# D.E word problem

#### bergausstein

##### Active member
1. radium decompose at the rate proportional to the amount itself. if the half life is 1600, find the percentage remaining after at the end of 200 years.

can you me go about solving this. thanks!

#### MarkFL

Staff member
Let $R(t)$ be the mass of radium in a given sample at time $t$. How can we mathematically state how this mass changes with time in general? Just look at the first sentence and use that to try to model this change with an initial value problem.

#### bergausstein

##### Active member
$R(t)=R_oe^{kt}$ where $k<0$

now the half-life is 1600 so,

$1600=3200e^{kt}$

#### MarkFL

Staff member
$R(t)=R_oe^{kt}$ where $k<0$

now the half-life is 1600 so,

$1600=3200e^{kt}$

You have the correct equation for the mass of a sample, but the half-life being 1600 years means that at time t=1600 then $R(t)=\dfrac{1}{2}R_0$. That is (I like to always use a positive constant):

$$\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0$$

Divide through by $$\displaystyle R_0$$ and the convert from exponential to logarithmic form to solve for $k$. Another way to look at it is:

$$\displaystyle R(t)=R_02^{-\frac{t}{1600}}$$

And then to find the percentage remaining after $t$ years, use:

$$\displaystyle \frac{100R(t)}{R_0}=100\cdot2^{-\frac{t}{1600}}$$

#### bergausstein

##### Active member
from here solving for k

$\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0$

dividing both sides by $R_o$

$\displaystyle e^{-1600k}=\frac{1}{2}$

taking the $\ln$ of both sides

$-1600k=\ln\frac{1}{2}$

dividing both sides by -1600

$\displaystyle k=-\frac{-\ln2}{1600}$ or $k=0.0004332$

now

$\displaystyle R(t)=R_0e^{-(0.0004332)t}$

how can I find the initial $R_o$?

Last edited:

#### MarkFL

Staff member
You don't need to know the initial amount, you need to know what percentage of the initial amount remains after 200 years.

#### bergausstein

##### Active member
You don't need to know the initial amount, you need to know what percentage of the initial amount remains after 200 years.
how can I do that if there's no given initial amount? can you tell how to go about it?

#### Prove It

##### Well-known member
MHB Math Helper
from here solving for k

$\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0$

dividing both sides by $R_o$

$\displaystyle e^{-1600k}=\frac{1}{2}$

taking the $\ln$ of both sides

$-1600k=\ln\frac{1}{2}$

dividing both sides by -1600

$\displaystyle k=-\frac{-\ln2}{1600}$ or $k=0.0004332$

now

$\displaystyle R(t)=R_0e^{-(0.0004332)t}$

how can I find the initial $R_o$?
Do you think it's a good idea to give a decimal approximation when the function simplifies greatly if kept exact?

\displaystyle \begin{align*} k &= \frac{1}{1600} \ln{ (2) } \end{align*}

and so

\displaystyle \begin{align*} R(t) &= R_0 \exp { \left[ \frac{t}{1600} \ln{(2)} \right] } \\ &= R_0 \exp{ \left[ \ln{ \left( 2 ^{ \frac{t}{1600} } \right) } \right] } \\ &= R_0 \cdot 2^{ \frac{t}{1600} } \end{align*}

and so after 200 years, what proportion of the initial amount do you have?

#### MarkFL

Staff member
how can I do that if there's no given initial amount? can you tell how to go about it?
This is what I posted before:

...to find the percentage remaining after $t$ years, use:

$$\displaystyle \frac{100R(t)}{R_0}$$

#### bergausstein

##### Active member
$\displaystyle \frac{100R(200)}{R_0}=100\cdot2^{\frac{200}{1600}}$

do you mean like this? how can I find the percentage of remaining here?

#### MarkFL

Staff member
$\displaystyle \frac{100R(200)}{R_0}=100\cdot2^{\frac{200}{1600}}$

do you mean like this? how can I find the percentage of remaining here?
Reduce the exponent, and then that is the exact percentage remaining, and then you can use a calculator to obtain a decimal approximation if you like.

#### Prove It

##### Well-known member
MHB Math Helper
It appears there was something wrong with your initial model.

You know that your model involves exponential decay, so after each unit of time passes, there is a certain constant amount multiplying through. In other words

\displaystyle \begin{align*} R_1 &= C\,R_0 \\ R_2 &= C^2\,R_0 \\ R_3 &= C^3\,R_0 \\ \vdots \\ R_t &= C^t\,R_0 \end{align*}

Since you know that when \displaystyle \begin{align*} t = 1600, R_{1600} = \frac{1}{2}R_0 \end{align*}, that means

\displaystyle \begin{align*} \frac{1}{2}R_0 &= C^{1600}\,R_0 \\ \frac{1}{2} &= C^{1600} \\ \ln{ \left( \frac{1}{2} \right) } &= \ln{ \left( C^{1600} \right) } \\ \ln{ \left( \frac{1}{2} \right) } &= 1600 \ln{(C)} \\ \frac{1}{1600} \ln{ \left( \frac{1}{2} \right) } &= \ln{(C)} \\ \ln{ \left[ \left( \frac{1}{2} \right) ^{\frac{1}{1600}} \right] } &= \ln{(C)} \\ \left( \frac{1}{2} \right) ^{ \frac{1}{1600} } &= C \\ 2^{ -\frac{1}{1600}} &= C \end{align*}

and thus your model is \displaystyle \begin{align*} R(t) = \left( 2^{-\frac{1}{1600}} \right) ^t \, R_0 = 2^{-\frac{t}{1600}}\,R_0 \end{align*}.

Now if \displaystyle \begin{align*} t = 200 \end{align*}, that gives

\displaystyle \begin{align*} R(200) &= 2^{-\frac{200}{1600}}\,R_0 \\ &= 2^{-\frac{1}{8}}\,R_0 \\ &\approx 0.917 \, R_0 \end{align*}

So what percentage of the original amount do you have?


# Probability.

#### M R

##### Active member
A spinner has three possible outcomes which occur with probabilities a, b and c where a+b+c=1.

What is the expected number of spins required until all three outcomes are seen?

There's an easy way and a harder way to do this. Guess which I did first.

#### Opalg

##### MHB Oldtimer
Staff member
A spinner has three possible outcomes which occur with probabilities a, b and c where a+b+c=1.

What is the expected number of spins required until all three outcomes are seen?

There's an easy way and a harder way to do this. Guess which I did first.
This is a variation on the coupon collector's problem. In the case when all the probabilities are equal ($a=b=c=1/3$), the expected number of spins is $3\bigl(1 +\frac12 + \frac13) = \frac{11}2.$

In the general case, I certainly don't see an easy way to approach the problem, and I don't get an easy-looking formula for the answer.

Write $A$, $B$, $C$ for the outcomes with probabilities $a$, $b$, $c$ respectively. If the first spin gives an $A$, then the expected number of spins until a $B$ or $C$ occurs is $\dfrac1{b+c}$. The probability that this outcome is a $B$ is $\dfrac b{b+c}$, in which case the expected number of further spins until a $C$ turns up is $1/c.$ And the probability that a $C$ occurs before a $B$ is $\dfrac c{b+c}$, in which case the expected number of further spins until a $B$ turns up is $1/b.$ Therefore the total expected number of spins for all three outcomes to occur (given that the $A$ appears first) is $$1 + \frac1{b+c}\Bigl(\frac b{b+c}\,\frac1c + \frac c{b+c}\,\frac1b\Bigr) = 1 + \frac{b^2+c^2}{(b+c)^2bc} = \frac1{bc} - \frac2{(1-a)^2}$$ (in the last step, I have written the $b^2+c^2$ in the numerator as $(b+c)^2 - 2bc$, and in the denominator $b+c = 1-a$).

Multiply that by $a$, which is the probability of the $A$ occurring first, add two similar terms for the probabilities of $B$ or $C$ occurring first, and you get the answer for the expected number of spins as $$1 + \frac{a^2+b^2+c^2}{abc} - 2\biggl(\frac a{(1-a)^2} + \frac b{(1-b)^2} + \frac c{(1-c)^2}\biggl).$$

That looks messy, not the sort of thing that you could find easily? But it does reduce to $11/2$ when $a=b=c=1/3$, which makes me think that it should be correct.

#### M R

##### Active member
Hi Opalg

Maybe it wasn't easy but it was much easier than the other method which I will post if no one else does.

Your formula does give the right answer for a=b=c but not for other possibilities.

For comparison purposes:

a=1/2, b=1/3, c=1/6 should give 73/10

and

a=9/20, b=9/20, c=1/10 should give 353/33.

#### Opalg

##### MHB Oldtimer
Staff member
Your formula does give the right answer for a=b=c but not for other possibilities.

For comparison purposes:

a=1/2, b=1/3, c=1/6 should give 73/10

and

a=9/20, b=9/20, c=1/10 should give 353/33.
Stupid stupid mistake! My method was correct but I left out a $+$ sign, converting a sum into a product. The expression $$1 + \frac1{b+c}\Bigl(\frac b{b+c}\,\frac1c + \frac c{b+c}\,\frac1b\Bigr)$$
(for the expected number of spins for all three outcomes to occur, given that the $A$ appears first) should have been $$1 + \frac1{b+c} + \Bigl(\frac b{b+c}\,\frac1c + \frac c{b+c}\,\frac1b\Bigr) = 1 + \frac{bc + b^2+c^2}{(b+c)bc} = 1 + \frac{(b+c)^2 - bc}{(b+c)bc} = 1 + \frac{b+c}{bc} - \frac1{b+c}.$$ The answer for the total expected number of spins then comes out as $$1 + \frac{a(b+c)}{bc} + \frac{b(c+a)}{ca} + \frac{c(a+b)}{ab} - \frac a{b+c} - \frac b{c+a} - \frac c{a+b}.$$ That gives values agreeing with your results for a=1/2, b=1/3, c=1/6 and for a=9/20, b=9/20, c=1/10.

#### M R

##### Active member
The history of this problem (for me personally):

On another forum someone posted a 'hard probability question'. It was the a=b=9/20, c=1/10 case. Looking back at that forum I see that I managed to get an answer six days later.

Months later I saw a post on MHF which taught me a better approach so I went back and did it again, the easy way. Unfortunately I can't view MHF to find out who my teacher was.

I think it's essentially the same as Opalg's method but here's the 'easy' method as I did it:

Events A, B and C have probabilities a, b and c.

Let the expected waiting time until A occurs be $$E(W_A)$$,

and the expected waiting time until A and B occur be $$E(W_{AB})$$ and so on.

$$E(W_{AB})=c(1+E(W_{AB}))+a(1+E(W_B))+b(1+E(W_A))$$

$$E(W_{AB})=c(1+E(W_{AB}))+a(1+1/b)+b(1+1/a)$$

$$E(W_{AB})=1 +cE(W_{AB})+a/b+b/a$$

$$E(W_{AB})=\frac{1+a/b+b/a}{1-c}$$

Similarly

$$E(W_{AC})=\frac{1+a/c+c/a}{1-b}$$

and

$$E(W_{BC})=\frac{1+b/c+c/b}{1-a}$$

Then $$E(W_{ABC})=a(1+E(W_{BC}))+b(1+E(W_{AC}))+c(1+E(W_{AB}))$$

and the method I used at first:

By thinking about how the sequence ends I knew I wanted all the ways to get As and Bs ending with C etc.

