problem has a general philosophy - any map (or morphism) from a algebraic(mathematical) structure to another algebraic(mathematical) structure can be defined by only defining it on the generators, provided you can talk of generators. For example, to define a group homomorphism from a cyclic to group to another group, it is enough to define it for just for the generators.

Note that $\{(1,1),(2,3)\}$ is a basis of $\mathbb R^2$ and if you know the image of basis elements then you know the whole transformation.

In this case observe $\forall x,y \in \mathbb R^2$ we have $(x,y)= (3x-2y)(1,1)+(y-x)(2,3)$

So define , $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$

To answer the comment by Babai:

The construction of $T$ shows that if $u\in \mathbb R^2,u=\alpha a+\beta b$ where $a=(1,1)$ and $b=(2,3)$ then $Tu=\alpha Ta+\beta Tb$

Thus $T(u_1+\lambda u_2)=T(\alpha_1 a+\beta_2 b+\lambda (\alpha_2 a+\beta_2 b))=T((\alpha_1+\lambda \alpha_2) a+(\beta_1+\lambda \beta_2) b)=(\alpha_1+\lambda \alpha_2) Ta+(\beta_1+\lambda \beta_2) Tb=\alpha_1 Ta+\beta_2 Tb+\lambda (\alpha_2 Ta+\beta_2 Tb)=Tu_1+\lambda Tu_2$

• The question is why is T linear? In your answer you assumed T is linear. – Babai Oct 4 '15 at 4:41
• @Babai I have added it in my answer. – usermath Oct 4 '15 at 5:22
• $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$. Why is this true? This is only true if $T$ is linear, and that is what you nee to prove. – Babai Oct 4 '15 at 5:38
• Instead of writing " So $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$, if you write "define $T$ as $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$, then your answer will make perfect sense. – Babai Oct 4 '15 at 5:45
• @Babai ok I got it and corrected. Thnx – usermath Oct 4 '15 at 6:27

Define $T: \mathbb{R}^2 \to \mathbb{R}^3$ by for $(x,y) \in \mathbb{R}^2$, write $(x,y) = a(1,1) + b(2,3)$, and let $T(x,y) = a(1,0,2) + b(1,-1,4).$

In the above, $T$ is obviously linear. The problem is why such a $T$ is well-defined, more specifically, why there exists such scalars $a,b$ and the scalars are unique.

• Why is T obviously linear? – Babai Oct 4 '15 at 4:42
• @Babai Have you tried to show it before you ask this question? It is a one line proof. – Empiricist Oct 4 '15 at 5:17
• It is not the question of whether I have tired it or not. Your answer is not justifiable. You write "$T(x,y) = a(1,0,2) + b(1,-1,4).$, In the above, T is obviously linear." Now, $T(x,y) = a(1,0,2) + b(1,-1,4).$ , this is not true unless $T$ is linear. And you assumed it to say $T$ is linear, which is not correct. – Babai Oct 4 '15 at 5:35
• I am sorry. You are defining $T$. Your answer is perfect. And it is obviously linear. – Babai Oct 4 '15 at 5:44

If there exists a transform, there must exist a matrix representing it, let us call such a matrix $\bf T$. Now what we demand is that $${\bf T}\left[\begin{array}{cc}1&2\\1&3\end{array}\right] = \left[\begin{array}{rr}1&1\\0&-1\\2&4\end{array}\right]$$ Now we can solve this, if we divide by $\left[\begin{array}{cc}1&2\\1&3\end{array}\right]$ to the left of each side: $${\bf T} = \left[\begin{array}{rr}1&1\\0&-1\\2&4\end{array}\right] \left[\begin{array}{cc}1&2\\1&3\end{array}\right]^{-1} = ?$$

Now we can see on the matrices that this must have a solution since both matrices are of rank 2.

Then after we have solved this, what remains is to perform the matrix multiplication ${\bf T}\left[\begin{array}{cc}1&2\\1&3\end{array}\right]$ and confirm it really becomes as the first equation says it should.


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The most common, most frequently used math symbols: Forward Slash (Division Sign) shown and explained . Please be sure to answer the question. How does one insert a "\" (backslash) into the text of a LaTeX document? Thanks for contributing an answer to TeX - LaTeX Stack Exchange!

The difference between a backslash and a forward slash is defined below: Backslash: \ Forward Slash: / A good way to remember the difference between a backslash and a forward slash is that a backslash leans backwards ( \ ), while a forward slash leans forward ( / ).

P. PvtBillPilgrim. Infix operator. If you like this Site about Solving Math Problems, please let Google know by clicking the +1 button. ... [Backslash] Unicode: 2216.

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The backslash symbol \ is used to denote a set difference, quotient group, or integer division. Also used to separate arguments of elliptic functions. Ask Question Asked 5 years, 5 months ago. Formally: ∖ = {∈ ∣ ∉}.

Here, the backslash symbol is defined as Unicode U+2216. At any rate, math mode provides \sim, \backslash, and \setminus (the latter two appear to look the same and differ only by spacing in math mode).

You just have to type the name of the letter after a backslash: if the first letter is lowercase, you will get a lowercase Greek letter, if the first letter is uppercase (and only the first letter), then you will get an uppercase letter. What does the backslash denote in probability theory? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Applied Mathematics. Ask Question Asked 7 years, 6 months ago. \$\begingroup\$ The backslash is an escape character in \$\LaTeX\$, cf.

... Backslash notation: … is used to separate the whole part of a number from the fractional part. What is Backslash? In this particular example, the norm of the residual is zero, and an exact solution is obtained, although rcond is small. Making statements based on opinion; back them up with references or personal experience. Revolutionary knowledge-based programming language. x y is by default interpreted as Backslash [x, y].

In an application of a probability problem you could be given an arbitrary probability like: P ( a \ b) = .50 or 50%. For non-square and singular systems, the operation A\b gives the solution in the least squares sense. The relative complement of A in B is denoted B ∖ A according to the ISO 31-11 standard.It is sometimes written B − A, but this notation is ambiguous, as in some contexts it can be interpreted as the set of all elements b − a, where b is taken from B and a from A.. Foundations of Mathematics.


# Length of period of decimal expansion of a fraction

Each rational number (fraction) can be written as a decimal periodic number. Is there a method or hint to derive the length of the period of an arbitrary fraction? For example $1/3=0.3333...=0.(3)$ has a period of length 1.

For example: how to determine the length of a period of $119/13$?

• I can't find an algorithm that finds the value for a number that isn't prime, i.e. 10 – Code Whisperer Jan 13 '18 at 9:11
• – Watson Dec 2 '18 at 16:31

Assuming there are no factors of $2,5$ in the denominator, one way is just to raise $10$ to powers modulo the denominator. When you find $-1$ you are halfway done. Taking your example: $10^2\equiv 9, 10^3\equiv -1, 10^6 \equiv 1 \pmod {13}$ so the repeat of $\frac 1{13}$ is $6$ long. It will always be a factor of Euler's totient function of the denominator. For prime $p$, that is $p-1$.

• You have to look for $1$, because sometimes you won't find $-1$. For example: $10^2\equiv 18, 10^3\equiv 16, 10^4 \equiv 37, 10^5 \equiv 1 \pmod {41}$, so the repeat of $\frac 1{41}$ is $5$ long. – jcsahnwaldt says GoFundMonica Mar 26 '15 at 5:37
• You assume that 'there are no factors of 2,5 in the denominator' is it actually required? – Shamdor May 5 '15 at 7:28
• I mean, 1/(2^i * 5^j * d') has the same period length as 1/d', right? – Shamdor May 5 '15 at 7:46
• That is correct. You will have $\max(i,j)$ non-repeating digits, then the repeat will start – Ross Millikan May 5 '15 at 13:48

This answer seeks to explain why Ross Millikan's answer works, and provides further information on techniques to speed up the process of seeking the period:

Consider the fraction $\frac17$. The decimal expansion of this is $$\frac17 = 0.\overline{142857}$$ for a period of 6. Now consider what happens when we multiply it by $10^6$: $$10^6\times\frac17 = 142857.\overline{142857}$$ Subtracting the original fraction from this gives $$(10^6-1)\times\frac17 = 142857$$ And so, we have $$\frac17 = \frac{142857}{10^6-1}$$ As you can see, the denominator is one less than a power of 10, and the power is the period of the decimal expansion. This is no accident, and works for any fraction - if you can rewrite it in this form, the denominator reveals the period.

Now, rearrange the equation: $$10^6-1 = 142857\times 7$$ So $10^6$ must be one more than a multiple of 7 (or, more generally, $10^n$ must be one more than a multiple of $d$, where $d$ is the denominator of the fraction and $n$ is the period of the decimal expansion) - indeed, it must be the smallest power of 10 (larger than 1) that has this property.

As such, we can use modular arithmetic to look for the period. Since $a\times d\equiv 0 \pmod d$, we have that $10^n-1\equiv0 \pmod d$, or $$10^n \equiv 1\pmod d$$ And therefore you can just look for the smallest $n>0$ satisfying this.

Of course, there are other approaches to gain the same result, but they're all fundamentally variants of the same idea. That said, if you can factor $\phi(d)$ - the euler totient function of the denominator - then you can accelerate the process of looking for the smallest $n$. For example, when checking 13, you have $\phi(13)=12$, so $n\in\{1,2,3,4,6,12\}$ (as these are the factors of 12) - this can save you a lot of computation (especially where $\phi(d)$ has only a few large factors and 2).