ABC
BAC

AABC
ABAC
BAAC

ABBC
BABC
BBAC

etc.

This led me to the sum

$$\displaystyle E(W)=\sum_{n=2}^\infty (n+1) \sum_{r=1}^{n-1} \binom{n}{r}(a^rb^{n-r}c+a^rc^{n-r}b+b^rc^{n-r}a)$$

$$\displaystyle =\sum_{n=2}^\infty (n+1) [ c((a+b)^n-a^n-b^n) +b((a+c)^n-a^n-c^n)+a((b+c)^n-b^n-c^n)]$$

This sum involves a number of geometric progressions and a number of sums of another type.

The other type is of the form $$\displaystyle S=\sum_{n=2}^\infty n x^n$$.

Multiplying by $$\displaystyle x$$ gives $$\displaystyle Sx=\sum_{n=2}^\infty n x^{n+1}$$ and so $$\displaystyle S-Sx = 2x^2+x^3+x^4+...$$

$$\displaystyle S(1-x)=x^2 + \frac{x^2}{1-x}$$ and $$\displaystyle S=\frac{x^2}{1-x}+\frac{x^2}{(1-x)^2}$$

Applying this formula, together with the formula for the sum of a GP we get

$$\displaystyle E(W)=$$

$$\displaystyle (a+b)^2-\frac{ca^2}{1-a}-\frac{cb^2}{1-b}$$

$$\displaystyle + (a+c)^2-\frac{ba^2}{1-a}-\frac{bc^2}{1-c}$$

$$\displaystyle + (b+c)^2-\frac{ab^2}{1-b}-\frac{ac^2}{1-c}$$

$$\displaystyle +(a+b)^2(1+1/c)-c\left(\frac{a^2}{1-a}+\frac{a^2}{(1-a)^2}+\frac{b^2}{1-b}+\frac{b^2}{(1-b)^2}\right)$$

$$\displaystyle +(a+c)^2(1+1/b)-b\left(\frac{a^2}{1-a}+\frac{a^2}{(1-a)^2}+\frac{c^2}{1-c}+\frac{c^2}{(1-c)^2}\right)$$

$$\displaystyle +(b+c)^2(1+1/a)-a\left(\frac{b^2}{1-b}+\frac{b^2}{(1-b)^2}+\frac{c^2}{1-c}+\frac{c^2}{(1-c)^2}\right)$$

What a mess!

This could be simplified using $$\displaystyle a+b+c=1$$ but having checked it against the first method I'm happy that it's correct and I'm not interested in doing the simplification.


# Surely You're Joking, Mr. Feynman! $\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx$ [duplicate]

Prove the following $$$$\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx=\frac{\pi}{4}+\frac{\pi}{4e^2}$$$$

I would love to see how Mathematics SE users prove the integral preferably with the Feynman way (other methods are welcome). Thank you. (>‿◠)✌

Original question:

And of course, for the sadist with a background in differential equations, I invite you to try your luck with the last integral of the group.

$$$$\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx$$$$

Source: Integration: The Feynman Way

• "The Feynman way" is just a fancy name for the good old differentiation under the integral? – Daniel Fischer May 21 '14 at 11:30
• @DanielFischer I know that Sir but I prefer naming "The Feynman Way" (ô‿ô) – Anastasiya-Romanova 秀 May 21 '14 at 11:33
• @DanielFischer: there seems to be a tendency on this site to label any differentiation of an artificially introduced parameter under the integral sign as "Feynman's <method, way, etc.>." I doubt Feynman invented it, but I see that loads of people read "Surely You're Joking...". – Ron Gordon May 21 '14 at 12:20
• @RonGordon Indeed. It's a good read, by the way. Now, will you have lemon or milk in your tea? – Daniel Fischer May 21 '14 at 12:23
• @V-Moy: Hey, I never said it was wrong; I just think it is a little funny. BTW it would not be the first time that a method or something else is named after its popularizer rather than its inventor. – Ron Gordon May 21 '14 at 12:38

This integral is readily evaluated using Parseval's theorem for Fourier transforms. (I am certain that Feynman had this theorem in his tool belt.) Recall that, for transform pairs $f(x)$ and $F(k)$, and $g(x)$ and $G(k)$, the theorem states that

$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k)$$

In this case, $f(x) = \frac{\sin^2{x}}{x^2}$ and $g(x) = 1/(1+x^2)$. Then $F(k) = \pi (1-|k|/2) \theta(2-|k|)$ and $G(k) = \pi \, e^{-|k|}$. ($\theta$ is the Heaviside function, $1$ when its argument is positive, $0$ when negative.) Using the symmetry of the integrand, we may conclude that

\begin{align}\int_0^{\infty} dx \frac{\sin^2{x}}{x^2 (1+x^2)} &= \frac{\pi}{2} \int_0^{2} dk \, \left ( 1-\frac{k}{2} \right ) e^{-k} \\ &= \frac{\pi}{2} \left (1-\frac1{e^2} \right ) - \frac{\pi}{4} \int_0^{2} dk \, k \, e^{-k} \\ &= \frac{\pi}{2} \left (1-\frac1{e^2} \right ) + \frac{\pi}{2 e^2} - \frac{\pi}{4} \left (1-\frac1{e^2} \right )\\ &= \frac{\pi}{4} \left (1+\frac1{e^2} \right )\end{align}

• @V-Moy: Yeah, I'm fast like that. – Ron Gordon May 21 '14 at 12:38
• Surely You're Joking, Mr. Gordon! （‐＾▽＾‐） – Anastasiya-Romanova 秀 May 21 '14 at 12:41
• @RonGordon Why did you answer the same question again since you knew it was a duplicate? You have already answered this question for me : math.stackexchange.com/questions/691798/…. – Jeff Faraci May 22 '14 at 14:15
• @Integrals: because I answered it three months ago and at my age, dementia sets in. I admit that sometimes I just answer the question without looking back through my vast database of answers. So, no, I did not know it was a duplicate. – Ron Gordon May 22 '14 at 14:21
• @Integrals: I would have if I had known. But so what - the worst thing in the world is that duplicate correct (and consistent) answers exist in the site? Vote to close as a duplicate if this bothers you. – Ron Gordon May 22 '14 at 14:27

Here is my attempt: \begin{align} \int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}dx&=\int_0^\infty\left[\frac{\sin^2x}{x^2}-\frac{\sin^2x}{1+x^2}\right]dx\\ &=\int_0^\infty\frac{\sin^2x}{x^2}dx-\frac{1}{2}\int_0^\infty\frac{1-\cos2x}{1+x^2}dx\\ &=\frac{\pi}{2}-\frac{1}{2}\int_0^\infty\frac{1}{1+x^2}dx+\frac{1}{2}\int_0^\infty\frac{\cos2x}{1+x^2}dx\\ &=\frac{\pi}{2}-\frac{1}{2}\frac{\pi}{2}+\frac{1}{2}\frac{\pi}{2e^2}\\ &=\frac{\pi}{4}+\frac{\pi}{4e^2} \end{align} where I use these links: $\displaystyle\int_0^\infty\frac{\sin^2x}{x^2}dx$ and $\displaystyle\int_0^\infty\frac{\cos2x}{1+x^2}dx$ to help me out.

Unfortunately, this is not the Feynman way but I still love this method.

• Thanks for fixing my answer @MycrofD. I'm too excited so I forgot about my English grammar. ≥Ö‿Ö≤ – Anastasiya-Romanova 秀 May 21 '14 at 11:30
• Next time put your attempt in the question. Besides, you got it right (>‿◠)✌. – Shahar May 21 '14 at 11:53
• @Shahar Thanks, +1 for your emo. ≧◠◡◠≦✌ – Anastasiya-Romanova 秀 May 21 '14 at 12:09
• @V-Moy You seem to be an expert in emo's and integrals. ✌ – Sawarnik May 21 '14 at 17:43
• @Sawarnik You might say I'm expert in emo but not in integral, not yet (✿◠‿◠) – Anastasiya-Romanova 秀 May 22 '14 at 8:17

Consider $$I(a)=\int_0^\infty\frac{\sin^2(ax)}{x^2(x^2+1)}dx$$ Differentiate it twice. Since $$\int_0^\infty\frac{\cos(kx)}{x^2+1}dx=\frac{\pi}{2e^k}$$ for $k>0$ we get $I''(a)=\pi e^{-2a}$. Note that $I'(0)=I(0)=0$, so after solving respective IVP we get $$I(a)=\frac{\pi}{4}(-1+2a+e^{-2a})$$ Now it only remains to substitute for $a=1$.

• Wait a second! Why never cross to mind to use Feynman method twice?? +1 Sir! ≥►.◄≤ – Anastasiya-Romanova 秀 May 21 '14 at 12:47
• @V-Moy Because you're not thinking like Feynman! Suppose the Feynman method just transforms your complicated integral into another complicated integral. Well, what tricks do you know for handling complicated integrals? That's right, the Feynman method! :) – David H May 21 '14 at 13:59
• Thanks for your advice Mr. @DavidH. I'll try to think like Feynman. I have a lot of time though. ᕙ(^▽^)ᕗ – Anastasiya-Romanova 秀 May 21 '14 at 16:17
• Why did you post this solution again? You knew this was a duplicate, this is a great solution but still. -1. math.stackexchange.com/questions/691798/… – Jeff Faraci May 22 '14 at 14:14
• @Integrals because it is not forbidden, and this answer perfectly fits to the question even more than the in the link you gave above. – Norbert May 22 '14 at 15:47


# Is the sequence $x_n=\dfrac{1}{\sqrt{n}}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{n}}\right)$ monotone?

Observe that $$x_1=1$$ and $$x_2=\dfrac{1}{\sqrt{2}}\left(1+\dfrac{1}{\sqrt{2}}\right)>\dfrac{1}{\sqrt{2}}\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right)=1$$.

Thus, $$x_2>x_1$$. In general, we also have $$x_n=\dfrac{1}{\sqrt{n}}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\ldots+\dfrac{1}{\sqrt{n}}\right)>\dfrac{1}{\sqrt{n}}\left(\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+\ldots+\dfrac{1}{\sqrt{n}}\right)=1$$.

Thus, $$x_n\geq 1$$ for all $$n\in \mathbb{N}$$. Also, we have,

$$x_{n+1}=\dfrac{1}{\sqrt{n+1}}\left(\sqrt{n}x_n+\dfrac{1}{\sqrt{n+1}}\right)=\dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}$$.

Is it true that $$x_{n+1}>x_n$$?