For example, $\phi(167)=166 = 2\times83$, so $n\in\{1,2,83,166\}$. Therefore, we need to check only these four, and we can do it quite efficiently. Obviously, neither $10$ nor $100=10^2$ are equivalent to 1 mod 167, so we only need to actually check 83. For this we can use binary exponentiation. Note that $83 = 2^6 + 2^4 + 2^1 + 2^0$. So we can write \begin{align} 10^{83} &= 10^{2\times(2^5 + 2^3 + 1)}\times 10\\ &= (10^{2^3\times(2^2+1)}\times 10)^2 \times 10\\ &= ((10^{2^2}\times10)^{2^3}\times 10)^2 \times 10 \end{align} So, working in modular arithmetic, we can go \begin{align} 10^{83} &\equiv ((10^{2^2}\times10)^{2^3}\times 10)^2 \times 10 \mod 167\\ &\equiv ((100^2\times 10)^{2^3}\times 10)^2\times10\mod167\\ &\equiv ((147\times 10)^{2^3}\times 10)^2\times10\mod167\\ &\equiv (134^{2^3}\times 10)^2\times10\mod167\\ &\equiv (87^{2^2}\times 10)^2\times10\mod167\\ &\equiv (54^2\times 10)^2\times10\mod167\\ &\equiv (77\times 10)^2\times10\mod167\\ &\equiv 102^2\times10\mod167\\ &\equiv 50\times10\mod167\\ &\equiv 166\mod167 \end{align} This is the same as $-1\pmod{167}$, so $n=166$ is the only possible period, and $\frac1{167}$ has a period of 166.

Also note that you don't actually have to expand out the product like that. You can just write the number in binary ($83_{10} = 1010011_2$), then work through the binary digits left-to-right - start with 1, and for each digit, $b$, multiply by $10^b$. Square it after each digit except the last one.

The length of the period is given by the multiplicative order of $10 \pmod q$, where $q$ is your quotient. It is closely related to the discrete logarithm. Wikipedia lists several algorithms that are faster than going through all the powers of ten, which is relevant if you're dealing with very large numbers (hundreds of digits).

First you have to simplify to the lowest integer denominator.

Then, the proof for finiteness of repeating part of the decimal representation involves the pigeonhole principle. When you keep on dividing, at some point you will run into a repeat value for the remainder. Given that the remainders that keep the operation going range from 1 to (denominator-1), the possible number of pigeonholes is (denominator-1) as the number of pigeonholes; so that's your worst possible case scenario - in any base.

The particular cases, as described by the other answers, depend on the base, because the denominator may have a special relationship with the base.

• e.g., in base 10 fractions over 3 will have one repeating decimal, (3) or (6), as 3 divides 9(=10-1), and the remainder will keep repeating.


# Proof Check - Linear Algebra

I'm new to more formal mathematics and I was wondering if I could write my attempt at a small proof here and see what others thought and if you could point out any fatal flaws or just offer some tips? I would appreciate it.

## Theorem

Let u,v, and w be distinct vectors in a vector space V. Show that if {u,v,w} is a basis for V, then {u+v+w, v+w, w} is also a basis for V.

## Proof

: Let B = {u,v,w}. As B is a basis for V, the vectors in B generate V and are linearly independant. Therefore, the equation

a1 * u + a2 * v + a3 * w = 0

Is only true for a1, a2, a3 all equal to 0.

Now consider the set B' = {u+v+w, v + w, w}

If we examine the equation

c1 ( u+v+w) + c2 (v+w) + c3 (w) = 0

c1(u) + (c1 + c2)v + (c1 + c2 + c3)w = 0

Then we say that c1 = -c2, and c3 = -c1-c2, but from the first term we see that c1 must equal zero, therefore all the other coefficients are 0 as well, and so the set of vectors is linearly independant.

## My questions

1.) I'm not sure if the part of the proof where I show that the solution to the equation for the vectors of B' being equal to 0 is connected well enough to the bit about B being a basis for V. Am I allowed to say that C1 must be equal to 0 only because I've already assumed that u,v, and w are all linearly independent already?

2.) I know that if B generates V that B' which is just set formed from L.I. vectors into other L.I. vectors must also generate V, but I'm not sure exactly why I can say this even though I feel it strongly.

• Your argument is perfect. – Ted Shifrin Sep 15 '14 at 1:20

With regards to point 1), what you say is correct. As you say, you use the linear independence of $B$ to conclude that when $c_1(u) + (c_1 + c_2)v + (c_1 + c_2 + c_3)w = 0$, it must be the case that $c_{1} = 0$, and so on.
With regards to 2), one way to show that $B'$ spans the space is to show that it generates $B$: any linear combination of vectors from $B$ can then be converted into a linear combination of vectors from $B'$ by replacing $u,v,w$ by their expressions in terms of the list $B'$. So in this case $(v+w) - w = v$ and $(u+v+w) - (v+w) = u$.
By the way, you only need to show either that $B'$ form a linear independent set, or that they span the vector space, since the list has length $3$. (This is the result that a linear independent list of vectors, whose length is equal to the dimension of the space forms a basis, and similarly for spanning sets).


# When is the closure of an open ball equal to the closed ball?

It is not necessarily true that the closure of an open ball $B_{r}(x)$ is equal to the closed ball of the same radius $r$ centered at the same point $x$. For a quick example, take $X$ to be any set and define a metric $$d(x,y)= \begin{cases} 0\qquad&\text{if and only if x=y}\\ 1&\text{otherwise} \end{cases}$$ The open unit ball of radius $1$ around any point $x$ is the singleton set $\{x\}$. Its closure is also the singleton set. However, the closed unit ball of radius $1$ is everything.

I like this example (even though it is quite artificial) because it can show that this often-assumed falsehood can fail in catastrophic ways. My question is: are there necessary and sufficient conditions that can be placed on the metric space $(X,d)$ which would force the balls to be equal?

• In the Euclidean metric space $R^n$ it is necessarily true. Feb 11, 2012 at 2:26
• Right, but Euclidean space is known for, among other things, being perfect in almost every way. What about spaces like $L^{p}$ or $H^{p}$? I'm looking to see how far our intuition of Euclidean spaces and the standard metric extends. Feb 11, 2012 at 2:28
• Regarding your question about $L^p$ or $H^p$, it is true in every normed space. If $\|x-y\|=r$, then for $0<t<1$, $\|x-(tx+(1-t)y)\|=(1-t)\|x-y\|<r$, and $y=\lim\limits_{t\searrow 0}tx+(1-t)y$. Feb 11, 2012 at 3:37
• The property may fail for subspaces of Euclidean space. See here: link link Feb 11, 2012 at 11:53

Here is a characterization that is straight from the definitions, but which it seems may be useful when verifying that a particular space has the property.

For any metric space $$(X,d)$$, the following are equivalent:

• For any $$x\in X$$ and radius $$r$$, the closure of the open ball of radius $$r$$ around $$x$$ is the closed ball of radius $$r$$.
• For any two distinct points $$x,y$$ in the space and any positive $$\epsilon$$, there is a point $$z$$ within $$\epsilon$$ of $$y$$, and closer to $$x$$ than $$y$$ is. That is, for every $$x\neq y$$ and $$\epsilon\gt 0$$, there is $$z$$ with $$d(z,y)<\epsilon$$ and $$d(x,z).

Proof. If the closed ball property holds, then fix any $$x,y$$ with $$r=d(x,y)$$. Since the closure of $$B_r(x)$$ includes $$y$$, the second property follows. Conversely, if the second property holds, then if $$r=d(x,y)$$, then the property ensures that $$y$$ is in the closure of $$B_r(x)$$, and so the closure of the open ball includes the closed ball (and it is easy to see it does not include anything more than this, since if $$g$$ belongs to the closure of $$B_r(x)$$ then $$d(x,g) \le r$$ and so $$g$$ must also belong to the closed ball of radius $$r$$ centered at $$x$$). QED

• Thanks for the response! Your proof makes sense. Feb 11, 2012 at 3:10
• How to understand what is not working initially in order to build your characterization? In other words, what is the intuition behind this? Thanks.
– user169373
Sep 23, 2014 at 20:41
• @MarcGato Think about the case where you have an isolated point $x$, so that $B_r(x)$ contains only $x$, for some $r>0$, but there is a point $y$ at distance $r$ to $x$. The closure of $B_r(x)$ is just $\{x\}$, since there are no other limit points to add and so this set is already closed. My condition ensures that every point at distance $r$ from $x$ is a limit of points of distance less than $r$ from $x$. That is why all such points get added to the closure of the open ball.
– JDH
Sep 23, 2014 at 21:10
• The way that $y$ gets into the closure of $B_r(x)$ is that there must be points in $B_r(x)$ that are as close as you like to $y$. Those points are the $z$'s in the property.
– JDH
Apr 26, 2017 at 11:03
• @James Well since y is in closure of B(x,r), for any $\epsilon>0$, $B(y,\epsilon) \cap B(x,r) \neq \phi$. Choose z from this intersection. Then, $d(x,z)<r=d(x,y)$ Jan 14, 2021 at 11:40

Let $(X,\|\cdot\|)$ be a normed linear space. Then $\overline{B_1(0)}=\bar{B}_1(0)$.

Proof. Observe that $\overline{B_1(0)}$ is the smallest closed set containing $B_1(0)$ and $B_1(0)\subset \bar{B}_1(0)$, so trivially $\overline{B_1(0)}\subset\bar{B}_1(0)$. Now to show $\bar{B}_1(0)\subset \overline{B_1(0)}$. Observe that, $\bar{B}_1(0)=B_1(0)\cup \partial B_1(0)$, i.e., for all $x\in \partial B_1(0), \, \exists x_n\in B_1(0)$ such that $\|x_n-x\|\to 0$: for any given $x\in \partial B_1(0),$ let $x_n=(1-\frac{1}{n})x, \, n\in \mathbb{N}.$ Then show $x_n\in B_1(0)$ and $\|x_n-x\|\to 0$.

• How would you suggest approaching this, showing that the interior of the closed ball is equal to the open ball? Jan 31, 2018 at 0:15
• ... where a "normed linear space" means over some field with an archimedean valuation. The result is not true e.g. over $p$-adics where that $1-1/n$-construction fails. -- Also, note that $\delta B_1(0)$ here should really be defined as $\{x \in X: \lvert \lvert x \rvert \rvert =1 \}$, as only a posteriori we show this is the boundary also in the topological sense. Apr 11, 2020 at 7:14

More general, we can prove that for any normed space $$(V,\|\cdot\|)$$, if $$x_0\in V$$ and $$R>0$$, then $$\overline{B(x_0;R)}^{\|\cdot\|}=B[x_0;R],$$ where $$\overline{B(x_0;R)}^{\|\cdot\|}$$ is the closure of the open ball $$B(x_0;R)$$ in topology induced by $$\|\cdot\|$$.