Edit : Thanks to the solution provided by a co-user The73SuperBug. Proving $$x_{n+1}-x_n>0$$ is equivalent to proving $$x_n<1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$$. This is explained below :

\begin{aligned} &x_{n+1}-x_n>0\\ \Leftrightarrow & \dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}-x_n>0\\ \Leftrightarrow & \left(1-\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)x_n-\dfrac{1}{n+1}<0\\ \Leftrightarrow & x_n<\dfrac{1}{(n+1)\left(1-\dfrac{\sqrt{n}} {\sqrt{n+1}}\right)}\\ \Leftrightarrow & x_n<1+\dfrac{\sqrt{n}}{\sqrt{n+1}}. \end{aligned}

• It's basically a Riemann sum for a convex function. The (good) accepted answer focuses on the specific function you have, but I think a generalization would not harm, so I've added an answer based on this alternative idea. May 17, 2020 at 12:17

The sequence is indeed increasing. Using what you have left off we need to prove: $$x_{n+1} - x_n > 0\iff ...x_n < 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$$. We prove this by induction on $$n \ge 1$$. Clearly $$x_1 = 1 < 1+ \sqrt{\frac{1}{2}}$$. Assume $$x_n < 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}$$, we show: $$x_{n+1} < 1+\dfrac{\sqrt{n+1}}{\sqrt{n+2}}$$. Using the recursive formula you had above: $$x_{n+1} = \dfrac{\sqrt{n}}{\sqrt{n+1}}x_n+\dfrac{1}{n+1}< \dfrac{\sqrt{n}}{\sqrt{n+1}}\left(1+\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)+\dfrac{1}{n+1}= 1+\dfrac{\sqrt{n}}{\sqrt{n+1}}< 1+\dfrac{\sqrt{n+1}}{\sqrt{n+2}}$$ which is clear because $$n(n+2) < (n+1)^2$$. Thus by induction $$x_n < 1 +\dfrac{\sqrt{n}}{\sqrt{n+1}}$$ and in turn implies $$x_{n+1} > x_n, \forall n \ge 1$$. Thus the sequence is increasing.
A generalization I couldn't pass by. Let $$\color{blue}{S_n=\frac1n\sum_{k=1}^{n-1}f\big(\frac{k}{n}\big)}$$ where $$f:(0,1)\to\mathbb{R}$$ is strictly convex: $$f\big((1-t)a+tb\big)<(1-t)f(a)+tf(b)\quad\impliedby\quad a If we put $$a=k/(n+1),b=(k+1)/(n+1),t=k/n$$ for $$0 here, we obtain $$f\Big(\frac{k}{n}\Big)<\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\frac{k}{n}f\Big(\frac{k+1}{n+1}\Big),$$ which, after summing over $$k$$ (to have "$$<$$" still, we must assume $$n>1$$), gives $$\sum_{k=1}^{n-1}f\Big(\frac{k}{n}\Big)<\sum_{k=1}^{\color{red}{n}}\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\sum_{k=\color{red}{1}}^{n}\frac{k-1}{n}f\Big(\frac{k}{n+1}\Big)=\frac{n-1}{n}\sum_{k=1}^{n}f\Big(\frac{k}{n+1}\Big),$$ i.e. $$\color{blue}{n^2 S_n<(n^2-1)S_{n+1}}$$. Returning to the question, if we put $$f(x)=1/\sqrt{x}$$, we get $$x_n=S_n+1/n$$ and $$n^2 x_n<(n^2-1)x_{n+1}+1$$; since $$x_{n+1}>1$$, the latter implies the needed $$x_n.


# Math Help - Midterm review

1. ## Midterm review

Agh, I've been working on this question for an hour and can't seem to make any headway. I seem to be completely lost on it.

a.) Show that the positive integers less than 11, except 1 and 10 can be split into pairs of integers such that each pair consists of integers that are inverses of each other modulo 11.

b.) use part (a) to show that 10! = -1(mod 11)

Any help would be appreciated

2. Originally Posted by Niotsueq
Agh, I've been working on this question for an hour and can't seem to make any headway. I seem to be completely lost on it.

a.) Show that the positive integers less than 11, except 1 and 10 can be split into pairs of integers such that each pair consists of integers that are inverses of each other modulo 11.

b.) use part (a) to show that 10! = -1(mod 11)
For a), you can find out by trial and error that, for example, $2\times6 = 12\equiv1\!\!\!\pmod{11}$. Therefore 2 and 6 are inverses of each other. So also are 7 and 8, because $7\times8 = 56\equiv1\!\!\!\pmod{11}$.

For b), $10! = 1\times 2\times3\times\cdots\times10 = 1\times(2\times6)\times\cdots\times(7\times8)\time s\cdots\times10$ (pair the numbers off as in part a)).

3. Hello, Niotsueq!

a) Show that the positive integers less than 11, except 1 and 10,
can be split into pairs of integers such that each pair consists of integers
that are inverses of each other modulo 11.
. . . . . $\begin{array}{ccc}2\cdot6 &\equiv & 1\text{ (mod 11)} \\3\cdot4 &\equiv & 1\text{ (mod 11)} \\ 5\cdot9 &\equiv & 1\text{ (mod 11)} \\ 7\cdot8 & \equiv& 1 \text{ (mod 11)} \end{array}$

b) use part (a) to show that 10! = -1 (mod 11)

$10! \;\equiv\;1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cd ot8\cdot9\cdot10\,\text{ (mod 11)}$

. . . $\equiv \;1\cdot(2\cdot6)\cdot(3\cdot4)\cdot(5\cdot9)\cdot (7\cdot8)\cdot10\,\text{ (mod 11)}$

. . . $\equiv\;1\cdot 1\cdot 1\cdot 1\cdot1\cdot10\,\text{ (mod 11)}$

. . . $\equiv\;10\,\text{ (mod 11)}$

. . . $\equiv\;\text{-}1\,\text{ (mod 11)}$

4. Thanks! That's a lot easier than I thought. But, I'm not quite sure why, for instance:

2*6 = 1 (mod 11)

How does this say that 2 and 6 are inverses? I think I may be confused on the definition.

Thanks alot to both of you.

5. Originally Posted by Niotsueq
Thanks! That's a lot easier than I thought. But, I'm not quite sure why, for instance:

2*6 = 1 (mod 11)

How does this say that 2 and 6 are inverses? I think I may be confused on the definition.

Thanks alot to both of you.

$2\cdot 6 \equiv 1 \mod 11$

this shows that they are inverses of each other.

remember what an inverse is in regular multiplication?

$3 \cdot \frac{1}{3}=1$

do you see how the original means they are inverses now?

6. Originally Posted by GaloisTheory1
$2\cdot 6 \equiv 1 \mod 11$

this shows that they are inverses of each other.

remember what an inverse is in regular multiplication?

$3 \cdot \frac{1}{3}=1$

do you see how the original means they are inverses now?

So two numbers are inverses of one another mod(n) if their product is congruent to one mod(n)?


# Prove that the sequence $(a_n)$ defined by $a_0 = 1$, $a_{n+1} = 1 + \frac 1{a_n}$ is convergent in $\mathbb{R}$

I will post the exercise below:

Prove that the sequence $(a_n)$ defined by $a_0 = 1$, $a_{n+1} = 1 + \frac 1{a_n}$ for $n \in \mathbb N$ is convergent in $\mathbb R$ with the Euclidean metric, and determine afterwards is limit. Can you intepret the limit geometrically (hint: Golden ratio)?

So I need to prove that the sequence is convergent in $\mathbb{R}$ with the Euclidean metric, and how do I prove that? The limit must be $1$, but how to interpret it geometrically?

• All I am willing to say is that "hint: Golden ratio" is a very strong hint. Oct 8, 2013 at 9:43
• The limit is not 1. @Duronman, the series is not monotonically increasing either. Oct 8, 2013 at 9:47
• @AdrianRatnapala true! I wrote too fast, sorry, I thought of $a_n + \frac{1}{a_n}$ because of the $1$. I will delete the comment to avoid confusion. Oct 8, 2013 at 9:54

Hint: If the limit $L$ exists, it must satisfy $L = 1 + \frac{1}{L}$, and so it cannot be 1. The solutions are the roots of the equation $L^2 - L - 1 = 0$, and so $L \in \{\frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2} \}$. That's where the golden ratio comes into play. Note also that the limit cannot be $\frac{1-\sqrt{5}}{2}$ since $a_n > 0$ for all $n$.

Let $\phi$ be the golden ratio, which is the only positive real such that $\phi = 1 + 1/\phi$. Now consider the sequence $\epsilon_n = a_n - \phi$. We seek to prove the invariant:

$$|\epsilon_n/\epsilon_0| \le \phi^{-n}.$$

For $n=0$, this is clear by inspection. With some algebra we see that:

$$\epsilon_{n+1} = 1 + {1\over a_n} - \phi = -{\epsilon_n\over a_n\phi}.$$

Now our invariant already implies that $a_n = \phi + \epsilon_n \ge \phi - \phi^{-n}\epsilon_0 \ge 1.$ Therefore

$$|\epsilon_{n+1}| \le \left|{\epsilon_n\over \phi}\right| \le \phi^{-(n+1)}.$$

So our invariant remains after $n \rightarrow n+1$. And therefore for all nonnegative integers $n$ by induction. But since $\phi > 1$, that invariant means that $\epsilon_n$ approaches zero or $a_n$ approaches $\phi$ as $n$ goes to infinity.

Writing out the first few terms will yield a very interesting pattern:

$a_0 = \frac{1}{1}, a_1 =\frac{2}{1}, a_2=\frac{3}{2}, a_3=\frac{5}{3} ...$

Do u see it now. Every term is a ratio of fibonacci nos. i.e. if $T_n$ is the nth fibonacci number starting from 1, then $a_n = \frac{T_{n+2}}{T_{n+1}}$. The limit of this is indeed the golden ratio. Try solving the difference equation to see it in case you are unaware of this result.

As for the proof that the observation is correct and not a coincidence, use induction. We verified the base cases above, so assume $$a_n = \frac{T_{n+2}}{T_{n+1}}$$

Now $$a_{n+1} = 1+ \frac{1}{a_n} = 1 + \frac{T_{n+1}}{T_{n+2}}= \frac{T_{n+2} + T_{n+1}}{T_{n+2}} = \frac{T_{n+3}}{T_{n+2}}$$

Hence proved. $\blacksquare$

• But you still need to prove that the ratios of consecutive Fibonnaci numbers actually converge to anything at all. Oct 10, 2013 at 8:44
• I told OP to solve the difference equation for $a_n = a_{n-1} + a_{n-2}$. From that you can evaluate the limit. Oct 10, 2013 at 8:48
• I am missing some knowledge about difference equations. The easiest method to find that limit (if it exists) I know is to find the roots of the iteration rule translated into a polynomial. But I did not know the limit was guaranteed to exist. Oct 14, 2013 at 18:07
• You are referring to the Z Transform method for solving difference equations. With it, you will get the solution for all n. Then you can proceed by my method. Oct 14, 2013 at 18:45
• Adrian, another way is to get the general solution to the recurrence relation and then just take limits. There will invariably be a dominating term which determines the limit. Dec 22, 2013 at 3:30

We will consider decreasing function $f:(0, \infty)\rightarrow(0,\infty), f(x)=1+\frac{1}{x}$. Function $fof:(0, \infty ) \rightarrow(0, \infty)$ is increasing and therefore subsequence $a_{2n},n\geq0$ is increasing, and subsequence $a_{2n +1},n\geq0$ is decreasing. Because $1\leq a_{2n}<a_{2n +1}\leq2$, it follows that both are converging subsequences with common limit the golden ratio, which is the limit sequence in question.


# How many different Tsuro tiles can exist?

The boardgame Tsuro consists of tiles, which each have 8 entry points. Each tile connects each point to exactly one other point. The game manual claims every tile is unique. The game consists of 35 such tiles.

How many unique such tiles could possibly exist?

My reasoning:

• You start with 8 free points. Choose any point, you now have 7 possibilities to connect.
• You now have 6 free points. Choose any point, you now have 5 possibilities to connect.
• You now have 4 free points. Choose any point, you now have 3 possibilities to connect.
• You now have 2 free points. Connect them (no choice possible).

This would lead to 7*5*3 : 105 possibilities. But I wouldn't know how to eliminate "doubles" caused by rotating a tile. Should I divide by 4, since 4 rotations are possible? That would be 26 tiles... but the game itself contains 35 and they are unique.

How should I reason?

• Well, here's why you can't just divide by $4$: Not all the rotations are unique - consider the tile in which each point is connected to the one directly opposite it on the other side - then this tile is rotationally symmetric. Another example in which not all rotations are unique is the bottom left tile in the picture you provided. – Peter Woolfitt Jan 2 '15 at 10:31

You can use Burnside's Lemma. I'll go through the calculation here, though you should read that page for this to make sense.

The set X is all 105 possibilities. The group G that acts on X is $\langle r | r^4 = 1 \rangle$, where r is a 90 degree rotation. Burnside's Lemma is applicable since each tile corresponds exactly to one orbit in X under G. The number of elements of X fixed by each element of G is

• $1: 105$
• $r: 5$
• $r^2: 25$
• $r^3: 5$

So the number of orbits (tiles) is $\frac{1}{4}(105+5+25+5)=35$. The hardest part is computing the $r^2: 25$ entry; here's how I did it:

• Case 1: No pair of antipodal points is connected. Choose any point: you have 6 legal possibilities to connect. Whatever you choose, $r^2$ fixes a second symmetrical connection, so you have 4 points left with 2 legal ways to connect them, for a total of 6*2=12 possibilities.
• Case 2: 2 pairs of antipodal points are connected. There are $\binom{4}{2}=6$ ways to pick the pairs and 2 ways to connect the reamining 4 points, for another 12 possibilities.
• Case 3: All 4 antipodal pairs are connected. There is only 1 such possibility.
• You'd probably also find this page interesting. – Benjamin Cosman Aug 28 '16 at 21:14
• $<abc>$ looks like this: $<abc>$, and $\langle abc\rangle$ looks like this $\langle abc\rangle$. #FriendsDoNotLetFriendsUseTheWrongAnglyThings – Mariano Suárez-Álvarez Aug 28 '16 at 21:19
• Interesting approach, thank you! – Konerak Aug 29 '16 at 7:07

One can reduce further the $105$ possible tiles, down to the minimum possible of $35$ topologically unique tiles via $mod\, 8$ arithmetic, $1+7=(8)=0$.

Modulo $8$ arithmetic (each tile has $8$ ports numbered $0123457$) allows $CW$ and $CCW 2$-dimensional rotations (can not flip a tile, as a flip would represent a $3D$ rotation). To keep this simple, there are no negative integers here, $0-1=7$ is being interpreted as $7+1=0$.

A $45^{\circ}$ rotation is represented by adding/subtracting 1 to/from the port #.

Rotations of $90^{\circ}, 180^{\circ}, 270^{\circ}$ are represented by adding/subtracting $2, 4, 6$ respectively. I just added $2, 4, 6$ used only addition.

Represent a pair of two connected ports of a tile by a $2$-digit number. For example $15$ means port#1 is connected to port#5, i.e. not decimal fifteen.

Represent each tile via a quad set of number pairs, non repeating digits,for example $\{01, 23, 45, 67\}$.

Note $\{01, 14, 26, 57\}$ is not a valid tile as digit "1" -port#1- appears twice.

Example of $180^{\circ}$ rotation:

Tile $\{04, 12, 36, 57\}$ is rotated $180^{\circ}$ by adding (modulo 8) $+4$ to each of its digits.

$$0+4=4$$

$$4+4=(8)=0$$

$$1+4=5$$

$$2+4=6$$

$$3+4=7$$

$$6+4=(10)=2$$

$$5+4=(9)=1$$

$$7+4=(11)=3$$

Thus rotated tile becomes $\{40, 56, 72, 13\}$ and will order/describe/ its ports from low-to-high (as each numbers-pair is commutative, $40=04$ and $72=27$).

Therefore this $180^{\circ}$ rotated tile becomes $\{04, 13, 27, 56\}$.

Rotate all $105$ tiles by $0^{\circ}, 90^{\circ}, 180^{\circ}, 270d^{\circ}$, (find & write all quad sets).

Use Excel, reorder the port# numbers from low-to-high, and eliminate all duplicate tiles.

End result = $35$ unique quad sets representing $35$ unique tiles.

May follow my analysis summary at:

There are 5 ways to make a tile that is invariant under $90^\circ$ rotations: Just connect one pint to any other point except its rotational predecessor or successor, then the rest is fixed. This would correct your count to $\frac{105-5}4+5=29$, which is more, but still too few. You should also count, how many tiles are possible with a $180^\circ$ symmetry. Of course, the $90^\circ$ symmetric tiles are among them, but there are also some "new" such tiles. If there are $k$ such tiles, then the total count gets corrected to $\frac{105-5-k}{4}+5+\frac k2$, so you can make an educated guess what $k$ is ;)


# An extrasensory perception strategy :-)

Inspired by classical Joseph Banks Rhine experiments demonstrating an extrasensory perception (see, for instance, the beginning of the respective chapter of Jeffrey Mishlove book “The Roots of Consciousness”), I consider the following experiment. A deck of cards is given to a magician John. Then John consecutively takes the cards from the deck, trying to guess suit of the taken card. He looks at the card after the guess for a feedback. The magician wishes to maximize the expected number $E$ of right guesses. For this purpose he devised the following Strategy: at each turn to guess a suit which has a maximal number of card left in the deck. As an easy exercise we can prove that for any sequence of cards in the deck Strategy ensures at least $n$ right guesses, where $n$ is the maximal number of cards with one suit in the deck. But we can consider a more interesting and complicated problem to calculate the expectation $E$ for Strategy (here we are assuming that the deck is so well shuffled such that all sequences of cards have equal probability). By the way, I conjecture that Strategy is the best for maximizing the expectation $E$, that is any other strategy yields not greater value of $E$. Now I wish to evaluate the expectation $E$ for Strategy. For the simplicity we shall consider only a case when there are only two suits ($m\ge 0$ cards of the first suit and $n\ge m$ cards of the second suit). Then $E(0,n)=n$ for each $n$ and we have the following recurrence $$E(m,n)=\frac{n}{m+n}(E(m,n-1)+1)+ \frac{m}{m+n}E(m-1,n)$$

for each $n\ge m\ge 1$.

The rest is true provided I did not a stupid arithmetic mistake.

I was interested mainly in asymptotics for the case $m=n$ and computer calculations suggested that $E(n,n)\sim n+c\sqrt{n}+o(\sqrt{n})$ for $c\approx 0.88\dots$.

Evaluating formulas for $E(m,n)$ for small values of $m\le 6$, I conjectured that there is a general formula

$$E(m,n)=n+m\sum_{i=1}^m\frac {c_{m,i}}{n+i}$$

for each $n\ge m\ge 1$, where $c_{m,i}$ are some integers satisfying the recurrence

$$(m-i)c_{m,i}+ic_{m,i+1}=(m-1)c_{m-1,i}$$

for every $1\le i\le m-1$.

Here are my values for $c_{m,i}$

i\m|  1   2   3   4   5   6
---+------------------------
1 |  1   2   4   8  16  32
2 |     -1  -4 -12 -32 -80
3 |          1   6  24  80
4 |             -1  -8 -40
5 |                  1  10
6 |                     -1

Then I discovered that for my data $c_{m,i}$ is divisible by $2^{m-i}$. After I did the division, I surprisingly obtained that $$c_{m,i}=(-1)^{i-1}2^{m-i}{m-1 \choose i-1}.$$ I expect that I can easily prove this equality by induction.

But I did not stop at this point because I observed that now the general formula for $E(m,n)$ can be compressed to the form

$$E(m,n)=n+m\int_0^1 x^n(2-x)^{m-1} dx.$$

All of above sounds nice for me and I spent a good time investigating the problem, but I am a professional mathematician, although I am not a specialist in the domain of the above problem. Therefore I care about the following questions. Are the above results new, good and worthy to be published somewhere? What another related problems are worthy to be investigated?

Thanks and merry Holidays.

• – Alex Ravsky Dec 21 '13 at 0:49
• Your expression for $E(m,n)$ looks a lot like Kolmogorov forward equation. – Alex Dec 21 '13 at 1:21
• Is your table of $c_{m,i}$ correct for $i=1$? The triangle and your expression for $c_{m,i}$ looks very like OEIS A013609 – Henry Dec 23 '13 at 22:15
• @Henry Thanks, I corrected the table. – Alex Ravsky Mar 9 '14 at 15:10
• Your integral expression for $E(m,n)$ is closely related to $\displaystyle B_{1/2}(n+1,m) = \int_{0}^{1/2} x^n (1-x)^{m+1} dx$, an incomplete Beta function. – Henry Mar 9 '14 at 16:58

This is to address your question about the asymptotics when $m=n$.

We only need to study the integral

$$I(n) = \int_0^1 x^n (2-x)^{n-1}\,dx$$

or, after we've made the change of variables $x = 1-y$,

\begin{align} I(n) &= \int_0^1 (1-y)^n (1+y)^{n-1}\,dy \\ &= \int_0^1 (1+y)^{-1} \exp\left[n \log\left(1-y^2\right)\right]\,dy. \end{align}

We'll proceed using the Laplace method. The quantity $\log(1-y^2)$ achieves its maximum at $y=0$, and near there we have

$$\log\left(1-y^2\right) \sim -y^2.$$

This motivates us to make the change of variables $\log(1-y^2) = -z^2$, so that

$$I(n) = \int_0^\infty \frac{z e^{-z^2}}{(1+\sqrt{1-e^{-z^2}})\sqrt{1-e^{-z^2}}} e^{-nz^2}\,dz.$$

Near $z=0$ we have

$$\frac{z e^{-z^2}}{(1 + \sqrt{1-e^{-z^2}})\sqrt{1-e^{-z^2}}} = 1 - z + \frac{z^2}{4} + \frac{z^4}{96} - \frac{z^6}{384} + \cdots,$$

and integrating term-by-term we obtain the asymptotic series

$$I(n) \approx \frac{\sqrt{\pi}}{2n^{1/2}} - \frac{1}{2n} + \frac{\sqrt{\pi}}{16n^{3/2}} + \frac{\sqrt{\pi}}{256n^{5/2}} - \frac{5 \sqrt{\pi}}{2048n^{7/2}} + \cdots.$$

Thus

$$E(n,n) \approx n + \frac{\sqrt{\pi}}{2} n^{1/2} - \frac{1}{2} + \frac{\sqrt{\pi}}{16} n^{-1/2} + \frac{\sqrt{\pi}}{256} n^{-3/2} - \frac{5 \sqrt{\pi}}{2048} n^{-5/2} + \cdots.$$

• Thanks, this is very good! :-D This correlates with my computer and analytical calculations. As I already wrote, my old computer calculations suggested that $E(n,n)\sim n+c\sqrt{n}+o(\sqrt{n})$ for $c\approx 0.88\dots$. Now I see that $c\approx \sqrt{\pi}/2$ with the precision up to two significant digits. I derived the integral formula only yesterday so I did not try to derive the asymptotics from it yet. Moreover, I did not know about Laplace method. – Alex Ravsky Dec 21 '13 at 6:55

If I understand right the game, and picturing the evolution as a path in a discrete grid -from $(0,0)$ to ($n,n$)- it's seen that each segment that goes towards the diagonal is a "win"; the other are misses, except for the ones that start from the diagonal itself, half of which are wins. Then, if the path of length $2n$ have $c$ diagonal-touchings (including the start, excluding the end), the total of wins is $n+c/2$.