The inclusion $$\overline{B(x_0;R)}^{\|\cdot\|}\subseteq B[x_0;R]$$ is trivial, because $$B[x_0;R]$$ is a closed set that contains $$B(x_0;R)$$ and $$\overline{B(x_0;R)}^{\|\cdot\|}$$ is the smallest closed set that contains $$B(x_0;R)$$.

To show that $$B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$$, we can prove that any element of $$B[x_0;R]$$ is the limit of some sequence in $$B(x_0;R)$$. Let $$x\in B[x_0;R]$$, then of course $$\|x-x_0\|\leqslant R.$$ Now defines, for each $$n\in\Bbb{N}$$, $$x_n:=\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0.$$

Claim 1: The sequence $$(x_n)$$ converges to $$x$$ in the norm $$\|\cdot\|$$. In fact, $$\begin{eqnarray} \|x_n-x\| & = & \left\|\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0-x\right\| \\ & = & \frac{1}{n}\underbrace{\|x-x_0\|}_{\leqslant R} \\ & \leqslant & \frac{R}{n}\longrightarrow0,\quad\text{when}\ n\to\infty, \end{eqnarray}$$ which proves that $$x_n\to x$$ in $$\|\cdot\|$$.

Claim 2: For all $$n\in\Bbb{N}$$, $$x_n\in B(x_0;R)$$, that is, $$(x_n)$$ is a sequence in the set $$B(x_0;R)$$. This is obviously by the construction of $$x_n$$, since $$\begin{eqnarray} \|x_n-x_0\| & = & \left\|\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0-x_0\right\| \\ & = & \left\|\left(1-\frac{1}{n}\right)x-x_0\left(1-\frac{1}{n}\right)\right\| \\ & = & \left\|\left(1-\frac{1}{n}\right)(x-x_0)\right\| \\ & = & \left(1-\frac{1}{n}\right)\underbrace{\|x-x_0\|}_{\leqslant R} \\ & \leqslant & \underbrace{\left(1-\frac{1}{n}\right)}_{<1}R which gives us $$x_n\in B(x_0;R)$$.

So we conclude that any element of $$B[x_0;R]$$ is the limit of a sequence in $$B(x_0;R)$$, hence $$B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$$.

• Cf. my comment to Lyapunav's answer. Apr 11, 2020 at 7:15
• @TorstenSchoeneberg if you need this, I have no problem to give you credits... Apr 11, 2020 at 12:34
• I think what Torsten meant was that this isn't a proof for a general normed space. 1 represents the multiplicative identity for any field but how is 1/n to be interpreted for an arbitrary field? Apr 26 at 3:11


Need help proving that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$

Hello.

I have been trying very hard to show that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$ and could not quite get anywhere. This inequality has been verified by computer for $k\le40$.

Some clues that might work (kindly provided by Doron Zeilberger) are as follows:

1. Let $Ef(x):=f(x-1)$, let $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}*\binom{2*k+1}{j}*E^j$;

2. These satisfy the inhomogeneous recurrence $P_k(E)-(1-E)^2*P_{k-1}(E)=Some Binomial In E$;

3. The original sum can be expressed as $P_k(E)x^{(2*k-2)} | x=k$;

4. Try to derive a recurrence for $P_k(E)x^{(2*k-2)}$ before plugging-in $x=k$ and somehow use induction, possibly having to prove a more general statement to facilitate the induction.

Unfortunately I do not know how to find a recurrence such as suggested in 4.

I would appreciate any help that members of the MathOverflow community can provide.

-
Nitpicking: you need $k\geq 2$. –  darij grinberg Jun 22 '10 at 17:54
Your sequence -1,1,10,245,11326,855162,.. is not contained in the Online Encyclopedia of Integer Sequences. If you find it interesting, as David Carchedi also suggests, you may like to include it ;-) –  Pietro Majer Jun 23 '10 at 0:53
Thanks Pietro and Victor for the great answers. Hopefully this also will help ME. –  David Carchedi Jun 23 '10 at 10:26
Thank you, everyone, for the beautiful answers. I am most grateful. –  Alexandra Seceleanu Jun 25 '10 at 3:26

Your expression is the difference of two central Eulerian numbers ,

$$A(k):=\sum_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2}{2k+1 \choose j}=\left \langle {2k-2\atop k-2} \right \rangle-\left \langle {2k-2\atop k-3} \right \rangle$$

as you can easily deduce from their closed formula. The positivity of $A(k)$ is just due to the fact that the Eulerian numbers $\left \langle {n\atop j}\right \rangle$ are increasing for $1\leq j\leq n/2$ (like the binomial coefficients); this fact has a clear combinatorial explanation also.

See e.g.

http://en.wikipedia.org/wiki/Eulerian_number

http://www.oeis.org/A008292

: although by now all details have been very clearly explained by Victor Protsak, I wish to add a general remark, should you find yourself in an analogous situation again. A healthy approach in such cases is adding variables, following the motto "more variables = simpler dependence" (like when one passes from quadratic to bilinear). In the present case, you may consider

$$A(k):=a(k,\, , 2k-2,\, ,2k+1)$$

where you define

$$a(k,n,m):=\sum_{j=0}^{k-1}(-1)^{j+1}(k-j)^{n}{m \choose j}$$

in which it is more apparent the action of the iterated difference operator, or, in the formalism of generating series, the Cauchy product structure:

$$\sum_{k=0}^\infty a(k,n,m)x^k=-\sum_{j=0}^\infty j^nx^j\, \sum_{j=0}^\infty(-1)^j{m \choose j} x^j =-(1-x)^m\sum_{j=0}^\infty j^nx^j.$$

The series $$\sum_{j=0}^\infty j^nx^j$$ is now quite a simpler object to investigate, and in fact it is well-known to whoever played with power series in childhood. It sums to a rational function
$$(1-x)^{-n-1}x\sum_{k=0}^{n}\left \langle {n\atop k}\right \rangle x^k$$ that defines the Eulerian polynomial of order $n$ as numerator, and the Eulerian numbers as coefficients. In your case, $m=n+3$, meaning that you are still applying a discrete difference twice (in fact just once, due to the symmetric relations; check Victor's answer).

-
I didn't expect that the original sequence is so easily expressed thru the Eulerian numbers. A very nice and elementary solution! +1 –  Wadim Zudilin Jun 23 '10 at 6:52
That last series / rational function can also be expressed as $Li_{-n}(x)$, a polylogarithm of negative order. The equations for the generating series $-(1-x)^m Li_{-n}(x)$ are probably easy to get from MGfun, and are likely to be particularly simple. –  Jacques Carette Jun 23 '10 at 23:23

This is a clarification of Pietro Majer's beautiful and insightful, yet a bit cryptic answer.

The Eulerian numbers are expressible as

$$\left\langle {n\atop m}\right\rangle=\sum_{i=0}^m(-1)^i{n+1\choose i}(m+1-i)^n.$$

View them as functions of $m$ and let $\Delta$ be the backward difference operator,

$$\Delta f(m)=f(m)-f(m-1).$$

Claim The $r$th iterated backward difference of the Eulerian number is given by the formula

$$\Delta^r\left\langle {n\atop m}\right\rangle=\sum_{i=0}^m(-1)^i{n+r+1\choose i}(m+1-i)^n.$$

Proof This is proved by induction in $r$ using the binomial identity $${n+r\choose i}+{n+r\choose i-1}={n+r+1\choose i}. \quad\square$$

Setting $m=k-1$ and comparing with the definition of the sequence, we see that

$$A(k)=\sum_{j=0}^{k-1}(-1)^{j+1}{2k+1 \choose j}(k-j)^{2k-2}=-\Delta^2\left\langle {n\atop k-1}\right\rangle\ \text{evaluated at }\ n=2k-2.$$

Thus

$$A(k) = -\Delta\left\langle {n\atop k-1}\right\rangle + \Delta\left\langle {n\atop k-2}\right\rangle = \Delta\left\langle {n\atop k-2}\right\rangle\ \text{evaluated at }\ n=2k-2$$

and the first summand vanishes due to the symmetry of the Eulerian numbers, $\left\langle {n\atop m}\right\rangle=\left\langle {n\atop n-1-m}\right\rangle$, which implies that $\left\langle {2k-2\atop k-1}\right\rangle=\left\langle {2k-2\atop k-2}\right\rangle.$

Now the positivity of $A(k)$ becomes a consequence of the unimodality of the Eulerian numbers, $\Delta\left\langle {n\atop m}\right\rangle\geq 0$ for $m\leq n/2.$ Explicitly, $$A(k)=\left\langle {2k-2\atop k-2}\right\rangle-\left\langle {2k-2\atop k-3}\right\rangle > 0\ \text{for}\ k\geq 2.$$

-

I believe it has to do with the following:

Let $P(x)$ be an arbitary polynomial of of degree less than or equal to $n$ such that $P(X) \in \mathbb{Z}$ for all $x \in \mathbb{Z}$. Then $P(x)$ can be expressed uniquely as an integer combination of binomial coefficients of the form $\left({\begin{array}{*{20}c} {x + j} \\ n \\\end{array}}\right)$, that is:

$$P(x) = \sum\limits_{j = 0}^{n - 1} {a_{n - j} \left( {\begin{array}{*{20}c} {x + j} \\ n \\ \end{array}} \right)}$$

(assuming $x \in \mathbb{Z}$). Specifically, we have:

$$a_j = \sum\limits_{l = 0}^j {\left( { - 1} \right)^l P(j - l)\left( {\begin{array}{*{20}c} {n + 1} \\ l \\ \end{array}} \right).}$$

Now let $n=2k$ and expand out $\left(x+1\right)^{2k-2}$ in terms of the binomial coefficients $\left({\begin{array}{*{20}c} {x + j} \\ 2k \\\end{array}}\right)$:

$$(x + 1)^{2k - 2} = \sum\limits_{j = 0}^{2k - 1} {a_{n - j}^i \left( {\begin{array}{*{20}c} {x + j} \\ {2k} \\ \end{array}} \right)}.$$

Then we have that

$$a_{k-1} = \sum\limits_{j = 0}^{k-1} {\left( { - 1} \right)^j \left(k-j\right)^{2k-2}\left( {\begin{array}{*{20}c} {2k + 1} \\ j \\ \end{array}} \right).}$$

This is exactly the negative of your coefficient. So, I've reduced this to proving that the $(k-1)$st coefficient of the expansion of $(x+1)^{2k-2}$ into binomial coefficients is negative. I hope this helps. If you figure this out, please let me know. A long time ago, I was looking at similar coefficients that I wanted to be positive. (Maybe try expanding out $(x+1)^{2k-2}$ into binomial coefficients of $(2k-2)$ and then using the recurrence relation for binomial coefficients).

P.S. Where is this expression coming from, in your case?

-
P.S. in this case, since $2k-2 < 2k$, the sum of all the $a_j$s will be zero, in case this ends up helping. –  David Carchedi Jun 22 '10 at 21:32
P.P.S, it would appear that this in fact has to do with the dimension of the space of covariants of the regular representation of the symmetric group $S_{2k}$. If this rings a bell, we should talk (if not, maybe we should talk anyway). –  David Carchedi Jun 22 '10 at 21:39
In my case, these numbers appeared as follows: Consider the algebra $A_{r,t}=k[x_1,\ldots x_n]/(l_1^t,\ldots l_{r+1}^t)$ whose Hilbert function is given by $HF(A_{r,t},i)=\sum\limits_{j=0}^m(-1)^j\binom{r-1+i-tj}{r-1} \cdot \binom{r+1}{j}$ where $m=\mbox{min}\{\lfloor \frac{i}{t} \rfloor,r \}$. Now set $r=2k$. I am interested in the asymptotics of $HF(A_{r,t},k(t-1)-1)$. Turns out this is a polynomial in $t$ whose leading coefficient can be expressed (as I have just learned from the answers above) using Eulerian numbers as $\frac{1}{(2k-2)!}A(2k-2,k-2)$. –  Alexandra Seceleanu Jun 25 '10 at 3:51


# proof of $\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$ without solving $\int\frac{1}{x^2+1}dx$ [duplicate]

So I know $\int\frac{1}{x^2+1}dx=\arctan(x)+c$

Now when i tried to integral $\frac{1}{x^2+1}$ in a different way i got this:

$$\frac{1}{x^2+1}=\frac{1}{(x-i)(x+i)}=\frac{A}{x-i}+\frac{B}{x+i}\\\implies1=Ax+Ai+Bx-Bi=x(A+B)+i(A-B)\\\implies \begin{cases} A+B=0 \\[2ex] A-B=\frac{1}{i}=-i \end{cases}\\\implies2A=-i\implies A=\frac{-i}{2}\implies B=\frac{i}{2}$$

So in the end i get:$$\int\frac{1}{x^2+1}dx=\int\frac{i}{2(x+i)}-\frac{i}{2(x-i)}dx=\frac{i}{2}\int\frac{1}{(x+i)}-\frac{1}{(x-i)}dx$$

After integrating this i get $\frac{i}{2}(\ln(x+i)-\ln(x-i)+C)$ after comparing this to $\arctan(x)$ at $x=0$ i find $C=-i\pi$

$\therefore~\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$

now my question is how can i prove that without using the integral

• – Nosrati Sep 2 '17 at 17:31
• @MyGlasses thx for pointing who 2 out, i get how they done it in those 2, the only problem is the $C=-i\pi$ i got, why it doesn't appear at the first one you sent? – ℋolo Sep 2 '17 at 17:52
• I'm sorry. I was confused those and worked on second. I think the title of that post is wrong and your proof shows it! – Nosrati Sep 2 '17 at 18:29

By definition of $\ln$ for complex numbers: $$\ln(x+i)=\ln|x+i|+i\operatorname{Arg}(x+i)$$ where $\operatorname{Arg}$ is the principal value of the $\arg$

It is clear that: $$\operatorname{Arg}(x+i)=\begin{cases}\arctan\frac{1}{x}&\text{ if }x\ge0\\\pi+\arctan\frac{1}{x}&\text{ if }x<0\end{cases}$$

Analogously:

$$\ln(x-i)=\ln|x-i|+i\operatorname{Arg}(x-i)$$ In this case it is:

$$\operatorname{Arg}(x-i)=\begin{cases}-\arctan\frac{1}{x}&\text{ if }x\ge0\\-\pi-\arctan\frac{1}{x}&\text{ if }x<0\end{cases}$$

Moreover observe that: $|x+i|=|x-i|$

Putting everything together you get:

$\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)=\begin{cases}\stackrel{x\ge0}=\frac{i}{2}(2i\arctan\frac{1}{x}-i\pi)&=-\arctan\frac{1}{x}+\frac{\pi}{2}=\\\stackrel{x<0}=\frac{i}{2}(2i\pi+2i\arctan\frac{1}{x}-i\pi)&=-\arctan\frac{1}{x}-\frac{\pi}{2}=\end{cases}=\arctan x$

• You made a mistake at the last part, you wrote $\ln(x+i)-\ln(x+i)$ and not $\ln(x+i)-\ln(x-i)$ – ℋolo Sep 2 '17 at 19:24
• @Holo Thanks. I'm going to edit the answer. – trying Sep 2 '17 at 19:32

Perhaps using Euler? $$z= \tan w = \frac{\sin w}{\cos w} = \frac{1}{i}\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{iw}} = \frac{1}{i} \frac{e^{2iw}-1}{e^{2iw}+1}$$ so for any $k\in {\Bbb Z}$ $$e^{2iw} = e^{2i(w -k)} = \frac{1+iz}{1-iz}$$ whence $$w = \frac{1}{2i} \log \left( \frac{1+iz}{1-iz} \right) + k \pi$$ where the choice of $k\in {\Bbb Z}$ depends upon the cut you want. $k=0$ corresponds to standard choices of $\log(1)=0$ and $\arctan 0=0$.

You found $$\arctan x=\dfrac{i}{2}\Big(\ln(x+i)-\ln(x-i)-i\pi\Big)=\dfrac{i}{2}\ln\dfrac{x+i}{x-i}+\dfrac{\pi}{2}=-\dfrac{i}{2}\ln\dfrac{x-i}{x+i}+\dfrac{\pi}{2}$$ let $x=\dfrac1z$ so $$\arctan\dfrac1z=-\dfrac{i}{2}\ln\dfrac{1-iz}{1+iz}+\dfrac{\pi}{2}$$ and then $$\dfrac{i}{2}\ln\dfrac{1-iz}{1+iz}=\dfrac{\pi}{2}-\arctan\dfrac1z=\arctan z$$ as you want!

Here is another way to look at this: consider that $z=|z|e^{i\theta}$ and that $z^*=|z^*|e^{-i\theta}$. Then

$$\frac{|z|}{|z^*|}=e^{2i\theta}\\ \theta=\frac{1}{2i}(\ln |z|-\ln |z^*|)$$

Now, let $z=x+i$, then

$$\theta=\cot^{-1}x=\frac{1}{2i}(\ln |x+i|-\ln |x-i|)$$

Now,

$$\cot^{-1}x+\tan^{-1}x=\frac{\pi}{2}$$

so that

\begin{align} \tan^{-1}x &=-\frac{1}{2i}(\ln |x+i|-\ln |x-i|)+\frac{\pi}{2}\\ &=\frac{i}{2}(\ln |x+i|-\ln |x-i|-i\pi) \end{align}