Hence, the problem is converted to the (probably simpler and already studied) problem of computing the expected numbers of diagonal touchings on a lattice path - or, in a fair ballot counting problem, compute the expected number of ties (or lead changes).

You ask "What (...) related problems are worthy (of investigation)?"

There is the question of what the expected score is if we use that strategy (which is the best strategy) with a Zener pack of 25 cards, namely which contains 5 cards of each of 5 different shapes. I asked that question here: the answer is 8.65.

Then there are misère problems where we try to mimimise the score. I have asked the misère version of the Zener card question here.


Why is $1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{n} \approx \ln(n) + \gamma$?

On StackExchange, I read that the harmonic series up to $\frac{1}{n}$ is approximately $\ln(n) + \gamma$, where $\gamma$ is the Euler-Mascheroni constant, which is close to $0.5772$. When I researched the Euler-Mascheroni constant, I only found it defined in terms of the difference between the harmonic series and $\ln(n)$.

Why is the series able to be approximated in this way, and what is the Euler-Mascheroni constant?

• Draw rectangles representing the sum (i.e. having height $1/n$ on the interval $[n, n + 1]$), and the area representing the integral. In the limit, these are almost the same, and the error is $\gamma$. – user61527 May 22 '14 at 0:16
• It's pretty easy to show, using $$\log n = \int_1^n\frac{dx}{x} = \sum_{i=1}^{n-1}\int_i^{i+1}\frac{dx}{x}$$ that $$\left(\sum_{i=1}^n\frac{1}{i}\right)-\log n$$ converges as $n\to\infty$. You can also easily prove that the limit is between $1/2$ and $1$. This limit, or some variant, is often the definition of the Euler-Mascheroni constant. – Thomas Andrews May 22 '14 at 0:21
• @ThomasAndrews I am sure you have a typo there. – chubakueno May 22 '14 at 0:23
• Fixed. @chubakueno – Thomas Andrews May 22 '14 at 0:24
• I like showing $\gamma$ is between one increasing sequence and one decreasing sequence that are also forced to come together; see math.stackexchange.com/questions/306371/… – Will Jagy May 22 '14 at 3:58

$$\log n = \int_1^n \frac{dx}{x} = \sum_{i=1}^{n-1}\int_{i}^{i+1}\frac{dx}{x}$$

So:

$$\left(\sum_{i=1}^n \frac 1 i\right)-\log n = \left(\sum_{i=1}^{n-1}\int_i^{i+1}\left(\frac 1i-\frac1x\right)dx\right) + \frac{1}{n}$$

Now, for $x\in[i,i+1]$, $0\leq\frac{1}i-\frac1 x\leq \frac{1}{i(i+1)}.$

So these terms are positive and $\sum_{i=1}^\infty \frac{1}{i(i+1)} = 1$. So as $n\to\infty$, this means:

$$\left(\sum_{i=1}^n \frac 1 i\right)-\log n$$ converges to a value less than $1$.

It's actually pretty easy to show, since $f(x)=1/x$ is concave, that:

$$\int_i^{i+1}\left(\frac 1i-\frac1x\right)dx>\frac{1}{2i(i+1)}$$

This means that the limit is between $1/2$ and $1$.

This is often the definition of the Euler-Mascheroni constant.

• Don't you need to show the series is increasing? Or have you shown that and I'm missing something. – Alex Zorn May 22 '14 at 0:49
• If $0<a_i<b_i$ and $\sum b_i$ converges, then $\sum a_i$ converges. Basically, $\sum_{i=1}^n a_i$ is increasing as $n\to\infty$ and is bounded above, therefore converges. – Thomas Andrews May 22 '14 at 1:14

As noted in the comments, the "computational" answer is that, comparing areas, we find

$$\int_1^{n+1}\frac{dx}{x}<1+{1\over 2}+\cdots +{1\over n}<1+\int_1^n\frac{dx}{x}$$ Evaluating the integrals gives $$\ln(n+1)<1+{1\over 2}+\cdots +{1\over n}<1+\ln n$$ so we have the very rough approximation that, denoting $$H_n=1+{1\over 2}+\cdots +{1\over n}$$ the sum is around $H_n\approx \ln n$.

A more insightful perspective is to compare the nature of the functions $H_n$ (often called the "harmonic sum") and $\ln x$. The former is discrete, and the latter continuous. In "discrete calculus" you often see $$\Delta f(n)=f(n+1)-f(n)$$ which is the discrete analogue of the derivative. Note that the "derivative" of the harmonic sum is $$\Delta H_n=H_{n+1}-H_n=\left(1+\cdots +{1\over n+1}\right)-\left(1+\cdots +{1\over n}\right)={1\over n+1}$$

Similarly, the derivative of the logarithm is $$\textrm{D}\ln x={1\over x}$$

(The analogy can be drawn further, depending on how much discrete maths you know) There is a sense then, in which the two functions are "companions" of each other - they live in different universes, but within those universes they share many of the same properties. This is a recurring theme in much of mathematics (you may be interested in researching the word "isomorphic", although it does not completely apply here).

• If you average $\ln (n + 1)$ and $\ln n + 1$ you get $\ln n + 0.5$ as an estimate for $H_n$ – vonbrand May 22 '14 at 4:36


# Solving Modulus-Equations

So, I've come across this modulus equation in my textbook: $$|2x-1|=|4x+9|$$

I looked at the solution to this equation and understand that in order for both sides to be equal, the quantities inside the brackets must either be the same or the negatives of each other.

The solution then uses the following theorem: if $|p| = b, b>0 \Rightarrow p = -b$ or $p = b$. $$2x-1 = -(4x+9)$$ or $$2x-1 = 4x+9$$ and solves both linear equations to get $x = -\frac{4}{3}$ or $x = -5$

I then asked myself why the solution didn't bother to find $$-(2x-1) = 4x+9$$ or $$-(4x-9) = -(2x-1)$$ and instead only found the two above. I then proceeded to calculate the above linear equations and got the exact same answers as above $x = - \frac{4}{3}$ or $x = -5$

I'd like to know why I achieved the same answers with this.

• Think at why $2x-1 = -(4x+9) \iff -(2x-1) = 4x+9\,$. – dxiv Feb 21 '17 at 23:02

The equation $-(2x-1)=4x+9$ has the same solution as $2x-1=-(4x+9)$ as the only difference between these two equations is it has been multipled by $-1$.
Similarly $2x-1=4x+9$ has the same solution as $-(2x-1)=-(4x+9)$ for the same reason.
So to solve the original equation you need to solve only one out of the pair of equations $-(2x-1)=4x+9$ and $2x-1=-(4x+9)$ and then solve one out of the pair of equations $2x-1=4x+9$ and $-(2x-1)=-(4x+9)$. Which one you do from each pair doesn't matter as they have the same solution.
I drew a graph of $$y = |4x+9| - |2x-1|.$$ Not to scale, for $x$ I made four squares equal $1,$ because the interesting $x$ values are $1/2$ and $-9/4.$ The graph is three lines joining up, but with different slopes (four times what I depicted). Had I checked the answer first, I would have moved it over a bit so as to see more of the negative $x$ axis.