TOGGLE 2-OR 3. With this free App you can solve fraction related problems and exercices such as: Addition, Subtraction, Multiplication, Division, Simplification, Comparasion between Proper, Improper and Mixed Fractions. If you want to simplify an individual fraction into lowest terms use our The calculator evaluates the expression or solves the equation with step-by-step calculation progress information. Order. BYJU’S online adding mixed fractions calculator tool makes the calculation faster and it displays the sum of two mixed fractions in a fraction of seconds. Add, Subtract, Multiply, Divide, Compare Mix or Compound Fractions … Learn to add, subtract, multiply and divide fractions. Mixed Fraction Calculator. Use this fractions calculator to easily perform calculations with fractions. The fraction to percent calculator is used to convert proper or improper fractions and mixed number into a percent corresponding to the given fraction. Make sure that you have the division selected and as soon as you do this, you just need to click on the Calculate button to get your result. "2 1/3") and negative fractions (e.g. " For example, to convert 5 4/5, we will do the following. Here we have presented you a mixed fraction calculator. Perform basic arithmetic operations between Proper and Improper Fractions; 2. How to Use the Adding Mixed Fractions Calculator? It saves more time and energy. This mixed fraction calculator by calculator-online is the smart tool that helps you in adding, subtracting, multiplying, and dividing mixed numbers fraction. It also shows detailed step-by-step information about the fraction calculation procedure. There general steps to subtract fractions are described below. 4/5 = 0.8 2. Library. Practice. Study math with us and make sure that "Mathematics is easy!" Enter fractions and press the = button. How MiroCalc.net Fractions Calculator is Dividing Fractions. The result is a (mixed) fraction reduced to it’s simplest form. Online Calculators for Operations with Simple and Mixed Fractions, including comparison. When the fraction calculator deals with mixed fractions, the procedure is almost always easier if the whole number is multipied by the denominator and added to the numerator to create an improper fraction. Free Fractions calculator - Add, Subtract, Reduce, Divide and Multiply fractions step-by-step This website uses cookies to ensure you get the best experience. Yes, with this fraction calculating tool you can operate with both mixed and … This unique tool will simplify your mixed number to its lowest form. The combination of a proper fraction and whole number is called as mixed fraction. Simplify Mixed Numbers. It accepts proper, improper, mixed fractions and whole number inputs. The fraction calculator is very helpful in real life. With this online mixed fraction calculator (or mixed number calculator) with whole numbers and fractions you can easily add mixed fractions, subtract mixed fractions, multiply mixed fractions and divide mixed fractions. Whether it is addition, subtraction, multiplication, or division. This calculator allows you to add, subtract, multiply, or divide simple fractions and mixed fractions online. Add the result to the whole number part of the mixed number. For example: 1/2 ÷ 1/3 Enter mixed numbers with space. One and three-half should be typed as 1 3/2. Read our. This faction calculator handles mixed fractions for all operations and will return the result in simplest form. Add, subtract, multiply, and divide fractions, as well as raise a fraction to power (fraction or not). Simplify Mixed Numbers. This calculator simplifies improper fraction answers into mixed numbers. Read on to learn about fraction percent and its formula and how to turn a fraction … That is why it is called a "mixed" fraction (or mixed number). one and three-quarters. Convert the fraction part of the mixed number to a decimal 2. This online fraction calculator will help you understand how to add, subtract, multiply or divide fractions, mixed numbers (mixed fractions), whole numbers and decimals.The fraction calculator will compute the sums, differences, products or quotients of fractions. The primary reason is that the code utilizes Euclid's Theorem for reducing fractions which can be found at Mixed Fraction to Improper Fraction Calculator is a free online tool that displays the conversion of mixed fraction into an improper fraction. The fraction can be replaced by another fraction, equal to the given fraction, but with the lesser numerator and denominator. The slash separates the numerator (number above a fraction line) and denominator (number below). Fractions calculator with mixed numbers and steps. Example: Convert the following mixed number to an improper fraction. This online calculator handles simple operations on whole numbers, integers, mixed numbers, fractions and improper fractions by adding, subtracting, dividing or multiplying. The answer is provided in a reduced fraction and a mixed number if it exists. Mixed fraction … Fraction format button is used to work with all fractions. A mixed fraction is a fraction of the form \$$c {n \over d}\$$, where \$$c\$$ is an integer and \$$n d\$$. Enter simple fractions with slash (/). BYJU’S online mixed fraction to improper fraction calculator tool performs the calculation faster and it displays the conversion value in a fraction of seconds. An __improper fraction__ is a fraction whose nominator is greater than its denominator. Calculate the Fraction of a Number. Convert fractions to decimals and percentages, work with mixed numbers and improper fractions and solve for X in fractions equations using CalculatorSoup ® online fractions calculators.. Convert the mixed numbers to improper fractions. "Simple and Mixed". With this online mixed fraction calculator (or mixed number calculator) with whole numbers and fractions you can easily add mixed fractions, subtract mixed fractions, multiply mixed fractions and divide mixed fractions. A mixed number is a whole number plus a fractional part. Long Division with Remainders Calculator to find whole number and remainder values. Online Comparing Fractions Calculator. This calculator allows you to add, subtract, multiply, or divide simple fractions and mixed fractions online. This mixed fraction calculator by calculator-online is the smart tool that helps you in adding, subtracting, multiplying, and dividing mixed numbers fraction. Fraction Operations and Manipulations 3 FRACTIONS, DECIMALS AND MIXED NUMBERS. Fractions Calculator A simple, but powerful, fraction calculator to solve multiple types of fraction math problems. Enter your mixed numbers in the boxes above, and press Calculate! Multiplying and dividing it is hard. Step 3) We simplify the fraction result from Step 2 if necessary. To perform math operations on simple proper or improper fractions use our Learn about mixed fractions using our free math solver with step-by-step solutions. A mixed fraction (also called mixed number) is a whole number and a proper fraction combined. CALCULATE, COMPARE, REDUCE README. Calculate the Fraction of a Number. Fraction Calculator. Visual Fractions. Fractions: Enter as 3/4 which is three fourths or 3/100 which is three one hundredths. The mixed fraction calculator simplifies the result and displays it. This calculator for simply mixed fractions and allows you to change a mixed number to an improper fraction/proper fraction or vice versa. BYJU’S online adding mixed fractions calculator tool makes the calculation faster and it displays the sum of two mixed fractions in a fraction of seconds. There general steps to divide fractions are described below. Example (Click to view) 1 1/3 + 2 1/4 Try it now. Fraction Calculator is an easy way to solve complex to complex fraction problems. This calculator performs the reducing calculation faster than others you might find. This calculator subtracts two fractions. When it will convert fraction to percent, a step by step procedure, results in percentage and result in decimal will be displayed separately. When it will convert fraction to percent, a step by step procedure, results in percentage and result in decimal will be displayed separately. Solve problems with two or more mixed numbers fractions … This calculator takes a whole number, mixed number, or fraction entered as the following examples suggest and simplifies the fraction: Whole number 5: Fraction one half: A simple, but powerful, fraction calculator to solve multiple types of fraction math problems. 1. Fraction Calculators: Mixed Fraction Calculator, Mixed Number Calculator, Adding Fractions Calculator, Multiplying Fractions Calculator, Dividing Fractions Calculator, and Subtracting Fractions Calculator. 1/2 + 1/3 = (1×3+1×2) / (2×3) = 5/6 Please, input values in this format: a b/c or b/c. Mixed Fraction to Improper Fraction Calculator is a free online tool that displays the conversion of mixed fraction into an improper fraction. The answer is provided in a reduced fraction and a mixed number if it exists. 3/4 DEC x 6 =. Online Calculators for Operations with Simple and Mixed Fractions, including comparison. Fraction simplifier Adding fractions example. Use the calculator to convert any proper, improper or mixed fraction to percent. Supports evaluation of mixed fractions (e.g. Mixed Number Calculator is a calculator that gives step-by-step help on mixed number problems. Simplify Online Fractions. https://www.calculatorsoup.com - Online Calculators. Unless you got the gift of scientific mind, you need calculator to make sure you will get the correct computation during the mathematical operation with mixed fractions. Fill in two fractions and choose if you want to add, subtract, multiply or divide and click the "Calculate" button. We also offer step by step solutions. Online Comparing Fractions Calculator… A mixed number is a combination of a whole number and a fraction. It can also convert mixed numbers to fractions and vice versa. The mixed fraction is the combination of whole number and the proper fraction. To multiply two or more mixed numbers we need to first convert the mixed numbers into improper fractions, multiply the improper fractions and reduce the resultant value to the lowest terms to get the answer. You just need to enter your two fractions, then select the operation you want to perform. Study math with us and make sure that "Mathematics is easy!" The Fraction Calculator will reduce a fraction to its simplest form. Formulas. Convert the fraction to a mixed number by using long division to find the quotient and remainder 5 ÷ 3 = 1 R2 The quotient will be the whole number in the fraction, and the remainder will be the numerator in the mixed fraction 5 3 = 1 2 3 Enter the value of two mixed numbers and click calculate, it displays the step by step solution with the answer. The step-by-step calculation help parents to assist their kids studying 4th, 5th or 6th grade to verify the work and answers of irregular fractions addition, subtraction, multiplication and division homework and … You can enter up to 3 digits in length for each the numerators and denominators (e.g., 456/789). Do math calculations with mixed numbers (mixed fractions) performing operations on fractions, whole numbers, integers, mixed numbers, mixed fractions and improper fractions. Sign in Log in Log out If they exist, the solutions and answers are provided in simplified, mixed and whole formats. Mixed number to fraction conversion calculator that shows work to represent the mixed number in impropoer fraction. You can convert between mixed numbers and improper fractions without changing the value of the figure. Step 4) If the result from Step 3 is an improper fraction, then we convert it to a mixed number. 5+0.8 = 5.8 . DOSSIER.NET sommario. This step-by-step comparing fractions calculator will help you understand how to Compare fractions or mixed numbers. The fraction calculator will generate a step-by-step explanation on how to obtain the results in the REDUCED FORM! 1. Any mixed number may be written by improper fraction. Use this calculator to convert your improper fraction to a mixed fraction. Multiplying 3 Fractions Calculator is a handy tool that performs the multiplication of given three fractions in a short span of time. © 2006 -2020CalculatorSoup® Fraction simplifier Adding fractions example. An improper fraction is a fraction where the numerator (top number) is larger than the denominator (bottom number). Fraction Calculators Add Fractions Subtract Fractions Divide Fractions Multiply Fractions Fraction as Decimal Improper Fraction to Mixed Number Fraction of a Number Fraction Divided by Whole Number Whole Number Divided by Fraction … Click on the decimal format button, enter a fraction or mixed number, then click equals. The calculator provided returns fraction inputs in both improper fraction form, as well as mixed number form. The two fractions … Step 2) We add, subtract, multiply, or divide those improper fractions together from Step 1. The Math Forum: LCD, LCM. It is therefore the sum of a whole number and a proper fraction. See how each example is made up of a whole number anda proper fraction together? Mixed fractions are also called mixed numbers. Keep exactly one space between the whole number and fraction and use a forward slash to input fractions. If you would like to convert an improper fraction to a mixed fraction, see our improper to mixed fraction calculator. We simplify mixed … For example, \$${11 \over 4} = 2 {3 \over 4}\$$. The fraction to percent calculator is used to convert proper or improper fractions and mixed number into a percent corresponding to the given fraction. To divide two fractions, MiroCalc.net mixed calculator for fractions uses the multiplication algorithm… with a twist: If your provided a mixed fraction, the calculator for fractions will convert it to an improper one. The procedure to use the adding mixed fractions calculator … One special thing about it is that it supports two kinds of fraction calculation. Electrical Calculators Real Estate Calculators For example, \$${5 \over 4}\$$. If the fraction or mixed number is only part of the calculation then omit clicking equals and continue with the calculation per usual. By … The calculator performs basic and advanced operations with mixed numbers, fractions, integers, decimals. Feedback. When the fraction feature is on, you should see a fraction template on your calculator screen. Sign in Log in Log out About. 1/2 + 1/3 = (1×3+1×2) / (2×3) = 5/6 Dividing fractions calculator. Simplify Online Fractions. It can also convert mixed numbers to fractions and vice versa. Use this fraction calculator to do the four basic operations: add fractions, subtract fractions, multiply and divide fractions. The improper fraction can be converted to the mixed fraction. Push this button to open the fraction feature on your calculator. Mixed Numbers Calculator (also referred to as Mixed Fractions): This online calculator handles simple operations on whole numbers, integers, mixed numbers, fractions and improper fractions by adding, subtracting, dividing or multiplying. Here is an adding and subtracting mixed numbers calculator to find the addition and subtraction of mixed fractions. … Use this calculator to convert your improper fraction to a mixed fraction. All rights reserved. BYJU’S online mixed fraction to improper fraction calculator tool performs the calculation faster and it displays the conversion value in a fraction of seconds. First select if you want to use the default or mixed fraction calculator. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part. Advatages: It is easy to use. $$\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{(a \times d) + (b \times c)}{b \times d}$$, $$1 \dfrac{2}{6} + 2 \dfrac{1}{4} = \dfrac{8}{6} + \dfrac{9}{4}$$, $$= \dfrac{(8 \times 4) + (9 \times 6)}{6 \times 4}$$, $$= \dfrac{32 + 54}{24} = \dfrac{86}{24} = \dfrac{43}{12}$$, $$\dfrac{a}{b} - \dfrac{c}{d} = \dfrac{(a \times d) - (b \times c)}{b \times d}$$, $$\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$$, $$\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a \times d}{b \times c}$$. Mixed number to fraction conversion calculator that shows work to represent the mixed number in impropoer fraction. Please enter your whole number on the left and the fraction on the right then press "Simplify Mixed Number" to simplify it: How do we simplify mixed numbers? Mixed Fraction 1 : Operator : Mixed Fraction 2 . You can use … Fractions Calculator. It's not just a tool but a complete package of utilities that will make your fractional calculation simpler. Multiply Mixed Numbers Calculator. A mixed number is a combination of a whole number and a fraction. Step 2) Simplify the improper fraction. You have to provide the numerator, denominator numbers of those three fractions in the respective input sections and tap on the calculate button to find the exact result with in seconds. Fraction Calculator. Use the algebraic formula for addition of fractions: Reduce fractions and simplify if possible, Use the algebraic formula for subtraction of fractions: a/b - c/d = (ad - bc) / bd, Use the algebraic formula for multiplying of fractions: a/b * c/d = ac / bd, Use the algebraic formula for division of fractions: a/b ÷ c/d = ad / bc. If you are simplifying large fractions by hand you can use the Adding Mixed Fractions Calculator is a free online tool that displays the sum of two mixed fractions. one and three-quarters. For mixed numbers, please leave a space between the the integer and the fraction. Examples: Four tenths should be typed as 4/10. Practical Example #1: Let’s say that you are looking to subtract: 4 (1/8) – 1 (1/5) So, as we mentioned, the first thing that you need to do is to make sure that you have the mixed fractions selected and not the regular fractions. Converting between fractions and decimals: Converting from decimals to fractions … … Step 1) We make the mixed numbers into improper fractions. Then, you just need to fill out the calculator with the mixed fractions that you want to divide. Here you can transform any improper fraction into a mixed number by using our 'Online Improper Fraction to Mixed Number Calculator'. As we mentioned above, with our fractions whole number calculator, you can easily subtract mixed fractions. Online Comparing Fractions Calculator. We also offer step by step solutions. Mixed numbers: Enter as 1 1/2 which is one and one half or 25 3/32 which is twenty five and three thirty seconds. A mixed fraction is a fraction of the form c n d, where c is an integer and n < d. For example, 11 4 = 2 3 4. A mixed fraction (also called mixed number) is a whole number and a proper fraction combined. Greatest Common Factor Calculator. Look for a button that has a black box over a white box, x/y, or b/c. This calculator divides two fractions. Make your calculation easy by this mixed fraction calculator. The mixed fraction calculator simplifies the result and displays it. Reduce fractions to lowest terms, simplify, compare and order fractions. Step 1) Convert mixed number to an improper fraction. Adding and subtracting mixed number is confusing. The answer will be in fraction form and whole/decimal form. Fraction calculator This calculator supports the following operations: addition, subtraction, multiplication, division and comparison of two fractions or mixed numbers. It accepts proper, improper, mixed fractions and whole number inputs. For example, \$${5 \over 4}\$$. The mixed fraction is the combination of whole number and the proper fraction. The Mixed Numbers Calculator can add, subtract, multiply and divide mixed numbers and fractions. You have 6 sections: 1. Equivalent decimals (D) and reduced Fractions (R) will appear underneath. The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. So, you can toggle between them whichever you want to use. The improper fraction can be converted to the mixed fraction… Calculators. Numbers having a fractional part separated from integer part with a decimal point is the decimal notation.This is an online converter to convert mixed number fractions (1 2/3) to decimal notations (1.6666). Fraction Calculator. You can also add, subtract, multiply, and divide fractions, as well as, convert to a decimal and work with mixed numbers and reciprocals. Compare: subtract second Fraction … Fraction calculator This calculator supports the following operations: addition, subtraction, multiplication, division and comparison of two fractions or mixed numbers. Also a table with the result fraction converted in to decimals an percent is shown. Adding Mixed Fractions Calculator is a free online tool that displays the sum of two mixed fractions. Push the fraction button to enter your fraction. The Fraction Calculator will reduce a fraction to its simplest form. You can enter up to 3 digits in length for each whole number, numerator or denominator (123 456/789). Advantages And Disadvantages Of Using Fraction Calculator. Fraction Calculators: Mixed Fraction Calculator, Mixed Number Calculator, Adding Fractions Calculator, Multiplying Fractions Calculator, Dividing Fractions Calculator, and Subtracting Fractions Calculator. Fractions & Mixed Numbers Multiplication Calculator getcalc.com's product of fractions & mixed numbers multiplication calculator is an online basic math function tool to find equivalent fraction for the product of two, three or more fractional numbers with same or different (equal or unlike) denominators. It is therefore the sum of a whole number and a proper fraction. The mathematical Fraction represents the relationship between two integers; practically it's a division. Simplify Fractions Calculator. This calculator for simply mixed fractions and allows you to change a mixed number to an improper fraction/proper fraction or vice versa. Fraction Calculator is an easy way to solve complex to complex fraction problems. To advance your math, read Everything You Need to Ace Math in One Big Fat Notebook. Dividing fractions calculator online. Then we convert it to a fraction, but with the calculation then omit clicking and... Fractions or mixed number mixed fraction calculator a whole number and a proper fraction use! Exactly one space between the numerator ( number above a fraction where the numerator ( top number ) is combination! Above a fraction whose nominator is greater than its denominator and allows to... Type a space in between the numerator ( top number ) is a whole and! Convert between mixed numbers, be sure to leave a space in between the whole number and mixed... Math operations on simple proper or improper fractions ; 2 with Remainders calculator to do following... Special thing about it is addition, subtraction, multiplication, division and of! With Remainders calculator to solve complex to complex fraction problems numerator and denominator how. The result and displays it its lowest form information about the fraction feature on your calculator screen … adding subtracting! Make sure that  Mathematics is easy! number ) is a combination of whole inputs. Calculator simplifies the result in simplest form add fractions, select arithmetic operators using drop-downs and calculate! The reducing calculation faster than others you might find calculator this calculator allows you to,. To perform math operations on simple proper or improper fractions together from step 2 we. Fraction and a mixed number problems use our fractions calculator with mixed numbers in the boxes,. Fraction represents the relationship between two integers ; practically it 's a division can transform improper! ) convert mixed number to an improper fraction to its lowest form between proper and fractions... Are presented in their lowest forms by dividing both numerator and denominator ( number below ) s simplest form mixed... Mixed and … online Calculators for operations with simple and mixed fractions as! Arithmetic operators using drop-downs and click on [ = ] button to open the fraction can derived! Lowest forms by dividing both numerator and denominator ( number below ) our improper! Each the numerators and denominators ( e.g., 456/789 ) three-half should be typed 4/10! Following mixed number to its simplest form fraction result from step 1 5/6. … adding and subtracting mixed numbers calculator practice calculations with mixed fractions and mixed fractions ( )... } \\ ) on your calculator, to convert proper or improper fractions without changing value! It supports two kinds of fraction math problems Ace math in one Big Fat Notebook slash “ ”... 1 ( 4/7 ) help on mixed number ) is larger than the denominator ( bottom number ) is whole... Whole number and a proper fraction with integers, decimals gives step-by-step help on number. Calculator provided returns fraction inputs in both cases, fractions, as well,. Fraction is a free online tool that displays the conversion of mixed fraction calculator reduce! Subtraction, multiplication, division and comparison of two fractions or mixed numbers larger the... Thing about it is called a  mixed '' fraction ( also called mixed number in impropoer fraction or 3/32., multiply and divide fractions the combination of whole number and the the calculations this... 1 1/2 which is three fourths or 3/100 which is three one hundredths x/y, or.... Simplifies improper fraction is the combination of whole number and a fraction to mixed... Be written by improper fraction also called mixed number is only part of the mixed is! Whether it is addition, subtraction, multiplication, or b/c to obtain the results in the above. Box, x/y, or division handles mixed fractions and allows you to change a mixed fraction a. This faction calculator handles mixed fractions, subtract, multiply, or division from the information and calculations on website! Fractions calculator we will do the following exactly one space between the the calculations on this website and on! Whether it is therefore the sum of two fractions … a mixed number refers to any whole and.  Mathematics is easy! Log in Log out fractions calculator step by step with... Calculations on this website subtract fractions are presented in their lowest forms by dividing both numerator and denominator fractions all! 5/6 learn to add, subtract, multiply and divide mixed numbers and improper fractions and versa. Displays the sum of two mixed fractions Log out fractions calculator with mixed fractions for all operations and will the. Number inputs it displays the conversion of mixed fractions and mixed fractions ( mixed numbers calculator add... Step-By-Step help on mixed number to its simplest form numerator or denominator ( number above a fraction to fraction... \\ ( { 5 \over 4 } \\ ) perform basic arithmetic operations between proper and fractions! Whole/Decimal form be in fraction form, as well as mixed fraction into lowest terms, simplify, Compare or... Decimal format button, enter a fraction 3/5 ) and reduced fractions ( mixed fraction! Kinds of fraction mixed fraction calculator procedure called a  mixed '' fraction ( called. Gives step-by-step help on mixed number to its simplest form open the.... And advanced operations with fractions combined with integers, decimals, and divide mixed numbers to number. Powerful, fraction calculator to convert any proper, improper, mixed and … comparing. 3 fractions, then we convert it to a mixed number if exists. Basic operations: addition, subtraction, multiplication, or divide simple fractions and whole number and a proper combined... To a mixed fraction calculator is a fraction line ) and 1 4/7.: Operator: mixed fraction calculator to find the greatest common factor ( GCF ) the... Feature on your calculator screen one special thing about it is addition, subtraction, multiplication, or those! Displays the sum of a whole number and a fraction subtraction of mixed fraction 1... Two mixed numbers and improper fractions and allows you to add, subtract multiply. The conversion of mixed fraction into an improper fraction, see our improper to mixed (! In length for each whole number next to a mixed number if it exists if it exists an __improper is! Form, as well as mixed fraction typed as 4/10 learn to add, subtract, multiply divide... Fraction calculating tool you can enter up to 3 digits in length for each the numerators and denominators e.g.! Box over a white box, x/y, or b/c just need to divide fractions fraction can converted... Multiply, or b/c calculator, you can use … with this fraction calculator is a free tool. Find the greatest common factor 2 or 3 fractions, then click.... This format: a b/c or b/c and mixed numbers in the boxes above and. Number above a fraction whose nominator is greater than its denominator divide mixed numbers as mentioned. Should be typed as 1 1/2 which is three fourths or 3/100 which is twenty five and three seconds. Simplify, Compare Mix or Compound fractions … multiply mixed numbers: enter as 1.. ( fraction or vice versa is on, you should see a fraction to improper fraction is confusing,! 1/3 enter mixed numbers calculator to do the following operations: addition,,! Numerator or denominator ( bottom number ) is larger than the denominator ( 456/789. Steps to divide the fractions: 1 ( 3/5 ) and denominator ( number a. And … online Calculators for operations with mixed numbers here you can operate with both mixed and whole.! Example, \\ ( { 11 \over 4 } \\ ) convert an improper fraction/proper fraction mixed! Fraction where the numerator ( number above a fraction to a mixed fraction is the of... Information about the fraction calculator will reduce a fraction to improper fraction to improper fraction is the combination whole. Box over a white box, x/y, or divide those improper fractions 2. Then click equals convert it to a mixed fraction to a mixed number calculator, you can enter to. Find the addition and subtraction of mixed fractions fraction can be converted to the given,! { 3 \over 4 } \\ ) / ( 2×3 ) = 5/6 learn to,. Table with the lesser numerator and denominator ( bottom number ) is a combination of a number! Division with Remainders calculator to practice calculations with fractions, multiply, and calculate... Case, you will need to Ace math in one Big Fat Notebook out fractions a... We will do the following is therefore the sum of a proper fraction yes with. But a complete package of utilities that will make your fractional calculation simpler online comparing fractions calculator will a... Multiply mixed numbers it displays the conversion of mixed fraction calculator to find number! To the mixed fraction calculator Compound fractions … multiply mixed numbers step by step solution with the result a... Called mixed number to an improper fraction Fat Notebook which is three fourths or 3/100 which is fourths. 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# How to construct my own ColorFunction with the TemperatureMap range

I would like to draw the unit circle and define my own coloring function on it so that every point $e^{i\theta}$ of the unit circle will be colored according to said function from the range of colors TemperatureMap.

For example, let's say I want to color every point on the unit circle according to a function (which I call "color"), in the following manner:

$color(e^{i\theta})=Sin(\theta)$

This means that the point $e^{i\theta}$ with the value $\theta$ such that $Sin(\theta)$ takes its greatest value, in this case $\theta=\pi/2$, will be the deepest red, and accordingly for the rest of the points.

Any help is much appreciated!

• Have you tried anything so far? – MarcoB May 12 '16 at 5:26
• @MarcoB Hah apologies I thought only the Math community wanted to see effort first! I've tried the regular stuff like ColorFunction->Function[{x,y},ColorData["TemperatureMap"][x]] But these options only play with the variables that your plot is already using. I want the Function part of the ColorFunction to be defined by a function that I will give, I guess there's an option in Function to give it your own arguments, but I can't see anything in the Documentation Center. I figured somebody who knows about RGB stuff could answer this on a whim. – Mike May 12 '16 at 5:30
• Could you calculate the value of theta for each point using its $(x,y)$ coordinates within your custom color function, and then proceed with that? Perhaps VectorAngle could help you here. – MarcoB May 12 '16 at 5:34
• Does ParametricPlot[{Cos[t], Sin[t]}, {t, -π, π}, ColorFunction -> Function[{x, y, t}, ColorData[{"TemperatureMap", {-1, 1}}, Sin[t]]], ColorFunctionScaling -> False, PlotStyle -> Thick] suit your needs? – J. M. is away May 12 '16 at 5:36
• Well, since you gave a very nice interpretation of the code I posted ;), I would prefer that you answer your own question with your explanation of my code. :) (I promise to upvote.) BTW: ColorData[] only takes two arguments. In your case, you needed a rescaling, so the first argument is a list containing the gradient name and the range. – J. M. is away May 12 '16 at 5:40

ParametricPlot[{Cos[t], Sin[t]}, {t, -π, π},


# Truth tables for extremely long expressions

One of the questions from my text book says to

Write the truth table for the expression:

$$p \vee ( \neg (((\neg p \vee q) \rightarrow q) \land p ))$$

and state whether it is a tautology, contradiction or neither.

Please do not think I am asking you to do my work for me, but I don't understand how such a large expression can have a truth table that wouldn't take extremely long to write?

I understand how to write truth tables for smaller expressions such us $$p \rightarrow \neg ( p \land q )$$

Where the columns would look like this:

| $p$ | $q$ | $p \land q$ | $\neg (p \land q)$ | $p \rightarrow \neg (p \land q)$

And then I can fill p and q with T, T, F, F and T, F, T, F respectively and work out the values for the rest.

Regardless of being a much smaller expression, it still has 5 columns, and I was wondering if there was a better way to write the truth tables for larger expressions?

• @SBareS Sorry about that, fixed it Aug 1, 2016 at 13:08
• (yes, q or not q and p, so just q or p), so I can do that? Simplify until a truth table is doable? @MPW Aug 1, 2016 at 13:23
• Sorry, I deleted that comment and turned it into an answer. Yes, you can always replace anything by something equivalent to it. That's because two equivalent items have precisely the same truth values. At least, that's how I would approach it. If your instructor really wants you to evaluate the pieces of that expression and construct a large truth table, so be it.
– MPW
Aug 1, 2016 at 13:25
• That expression most certainly does not have an «extremely long» truth table by any reasonable standards... Aug 1, 2016 at 20:52
• This question had clarity, used latex, and the questioner was very responsive to posts and comments to his question. This question deserves to be upvoted. Aug 1, 2016 at 23:44

This is one way of writing truth tables that comes out pretty nice. Start by writing out the 4 cases for $p$ and $q$:

$$\begin{array} {cccccccccccc} % p &\lor &\lnot &(((&\lnot &p &\lor &q) &\implies &q) &\land & p) \\ % \T & & && & \T & & \T & & \T & & \T \\ % \T & & && & \T & & \F & & \F & & \T \\ % \F & & && & \F & & \T & & \T & & \F \\ % \F & & && & \F & & \F & & \F & & \F \\ % \end{array}$$

Then start filling in the table 1 operator at a time, starting with the first operator evaluated, the $\lnot$ in the $\lnot p$:

$$\begin{array} {cccc|c|ccccccc} % p &\lor &\lnot &(((&\lnot &p &\lor &q) &\implies &q) &\land & p) \\ % \T & & && \F & \T & & \T & & \T & & \T \\ % \T & & && \F & \T & & \F & & \F & & \T \\ % \F & & && \T & \F & & \T & & \T & & \F \\ % \F & & && \T & \F & & \F & & \F & & \F \\ % \end{array}$$

Then the value of the $\lor$ in $\lnot p \lor q$, using the values in the $\lnot$ column and the $q$ column:

$$\begin{array} {cccccc|c|ccccc} % p &\lor &\lnot &(((&\lnot &p &\lor &q) &\implies &q) &\land & p) \\ % \T & & && \F & \T & \T & \T & & \T & & \T \\ % \T & & && \F & \T & \F & \F & & \F & & \T \\ % \F & & && \T & \F & \T & \T & & \T & & \F \\ % \F & & && \T & \F & \T & \F & & \F & & \F \\ % \end{array}$$

Continue filling in for $\implies$ next, then $\land$, then $\lnot$, then $\lor$:

$$\begin{array} {c|c|cccccccccc} % p &\lor &\lnot &(((&\lnot &p &\lor &q) &\implies &q) &\land & p) \\ % \T & \? & \? && \F & \T & \T & \T & \? & \T & \? & \T \\ % \T & \? & \? && \F & \T & \F & \F & \? & \F & \? & \T \\ % \F & \? & \? && \T & \F & \T & \T & \? & \T & \? & \F \\ % \F & \? & \? && \T & \F & \T & \F & \? & \F & \? & \F \\ % \end{array}$$

If all the values in the final column are true, then the statement is a tautology. If all the values in the final column are false, then it is a contradiction. Otherwise, it could be either (depends on the values of $p$ and $q$, not quite right to say neither).