In fourth case, a binary addition is creating a sum of 1 + 1 = 10 i. Multiplication in abstract algebra, as between vectors or other mathematical objects, does not always obey these rules. Multiplication Example Multiplicand 1000ten Multiplier x 1001ten-----1000 0000 0000 1000-----Product 1001000ten In every step • multiplicand is shifted • next bit of multiplier is examined (also a shifting step) • if this bit is 1, shifted multiplicand is added to the product. The value of a particular N-bit binary number x in a U(a,b)representation is given by the expression x. A multiplication algorithm is an algorithm (or method) to multiply two numbers. The rule for addition of two binary coded decimal (BCD) numbers is given below. However, each bit of implies different algorithmic path during each iteration, that is, if , only a point doubling is necessary. t= 0, then ˆ Q(xt) ←Q(xt−1) P(xt) ←P(xt−1)(1−p), (2) if x. Following are the procedure for multiplying binary numbers. (e) G = (R×. Binary signed 2's complement: 2. Thus the product of a3 into b2 is a3b2 or aaabb. Only valid for independent events. Binary multiplication is one of the four binary arithmetic. Thus, the implementation uses CC to represent the number 12x10 + 12 = 132. Binary multiplication is actually much simpler to calculate than decimal multiplication. 0+0=0 0+1=1 1+0=1 1+1=10. In mathematics when you perform computational actions, you must have in mind that there is a sequence that need to be respected in. There are two types of multiplication for matrices: scalar multiplication and matrix multiplication. The sub-module numpy. Rewrite the smaller number such that its exponent matches with the exponent of the larger number. The most widely known binary operations are those learned in elementary school: addition, subtraction, multiplication. The broad perspective taken makes it an appropriate introduction to the field. Distributivity of Binary Operations. an old video where sal gives several examples of polynomial multiplication. 3rd: Flip the second fraction. Two's complement converter calculator is used to calculate the 2's complement of a binary or a decimal number. The pairs of 3 are sent to a carry-save adder, which takes a sum of 3 inputs and reduce it to a sum of 2 inputs (x + y + x => c + s). Ive watched some basic binary multiplication video's on youtube and i understand the logic and i can probably make it in redstone, but it will be huge and will be a mess. Excellent! Thank you for this explanation of binary multiplication. What is the modern slide rule? 4. Definition of multiplication rule, from the Stat Trek dictionary of statistical terms and concepts. I'd be happy to share it with anyone once I figure out how to place it on this site. Sparse matrices, which are common in scientific applications, are matrices in which most elements are zero. write sin x. Binary Multiplication. The binary system is a way of representing numbers in base 2, i. So, the result became 0. ), with steps shown. Binary Multiplication •Sizing •In binary addition -we are generally representing something that ultimately is to be executed in hardware •Our hardware cannot change the number of bits (wires) it can hold •We must establish a maximum number size •For multiplication the size of the result must be the sum of the. Some compilers ignore this rule and detect the invalidity semantically. Lesson 1: Introduction to Multiplication. we add from right to left and the carry over get's added to the digits in the next column. A fun little trick, really, about how to quickly extract the cube roots of large integers mentally. Are division and multiplication inverse operations? Any operation is called a binary operation when defined by a decimal or binary expansion like (3; 3. There is no overflow, so correct result is zero. Multiplication Example Multiplicand 1000ten Multiplier x 1001ten-----1000 0000 0000 1000-----Product 1001000ten In every step • multiplicand is shifted • next bit of multiplier is examined (also a shifting step) • if this bit is 1, shifted multiplicand is added to the product. So to do 11+11 note that 1+1=10 so write zero and carry the one. Binary fractions work like decimal fractions: 0. the operations are done with algorithms similar to those used on sign magnitude integers. Each digit in a binary number is called a bit. Any boolean expression always follow closure with respect to binary addition (+ operator) and binary multiplication (. No strict rules are defined but just easy to read but precise. , 0s and 1s) to physical information, because only two kinds of physical objects or states are needed. This table includes all of the possible multiplications between the numbers 1 to 12. So, 1111 (in binary) = 8 + 4 + 2 + 1 = 15 (in decimal) Accounting for 0, this gives us 16 possible values for four binary bits. The operation of multiplication is extended to other real numbers according to the rules governing the multiplicative properties of. I want to make the binary class fun for my students, and I would like to apply activities to make it easier. Infix, Prefix and Postfix Expressions. Binary, or base 2, is one of many possible Number Systems. use the General Multiplication Rule. aaaaa x bbb. Binary multiplication is implemented using the same basic longhand algorithm that you learned in grade school. In fact it's even easier since each digit can only be a 1 or a 0. 5^2)^5)^5 (9 multiplication steps) Doing that requires factorisation, which may make it not worthwhile. Matrix elements should be numerical. A (B+C) = (AB) + (AC) This is what is known as the AND distributes over OR. 1 + 1 =10. This video con. Intro to Algebra: Exponent Rules (Multiplication) Students' understanding of exponent rules for multiplication is put to the test in this assessment which provides an introduction to algebra. Tree 149 Depth-first Search 136 Hash Table 134 Greedy 103 Binary Search 93 Breadth-first Search 75 Tree 14 Ordered Map 12 Geometry 9 Queue 9 Minimax 8 Brainteaser 7 Binary Indexed Tree 6. The broad perspective taken makes it an appropriate introduction to the field. Practice math problems like Extend Number Patterns (with Rule Mentioned) with interactive online worksheets for 4th Graders. In this case the horizontal. The number of bits required to represent the product is at least (n+m) for unsigned multiplication and (n+m+1) for signed (twos complement) multiplication. In the case of decimal multiplication, we need to remember 3 x 9 = 27, 7 x 8 = 56, and so on. Problem Solution. The binary number system uses only two digits 0 and 1 due to which their addition is simple. Since they were not interested in finding proofs justifying the used procedures, they were. With three or more, it is also straightforward, but you use the Even-Odd Rule. This application encodes and decodes ASCII and ANSI text. We describe three algorithms for multiple-point multiplication on elliptic curves over prime and binary ﬁelds, based on the representations. What if the digits in each number are not even, or the same? What is the rule in dividing it into two parts? Example: x = 12345 y = 2478 or. Let’s add two binary numbers to understand the binary addition. This video con. How to use this calculator: In the calculator, there are two input fields intended for entry of binary. Here the steps took place are. The main rules of the binary division include: subtraction, multiplication and division operations register with BYJU'S -The Learning App and also watch exciting videos to learn with ease. valuation functions Abstract Syntax of Binary Numerals A simple BNF grammar will suffice: B ∈ Binary-numeral. Commercial applications like computers, mobiles, high speed calculators and some general purpose processors require […]. The maximum value we can have with three binary digits is 111 = decimal 7 calculated as follows-1*1 + 1*2 + 1*4. Starting from the right, the first "A" represents the "ones" place, or 16 0. Basic Arithmetic Operation-(lecture#1 basic rules of addition,multiplication,subtraction,division). In MQL4, there are some natural limitations to the rules of using operations in expressions. The positions in a binary number (called bits rather than digits) represent powers of two rather than powers of ten: 1, 2, 4, 8, 16, 32, and so on. That is, a+ (b+ c) = (a+ b) + c for all a;b;c 2F. A complex number is an expression of the form a + bi , where a and b are real numbers, and i has the property that i 2 = –1. a × e = e × a = a This is possible if e = 1 Since a × 1 = 1 × a = a ∀ a ∈ R 1 is the identity element for multiplication on R Subtraction e is the identity of * if a * e = e * a = a i. Long multiplication can be carried out with binary numbers and is explored in this section. Binary Addition. Binary exponentiation (also known as exponentiation by squaring) is a trick which allows to calculate Most obviously this applies to modular multiplication, to multiplication of matrices and to other. Multiplication: multiplying, multiply. Though binary division not too difficult, it can initially be a bit harder to understand than the other binary operations as they shared similarities. Add 1 to the result using unsigned binary notation c. (25 ends up with 0010 0101) Now another thing to remember, there is a rule established in the multiplication of the values as written in the book, Cryptography and Network Security, that multiplication of a value by x (ie. The first two round to a nearest value (ties to even and ties away from zero); the others are called directed roundings: towards zero, towards positive infinity and towards negative. Multiplication of pure imaginary numbers by non-finite numbers might not match MATLAB. For any elements a and b of {-1, 1}, (a+b) is not an element of {-1, 1}. Then, the program multiplies these two matrices (if possible) and displays it on the screen. Definition Let R be a set on which two binary operations are defined, called addition and multiplication, and denoted by + and ·. 1 * 1 = 1 1 * 0 = 0 * 1 = 0. The rules for binary multiplication are: In truth table form, the multiplication of two bits, a x b is: Observe that a x b is identical to the logical and operation. Step 1: When we multiplying 0 and 0, we get 0. It consists of only 0, 1 digit and rules for addition, subtraction, and multiplication are the same as decimal numbers. INTEGER BINARY 0 000 1 001 2 010 3 011 4 100 5 101 6 110 7 111. Applying the rules from the previous section, the multiplication follows these steps: The first two numbers (5. To do it we set the value range from 0 to 1, and choose the symmetric matrix type. The fastest way of multiplying two big integers. Algorithm: To multiply two decimal numbers x and y, write them next to each other, as in the example below. Converting from hex to binary is a lot like converting binary to hex. However, each bit of implies different algorithmic path during each iteration, that is, if , only a point doubling is necessary. , digits) is performed in a manner similar to decimal multiplication. The multiplication of an n-bit binary number with an m-bit binary number results in a product that is up to m + n bits in length for both signed and unsigned words. mlt": nn = na if n = 2 then f$= "b" + ". That is all the way from 1 x 1 to 12 x 12. How to use this calculator: In the calculator, there are two input fields intended for entry of binary. This online calculator for addition and subtraction multiplication and division of binary numbers online. For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. This is the first step in this step the least significant bit or the right most bit of B is multiplied with all the digits of A from the right side and the result is written. Each cell should contain a zero or a one. Multiplication of binary numbers can be achieved either combinationally or sequentially. Therefore, multiplication must be explicitly denoted by using the * operator, as in a * b. Two dots in the right-most box is worth one dot in the next box to the left. This gives signal of overflow-7+7=0 00111 binary 7 01001 two's complement of 7 11110 carries 10000 result of addition 16 - but fifth bit can be ignored, real result is 0. There are various number systems in the world. In fourth case, a binary addition is creating a sum of 1 + 1 = 10 i. They follow their own set of simple rules and trade on their own special exchange that has been set up just for them. abstract syntax 2. Lattice multiplication is also known as Italian multiplication, Gelosia multiplication, sieve multiplication, shabakh, Venetian squares, or the Hindu lattice. The magnitude can be interpreted as a +01 Id, the remainder from a carry out of I 28d. Learn multiplication table with easy to memorize, helpful times tables are simple to read. Watch this illustration and play with the code. These rules are applied for each operation, such that nested calculations imply the precision of each component. en: binary numbers adding substraction multiplication division. Message fields can be one of the following For example, when an old binary parses data sent by a new binary with new fields, those new fields become unknown fields in the old. (Where there is only one number above, you just carry down the 1. • More general rule: – The principle of inclusion and exclusion. Let's illustrate the multiplication rules for 3-bit binary numbers A = 111 and B = 101, beginning with the lowest and highest digits of the multiplier (Figure 2. For all a b b a ba. reflexive: if for every x X, xRx holds, i. Commute= travel (move) If you move swap the numbers ina multiplication and addition the anwser will still be the same! Rule: a+b = b+a Sample: 1+2= 3, 2+1=3. Same rule does not hold for natural numbers because if we take two numbers such as x and y and perform binary. 1 The Multiplication Rule. 1 * 1 = 1 2. According to the binary multiplication rules, the numbers in the bracket give the decimal equivalents of the binary numbers. Observations on Multiply Version 3. MUL Multiply; DEC VAX; signed multiplication of scalar quantities (8, 16, or 32 bit integer) in general purpose registers or memory, available in two operand (first operand multiplied by second operand with result replacing second operand) and three operand (first operand multiplied by second operand with result placed in third operand) (MULB2 multiply byte 2 operand, MULB3 multiply byte 3 operand, MULW2 multiply word 2 operand, MULW3 multiply word 3 operand, MULL2 multiply long 2 operand. The examples of valid hexadecimal numbers are 4E1or CD3A01. (25 ends up with 0010 0101) Now another thing to remember, there is a rule established in the multiplication of the values as written in the book, Cryptography and Network Security, that multiplication of a value by x (ie. However, it is not guaranteed to be compiled using efficient routines, and thus we recommend the use of scipy. Note that in each subsequent row, placeholder 0's need … As an example of binary multiplication we have 101 times 11, 101 x 1 1. Here's the demo for subtraction. 