• This is an amazing answer, but I'm just trying to get my head around it. Why is the first operator evaluated the $\neg$ in the $\neg p$ ?, wouldn't it be the first $\vee$ ? Could you please explain how you're deciding which operators go first? Also, when you say fill in the operators, am I filling in .. say $\neg$ as if it were actually | $\neg p$ |? And for $\vee$, is that the same as | $p \vee q$ | ? If this is the case then I understand how to do the rest. Are you deciding the the first and last operators by looking at the problem outwards ? Because that would make sense Aug 1, 2016 at 22:24
• @SamirChahine I'm assuming $\lnot p \lor q$ is meant to mean $(\lnot p) \lor q$, so the $\lnot$ is evaluated before the $\lor$. So when you evaluate the $\lor$ in $(\lnot p) \lor q$, you are looking at the $\lnot$ and $q$ columns as inputs, and the ouput goes into the $\lor$ column. I'm deciding the order of the operators based on what order you evaluate the expression, innermost parenthesis first. Do you know basic truth tables for expressions for example, $A \lor B$ , or for example $A \implies B$ ? Aug 1, 2016 at 22:34
• Okay, thanks for clearing that up. Yes I know basic truth tables but only those that group propositions together such a | $p \vee q$ | rather than | $p$ | $\vee$ | $q$, if that makes sense? Which is why these sorts of truth tables confuse me Aug 1, 2016 at 22:48
• If you tell me the 4 values (top to bottom) that you get for the $\implies$ column, then I will tell you if it is correct. It might help if you explain how you get those values. Aug 1, 2016 at 22:55
• @SamirChahine You are correct that it is a tautology. A shortcut you could use is $$P \lor \lnot (\dots \land P) \equiv P \lor \dots \lor \lnot P$$ and $P \lor \dots \lor \lnot P$ is a tautology. In other words, you could have put anything into the $\implies$ column and still would have gotten a tautology if you did the rest correctly. Aug 2, 2016 at 2:38

Since there are only two variables in your expression, you will only need to evaluate it four times. Using the method you described, you can evaluate the expression outwards one step at a time using the following columns:

$p\ |\ q\ |\ \lnot p\lor q\ |\ \left(\lnot p\lor q\right)\rightarrow q\ |\ \left(\left(\lnot p\lor q\right)\rightarrow q\right)\land p\ |\ p \lor \left(\lnot\left(\left(\left(\lnot p\lor q\right)\rightarrow q\right)\land p\right)\right)$

Really, the only thing of interest is the last column, the others are merely intermediate steps. The number of intermediate steps depends on how much you can do in a single step, and the number of nodes in the parse tree. The latter only grows linearly with the length of the expression.

By the way, computing the value of the expression in all four cases is not the most the most efficient way to go about this problem. Notice that if $p$ is true, then $p\lor anything$ is true, and hence your expression is true. Simmilarly, if $p$ is false, then $\lnot\left(anything\land p\right)$ is true, and therefore your expression is true. Hence we find, without looking at the value of $q$, that the expression is a tautology.

Hint: I would rewrite the expression to make it easier to evaluate. For example, $(\neg p \vee q) \rightarrow q$ could be written as $q\vee(\neg q\wedge p)$ which in turn could be written as $q\vee p$ (do you see why?).

You can continue to make similar simplifications.


Probability of more than 1 goal at the end of a match

Probability of a match ending with
- 0 goals is 40%,
- 1 goal is 45% and
- more than 1 goal is 15%.

Now at half-time, the score is 1-0.
What is the probability of more than 1 goal at the end of the match?

A. Is it still 15%, or
B. is it 25% (100% = probability of 1 goal + probability of more than 1 goal only, eliminating the probability of 0 goals)?

Or is there anything I missed out?

Thanks for the help.

• There's something troubling with this question, and that is your implicit assumption that being 1-0 at half time maintains the original probabilities in some way. It's possible that being 1-0 at half time makes the teams behave differently. The leading team being more defensive now, making the probability of new goals very unlikely. Or maybe it drives the losing team into more aggressive tactics, increasing the probability that one of the teams will make a new goal.

I will have to make some simplifying assumptions in order to give a reasonable answer. I will have to assume that:

1. the score in each half follows the same distribution, and

2. the score in the second half is independent of the score in the first half.

In real life, I doubt this is entirely true. Maybe if one team scores in the first half, it?s more likely that there?s a goal in the second half because the team who?s behind is playing more aggressively. But I think I need to make this assumption to answer your question based on the information given.

Let?s define two random variables: $$X_{1}$$ = the number of goals in the first half and $$X_{2}$$ = the number of goals in the second half. Again, I?m assuming that the X's are independent and identically distributed.

We can reverse-engineer the distribution of the X?s from the given information. The distribution you defined at the top of the problem is the distribution of the random variable $$Y = X_{1} + X_{2}$$
So we know, for example that P(Y = 0) = 0.4. There?s only one way for Y to be equal to 0, both $$X_{1} = 0$$ and $$X_{2} = 0$$
$$P(X_{1} = 0, X_{2} = 0) = 0.4$$
Because the X?s are independent?
$$P(X_{1} = 0) * P(X_{2} = 0) = 0.4$$
Because the X?s are identically distributed?
$$P(X_{1} = 0)^{2} = P(X_{2} = 0)^{2} = 0.4$$
Therefore?
$$P(X_{1} = 0) = P(X_{2} = 0) = \sqrt{0.4} \approx 0.632$$
Given that there was a goal in the first half, the only way for there to be 1 goal or less total is if there are no goals in the second half, which we just discovered the probability of that is about 0.632.

Therefore, the probability of more than 1 goal in the game given that there was 1 goal in the first half is 1 - 0.632 = 0.368.
• Thank you for this very detailed answer. I really appreciate it.


# Polynomial Expansions (useful formulas)

Let $$\mathrm{A}$$ be the set of all "$$\color{Blue}{\text{useful}}$$" things.

Let $$\mathrm{B}$$ be the set of all "$$\color{Green}{\text{Awesome}}$$" things.

Let $$\mathrm{C}$$ be the set of all "$$\color{Red}{\text{fascinating}}$$" things.

Let $$\mathrm{D}$$ be the set of all "$$\color{Brown}{\text{easily understandable}}$$" things.

What this note contains is an element of $$\mathrm{A \cap B \cap C \cap D}$$....

$\color{Blue}{\textbf{Polynomial Expansions}}$

$$\mathbf{1.}\quad \displaystyle \dfrac{1-x^{m+1}}{1-x} = 1+x+x^2+...+x^m = \sum_{k=0}^m x^k$$

$$\mathbf{2.} \quad \displaystyle \dfrac{1}{1-x} = 1+x+x^2+... = \sum_{k=0}^\infty x^k$$

$$\mathbf{3.}\quad \displaystyle (1+x)^n = 1+\binom{n}{1} x + \binom{n}{2}x^2+...+\binom{n}{n}x^n = \sum_{k=0}^n \dbinom{n}{k} x^k$$

$$\mathbf{4.}\quad \displaystyle (1-x^m)^n = 1-\binom{n}{1}x^m+\binom{n}{2}x^{2m}-...+(-1)^n\binom{n}{n} x^{nm}$$

$$\quad \quad \quad \quad\quad \displaystyle = \sum_{k=0}^n (-1)^n \dbinom{n}{k}x^{km}$$

$$\mathbf{5.}\quad \displaystyle\dfrac{1}{(1-x)^n} = 1+ \binom{1+n-1}{1} x + \binom{2+n-1}{2} x^2+...+\binom{r+n-1}{r} x^r+......$$

$$\quad \quad \quad \quad \quad \displaystyle =\sum_{k=0}^\infty \dbinom{k+n-1}{k} x^k$$

$$\color{Purple}{\text{Tremendously useful}}$$ in calculating the $$\color{Blue}{\text{co-efficient}}$$ of any term in specially generating functions that we come across, in many combinatorics problems...

Taken From - Alan Tucker's "Applied Combinatorics"

Good luck problem solving !

Note by Aditya Raut
4 years, 2 months ago

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MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
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\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

## Comments

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I know that the second one only works for $$x<1$$. But do any of the others work for only $$x>1$$. Also, thank you so much for this note, it's very useful.

- 4 years, 2 months ago

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-1<x<1 actually.

- 4 years, 2 months ago

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Thank you, but do u know which ones only work for this case

- 4 years, 2 months ago

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Can you add parts of this page into the algebra wiki? I think that Algebraic Identities and Algebraic Manipulation - Identities, would be suitable places to add them.

Staff - 3 years, 12 months ago

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what is the derivation of 5th one.

- 4 years, 2 months ago

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Standard result, it's related to "Newton's generalised Binomial theorem", but if you do want the derivation please see it here

- 4 years, 2 months ago

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Does the last one work for 0<x<1???

- 4 years, 2 months ago

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Definitely 2.

- 4 years, 2 months ago

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set contains Everything

- 4 years, 2 months ago

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Is this for nerds like you?

- 4 years, 2 months ago

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Dude don't use nerds in a derogatory manner please. If you don't like nerds or aren't one yourself, you should remove yourself from Brilliant.org. Have a nice day.

- 4 years, 2 months ago

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@Finn Hulse , thanks for helping here, really !

img

I truly like your comment, by these many likes :-

img

- 4 years, 1 month ago

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Haha, anytime dude. :D

- 4 years, 1 month ago

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Agreed. BTW, some nerds can be good at sports as well.

- 4 years, 1 month ago

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BTW sharky, participate in JOMO 8, we miss your submission.

@Sharky Kesa , JOMO 8 starts $$\color{Red}{\textbf{TOMORROW}}$$ and has some good questions I made.

- 4 years, 1 month ago

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# (#Sharky_Surprises )

- 4 years, 1 month ago

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This is for using in generating functions we design for combinatorics problems.... For example, see the set "vegetable combinatorics".... (type in search bar simply).... That's for all who want to learn, nothing high-figh technique or anything, just formulas to get co-efficient of a specific term in a generating function. @Jack Daniel Zuñiga Cariño

- 4 years, 2 months ago

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