0 is written in the given. Only valid for independent events. Binary Conversion Practice! ! ! !Binary Places: 32, 16, 8, 4, 2, 1 Convert these binary numbers to decimal: 1 10 11 100 101 1000 1011 1100 10101 11111 Convert these decimal numbers to binary:. It can be calculated easily if we know the following rules. I'd be happy to share it with anyone once I figure out how to place it on this site. Sponsored Links. Two's complement converter calculator is used to calculate the 2's complement of a binary or a decimal number. So, the result became 0. Quantities with unlike units may sometimes be multiplied, resulting in such units as foot-pounds, gram-centimeters, and kilowatt-hours. Binary literals can make relationships among data more apparent than they would be in hexadecimal or octal. To multiply in binary, you multiply the first number by each of the digits of the second number in turn starting from the right-hand side (in the same way that you would do multiplication in decimal). This is illustrated below. This means 3 groups of 2 equals 6 in all. C program to convert decimal number to roman. Existing Algorithms: • The naive binary multiplication algorithm has a time complexity O(n^2) where n is the number of bits of the numbers being multiplied. Arrays, multiplication and division Jennie Pennant, with the help of Jenni Way and Mike Askew, explores how the array can be used as a thinking tool to help children develop an in-depth understanding of multiplication and division. Other :: Binary Classification. ) Keep going, always adding pairs of numbers from the previous row. Additionally, the output is restricted to a 10-bit word with binary-point-only scaling of 2-4. Binary Card Game Explained ; Binary Card Game, the computer plays against. Instead of needing to know your multiplication tables you only really need to know three things. Power rule with positive exponents; Power rule with negative exponents: Problem type 1; Simplifying a sum or difference of three univariate polynomials; Product rule with positive exponents; Multiplying binomials with leading coefficients of 1; Solving a two-step equation with integers; Solving an equation to find the value of an expression. binary calculator as a 1 and 0. Binary Multiplication Calculator is an online tool for digital computation to perform the multiplication between the two binary numbers. }\) We use symbols to represent binary operations instead of function names, just as we do with addition and multiplication of integers. A multiplication algorithm is an algorithm (or method) to multiply two numbers. Converting Decimal Numbers to Binary Numbers. It is enough to show only one rule-break to prove that {-1, 1} is not a group with respect to addition. 24 Addition and Multiplication in Q The following rules define binary operations on Q. So in the example above: Relative uncertainty = (0. You will be able to use it both as a binary to decimal converter and as a decimal to binary calculator. Binary multiplication is implemented using the same basic longhand algorithm that you learned in grade school. 2 Unary Operations Up: 1 Binary Operations Previous: 1 Addition and subtraction Index 2 Matrix multiplication Matrices are conformable for multiplication if and only if the number of columns in the first matrix equals the number of rows in the second matrix. A binary operator; the result is $$\binary{1}$$ if at least one of the two operands are $$\binary{1}\text{,}$$ otherwise the result is \(\binary{0}\text{. The general rule is that if the OV flag is set to I, then complement the sign bit. • Binary multiplication of unsigned integers reduces to "shift and add". The process in the binary system is easier than it is in the Binary multiplication calculator is a simple method to multiply binary values without using manual mehtods. When a string contains single or double quote characters, however, use Function annotations should use the normal rules for colons and always have spaces around the. basis is used in all matrix and matrix-vector multiplication algorithms, which are considered in this and the following sections. Rule 3: Lastly, perform all additions and subtractions, working from left to right. Binary Addition, Subtraction , Multiplication and division. + 01 0 01 1 10 · 01 0 00 1 01 Table C. Binary integer multiplier 210 first receives binary coefficients (C1 and C2) of two input operands and performs a binary multiplication, C' = C1 · C2. Definition: Binary operation. If we know the rules of binary multiplication & division, the solution of t. Note that binary operators work on vectors and matrices as well as scalars. 1415; …) converges to a unique. That means that every piece of binary code in a computer must be converted into a physical object or state. Binary Multiplication: Binary multiplication is exactly as it is in decimal, i. When a string contains single or double quote characters, however, use Function annotations should use the normal rules for colons and always have spaces around the. The rules for binary multiplication are: In truth table form, the multiplication of two bits, a x b is: Observe that a x b is identical to the logical and operation. Multiplication by an Integer Constant 7 The average time complexity seems to be exponential. The rule for addition of two binary coded decimal (BCD) numbers is given below. MUL Multiply; DEC VAX; signed multiplication of scalar quantities (8, 16, or 32 bit integer) in general purpose registers or memory, available in two operand (first operand multiplied by second operand with result replacing second operand) and three operand (first operand multiplied by second operand with result placed in third operand) (MULB2 multiply byte 2 operand, MULB3 multiply byte 3 operand, MULW2 multiply word 2 operand, MULW3 multiply word 3 operand, MULL2 multiply long 2 operand. Addition Worksheet A :. Binary Options Trading the simplest form of Best Trading Platform in the financial sector, that has become the traders best choice recently in Option Trading. Starting from the right, the first "A" represents the "ones" place, or 16 0. C program for fractional decimal to binary fraction conversion. (c) G= (N,+), the natural numbers in Z with addition as the binary operation. The evaluation using the and and or operators follow these rules: and and or evaluates expression Bitwise operators are used to compare integers in their binary formats. This means you are either multiplying each digit by 0 or 1, which will give you either a 0 or 1 as the answer. Test your method out on these numbers, and find a way to check your work. Binary Multiplication Binary multiplication is performed in same way as decimal numbers. The group operation is multiplication of complex numbers. E A B C E E A B C A A B C E B B C E A C C E A B T Combination order is "top" then "side"; e. 4 cm) × 100% = 5. See full list on byjus. Binary multiplication is arguably simpler than its decimal counterpart. An example is the unary and binary minus, (). Let’s add two binary numbers to understand the binary addition. If you want to build up one for yourself, you can look at this scheme of 2 bits times 2 bits multiplier -. What is the result of multiplying the binary number 10010 by 101? 1011010. Then R is called a commutative ring with respect to these operations if the following properties hold: (i) Closure: If a,b R, then the sum a+b and the product a·b are uniquely defined and belong to R. What if the digits in each number are not even, or the same? What is the rule in dividing it into two parts? Example: x = 12345 y = 2478 or. Translate the number 22810 в binary like this: the Integer part of the number is divided by the base of the new. In mathematical operations involving significant figures, the answer is. Binary Numbers Toggle the 1s and 0s by clicking on them to reveal dots and make binary numbers. semantic algebras 3. Multiplying a binary term (multiplicand m) by 2 simply shifts mone power level up, for example, 2 · K = P or 2 · P = T. Generally you won’t need to use this operator since it’s redundant. The first two numbers (5. The Multiplying Binary Numbers (Base 2) (A) Math Worksheet from the Multiplication Worksheets Page at Math-Drills. For example, an int data type can hold 10 digits, is stored in 4 bytes, and doesn't accept decimal points. Let A = {1, 3, 5, 9, 11, 13} and let$\odot$define the binary operation of multiplication modulo 14. Note: In the Chain Rule, we work from the outside to the inside. There are four rules of binary addition. (Where there is only one number above, you just carry down the 1. C program to convert decimal number to roman. Booth's Multiplication Algorithm is used to multiplication of two signed binary numbers. So basically all you need to do is multiply the powers. unambiguouslywrite a⋆b⋆c to denote either of the iterated products. 1 The Number of Elements in a List. FrankenPC makes a fair start showing binary addition. Let's illustrate the multiplication rules for 3-bit binary numbers A = 111 and B = 101, beginning with the lowest and highest digits of the multiplier (Figure 2. If we only want to invest in a single N-bit adder,. Complex multiplication is a more difficult operation to understand In the above formula for multiplication, if v is zero, then you get a formula for multiplying a complex. Binary division and multiplication are both pretty easy operations. for overloading binary operators, addition, subtraction, multiplication, division and comparison". " The Multiplicand is the number taken or multiplied. Fast multiplication. For each number of nodes, n, there is a certain number of possible binary tree configurations. Note that in each subsequent row, placeholder 0's need … As an example of binary multiplication we have 101 times 11, 101 x 1 1. We see what are the rules of binary multiplication and division and how to solve. Two last carries are 11. How to use this calculator: In the calculator, there are two input fields intended for entry of binary numbers. The unary minus operator returns the operand multiplied by -1. The Standard Multiplication Algorithm. There are four rules of binary addition. abstract syntax 2. Example 7. For example, an int data type can hold 10 digits, is stored in 4 bytes, and doesn't accept decimal points. Enter the two numbers that you want to implement the. Just as we get a number when two numbers are either added or subtracted or multiplied or are divided. Binary options are a different kind of option than any you're likely familiar with. When performing a binary. 1 Lecture 8: Binary Multiplication & Division Today s topics: Addition/Subtraction Multiplication Division Reminder: get started early on if signs disagree These rules fulfil the equation above 19. For example, in mathematics and most computer languages, multiplication is granted a higher precedence than addition, and it has been this way since the. 1415; …) converges to a unique. If neither value is a double but one is a float, then Java treats both values as floats. 1 The Number of Elements in a List. Binary numbers can represent two states: 0, 1, and 2-bit binary numbers can represent (2=) 4 states: 00, 01, 10, and 11; by analogy, 7-bit binary numbers can represent (2=) 128 species. The arithmetic of binary numbers means the operation of addition, subtraction, multiplication and division. Beginners introduction to binary, hexadecimal and octal numbers. It's still a base-10 representation, but the algorithm doesn't need the digits to be between 0 and 9. Home Work. Stanford University. The relative uncertainty gives the uncertainty as a percentage of the original value. C program for addition of binary numbers. Conversion rules for converting decimal to other number system First step is to identify the base of target number system i. The Manchester Baby, operational in 1948, had no multiplication hardware. Post navigation. b) Whenever b(i,j) =1 , a(i,t) should also be 1, but for both alias sets i, t ; alias (i,j); Binary Variables b(i,j. For a node's world transformation: W = P. A simplistic way to perform multiplication is by repeated addition. There are four basic operations for binary addition, as mentioned above. It uses "engine" of Mathematical calculator. There are no ads, popups or other garbage. The magnitude can be interpreted as a +01 Id, the remainder from a carry out of I 28d. See General Rules for Operator Overloading for more information. 2 * 10 = 20 and 3 * 100 = 300. Floating Point Addition. Let us take two binary numbers A = 1001 and B = 101 we want to find out A × B. The 1←2 Rule Whenever there are two dots in single box, they “explode,” disappear, and become one dot in the box to the left. Dividing Binary Numbers. Robot Multiplication for Kids the 11s Master your Multiplication Facts by byrih. The general rule is: (x m) n = x mn. + Addition. The precedence order for arithmetic operators places multiplication and division above addition and. There are many applications of matrices in computer programming; to represent a graph data structure, in solving a system of linear equations and more. L = local transformation matrix. 2nd: Change the division sign to a multiplication sign. Binary Multiply - Repeated Shift and Add. The binary multiplication is much easier as it contains only 0s and 1s. The number can be converted to decimal by multiplying out as follows: 1*1 + 0*2 + 1*4 = 5. For example, multiplication of two 4-bit numbers requires a ROM having eight address lines, four of them, X 4 X 3 X 2 X 1 being allocated to the multiplier, and the remaining four, Y 4 Y 3 Y 2 Y 1 to the multiplicand. Then, b is called inverse of a. abstract syntax 2. The 3 basic. More Topics Related. More Examples:. Binary addition/subtraction is similar to regular (daily life) addition/subtraction, but here addition/subtraction performs only two digits those are 0 and 1, these are binary digits hence such kind of addition/subtraction is called binary addition/subtraction. Find the missing multiple or product for any multiplication fact You can use the printable multiplication table below to check your answers or as a study guide. With three or more, it is also straightforward, but you use the Even-Odd Rule. 741 Op-Amp Circuit Binary Addition Binary Multiplication Section 4. Last but not the least, a major reason computers use the binary system is that the two-state system is the number system best suited to the optical and magnetic storage components of the computer. How to do Addition, Subtraction, Multiplication and Division in Android Studio. However, it is not guaranteed to be compiled using efficient routines, and thus we recommend the use of scipy. Point multiplication is usually computed by a sequence of point doublings (triplings, halvings) and additions, whose numbers depend on the length and the number of non-zero digits of k. Binary literals can make relationships among data more apparent than they would be in hexadecimal or octal. Example − Addition Binary Subtraction. To start practising, just click on any link. To convert 0xDEADBEEF (a commonly used code to indicate a system crash), begin by sorting the digits into "bins":. A lower precedence number means tighter binding Other numeric operations are similar: subtraction (-), and multiplication (*). binary calculator as a 1 and 0. In Binary Arithmetic, 1 + 1 = 10, as per the rules of binary addition. Leibniz pointed out in 1703 that to do simple arithmetic in binary, such as addition and multiplication, you don’t need to memorize such rules as 5 + 4 = 9, or 6 × 7 = 42. But we do not know much more. Assume the multiplicand (A) has N bits and the multiplier (B) has M bits. How to convert letters (ASCII characters) to binary and vice versa. eve Engels, 2006 Slide 7 of 10 –Example: two-digit multiplication a 1 a 0 x b 1 b 0 a 1 b 0 a 0 b 0 a 1 b 1 a 0 b 1 + p 2 p 1 p 0 p 0 = a 0 b 0 p 1 = a 1 b 0 + a 0 b 1 p 2 = a 1 b 1 + a 0 b 1 a 1 b 0 (addition here, not OR). Binary Addition, Subtraction , Multiplication and division. Multiplication And Division of Octal Numbers Calculator, Octal Multiplication And Division: Required Data Entry Enter A Octal Value. Addition, subtraction, multiplication, and division are all binary operators with which we are familiar from grade school. (b) G= (N,·), the natural numbers in Z with multiplication as the binary oper-ation. flipped over). For example, the real numbers form a ﬁeld, with ‘+’ and ‘·’denoting ordinary addition and multiplication. " Every digit must be either 1 or 0. Binary Arithmetic 1: Adding binary numbers. Multiplication done algebraically. All it includes are addition of binary. Binary arithmetic is essential part of all the digital computers and many other digital system. What is the associative property of Multiplication? Associative property of multiplication is the rule that says that the way that factors are grouped in a multiplication problem will not affect the answer. In modern terms it employs the decomposition of one number into its binary components, addition to produce "doubles" or multiples of the other number, and computing a total of doubles identified by odd divisors or the powers of two. Overflow rule:- If two number are added, and they are both positive or negative than overflow occurs if and only if the result has opposite sign. Let us understand the binary multiplication on natural numbers and real numbers. 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0. Such models can be linearized quite easily and are in most cases better solved as a linear. Lesson 1: Introduction to Multiplication. The rules of differentiation (product rule, quotient rule, chain rule, …) have been implemented in JavaScript code. Multiplication Tables L All possible binary combinations of symmetry operations can be summarized in a multiplication table. number system to which we have to convert the decimal number system. In mathematics, a binary operation or dyadic operation is a calculation that combines two elements (called operands) to produce another element. 17 * 5 = 85 / (Division) By using this operator we can perform division of numbers. Binary Triple Octal Decimal Hexadecimal Binary-decimal. If ⋆ is associative we can. Assigning a precedence to grammar rules. If you skip parentheses or a multiplication sign, type at least a whitespace, i. No strict rules are defined but just easy to read but precise. Use as an example 18 or 10010: 18 = 16 + 2 = 2 4 + 2 1. (2) In each part below, a rule is given that determines a binary operation on Z. Sponsored Links. Enter the two numbers that you want to implement the. A carry-save adder's propagation time is constant, while traditional a 2 input adder's propagation time is proportional the the width of the operands. C program to convert decimal number to roman. One technique for performing multiplication in any number base is by repeated addition; for example, 6×4=6+6+6+6=24 in decimal (similarly, 4×6=4+4+4+4+4+4=24). When a multiplication problem is given abstractly, there is no such distinction, so we prefer to use the symmetrical term "factor. This is the currently selected item. mlt": nn = nb open f$ for binary as #1 for n2 = 1 to nn randomize timer x$= ltrim$ (str$(int (rnd * 10))) seek #1, n2: put #1, n2, x$ next n2 seek #1, n2 close (1) next n ' open "a" + ". This is, of course, 10 times the starting value - that is how you multiply things by 10. Decitrig: A rule with the trigonometric scales with decimal degrees (S, ST, T) [0. So, the result became 0. multiplication of any element by 0 yields 0; 3. The number 1010110 is represented by 7 bits. Therefore, multiplication must be explicitly denoted by using the * operator, as in a * b. The value of a particular N-bit binary number x in a U(a,b)representation is given by the expression x. , E A B C E EE = E EA = A EB = B EC = C A AE = A AA = B AB = C AC = E B BE = B BA = C BB = E BC = A C CE = C CA = E CB = A CC = B. Commercial applications like computers, mobiles, high speed calculators and some general purpose processors require […]. Binary Multiplication Calculator is an online tool for digital computation to perform the multiplication between the two binary numbers. In Karatsuba algorithm for multiplying two numbers, we divide each number into two. A binary multiplier is a combinational logic circuit used in digital systems to perform the multiplication of two binary numbers. Unit-2: Binary Multiplication-Booth's Algorithm. As an example of binary multiplication we have 101 times 11, 101 x 1 1. eve Engels, 2006 Slide 7 of 10 –Example: two-digit multiplication a 1 a 0 x b 1 b 0 a 1 b 0 a 0 b 0 a 1 b 1 a 0 b 1 + p 2 p 1 p 0 p 0 = a 0 b 0 p 1 = a 1 b 0 + a 0 b 1 p 2 = a 1 b 1 + a 0 b 1 a 1 b 0 (addition here, not OR). Read about Binary Addition (Binary Arithmetic) in our free Electronics Textbook. The unary minus operator returns the operand multiplied by -1. Multiplication of a binary number n by the binary number 100 (=4) is done by adding two bits 0 to the right of the binary representation of n. The exact same rule exists in binary. The binary number system uses only two digits 0 and 1 due to which their addition is simple. A, 1 is entered in its position that yields 1000 since 2 3 = 8. The interaction of the parts makes the dinner range from good to great. There are four basic operations for binary addition, as mentioned above. The form calculates the bitwise exclusive or using the function gmp_xor. One technique for performing multiplication in any number base is by repeated addition; for example, 6×4=6+6+6+6=24 in decimal (similarly, 4×6=4+4+4+4+4+4=24). to multiply and divide polynomials, use conv and deconv on. The result of binary arithmetic for A with and B with gives the worst case p:. The multiplication method is sometimes called doubling and halving or called Russian peasant multiplication. Arithmetic operations in the binary system are extremely simple. (Like multiplying by 10 in our normal notation. Binary Option trading platforms allows e-Traders to make investments by predicting the future direction of an asset and make up to 85% profits in a span of few minutes. 0026), with the final result having 8 decimal places (0. Instead, you need. To convert fraction to binary, start with the fraction in question and multiply it by 2 keeping notice of the resulting integer and fractional part. Multiplying and dividing positive and negative numbers is a simple operation with two numbers. " The Multiplicand is the number taken or multiplied. As a result of multiplication you will get a new matrix that has the same quantity of rows as the 1st one has and the same quantity of columns as the 2nd one. 11 ⋅ 2 − 3 1 01. FREE holiday, seasonal, and themed multiplication worksheets to help teach the times tables. Starting from the right, the first "A" represents the "ones" place, or 16 0. Specifying Field Rules. To perform addition, subtraction, multiplication and division in python, you have to ask from user to enter any two number, and then ask again to enter the operator to perform the desired operation. Binary Card Game Explained ; Binary Card Game, the computer plays against. Binary code, as it turns out, is easy to convert from electronic information (e. The numbers we are multiplying together are called factors, and the result of multiplication is called the. multiply numbers right to left and multiply each digit of one number to every digit of the other number, them sum them up. Draws binary random numbers (0 or 1) from a Bernoulli distribution. Each row and each column is unique and contains as many zeros as ones. (b) Using binary representa-tion 10011 1101 10011 10011 1011111 10011 11110111 (c) Adding immediately Figure 7. The magnitude can be interpreted as a +01 Id, the remainder from a carry out of I 28d. them sum them up. (LL 0, LL 1,LL 01, etc. For a node's world transformation: W = P. Most processors perform n-bit by n-bit multiplication and produce a 2n-bit result (double bits) assuming there is no overflow condition. Here Binary number is a number that can be represented using only two numeric symbols - 0 and 1. " Every digit must be either 1 or 0. Binary, or base 2, is one of many possible Number Systems. Multiplication of pure imaginary numbers by non-finite numbers might not match MATLAB. Suppose for instance that we want to multiply 13 11, or in binary notation, x= 1101 and y= 1011. In mathematics when you perform computational actions, you must have in mind that there is a sequence that need to be respected in. Distributivity of Binary Operations. Learn Java program multiplication starting from its overview, How to write, How to set environment , How to run, Example like Addition, Subtraction , Division, Multiplication etc. C++ Program to Multiply Two Matrix Using Multi-dimensional Arrays This program takes two matrices of order r1*c1 and r2*c2 respectively. Note that in each subsequent row, placeholder 0's need … As an example of binary multiplication we have 101 times 11, 101 x 1 1. The representations of the multiplicand and product are not specified; typically, these are both also in two's complement representation, like the multiplier, but any number system that supports addition and subtraction will work as well. Matrix multiplication falls into two general categories: Scalar: in which a single number is multiplied with every entry of a matrix. A general rule when multiplying a Qm format number by a Qn format number, is that the product will be a Q(m+n) number. In binary, 2 * 3 = 6 is 10 * 11 = 110, and 4 * 3 = 12 is 100 * 11 = 1100. For example: x= 1234 y= 2456 Then a = 12, b = 34, c = 24 , d = 56. That is binary numbers can be represented in general as having p binary digits and q fractional digits. Multiplying and dividing positive and negative numbers is a simple operation with two numbers. * (Multiplication) By using this operator we can perform multiplication of numbers. Other arithmetic operators work only with numbers and always convert their operands to numbers. Two dots in the right-most box is worth one dot in the next box to the left. resample(rule[, axis, closed, label, …]) Resample time-series data. Convert number. The balanced ternary multiplication table is depicted in Fig. If the array isn't sorted, you must sort it using a sorting technique such as merge sort. By using this website, you agree to our Cookie Policy. Note that since binary operates in base 2, the multiplication rules we need to remember are those that involve 0 and 1 only. The question is the extent to which we can unify addition and multiplication, realizing them as terms in a single underlying binary operation. Multiplication by an Integer Constant 7 The average time complexity seems to be exponential. A 4-bit, 2's complement example:. Binary multiplication can be achieved by using a ROM as a look-up’ table. Browse other questions tagged rsa algorithm-design montgomery-multiplication or ask your own question. Note that in each subsequent row, placeholder 0's need to be added, and the value shifted to the left, just like in decimal multiplication. (25 ends up with 0010 0101) Now another thing to remember, there is a rule established in the multiplication of the values as written in the book, Cryptography and Network Security, that multiplication of a value by x (ie. Our source code has been written in characters we can read, but it needs to be. To find the probability of the two dependent events, we use a modified version of Multiplication Rule 1, which was presented in the last lesson. In the case of decimal multiplication, we need to remember 3 x 9 = 27, 7 x 8 = 56, and so on. Linear algebra. com binary converter online for free. It gives a range of activations, so it is not binary activation. This online calculator for addition and subtraction multiplication and division of binary numbers online. So in the example above: Relative uncertainty = (0. In this tutorial, we will create a custom pipe so that the currency format can run well. According to all the rules above, x will be converted to int first. Leibniz pointed out in 1703 that to do simple arithmetic in binary, such as addition and multiplication, you don’t need to memorize such rules as 5 + 4 = 9, or 6 × 7 = 42. If an operation consists of k steps and. Also, we'll perform addition and subtraction on them. The first two numbers (5. To help me remember I wrote some fully commented code for an 8 x 8 bit unsigned multiplier. Stats: Probability Rules. The code works fine when I give an integer input, but doesn't give correct result for floating point input. 5^2)^25 (25 multiplication steps) = ((3. For K-12 kids, teachers and parents. 0015 to 1,000,000), two binary scales for adding and subtracting fractions, a scale of drill sizes, a scale of thread sizes, and millimeters. 70 × 10-1 with 9. The arithmetic unit performs parallel one-step addition (subtraction), multiplication and division. Converting Decimal Numbers to Binary Numbers. Binary Addition, Subtraction , Multiplication and division. In binary multiplication, we only need to remember the following. Similarly, whenever we would like to sum two binary numbers, only we will have a carry if the product is bigger than 1 because, in binary numbers, 1 is the highest number. To find the probability of the two dependent events, we use a modified version of Multiplication Rule 1, which was presented in the last lesson. What is the rules of binary in addition? As with decimal numbers, you start by adding the bits (digits) one column, or place. Other arithmetic operators work only with numbers and always convert their operands to numbers. It is enough to show only one rule-break to prove that {-1, 1} is not a group with respect to addition. Test yourself on multiplication facts. The difference being the "point" bookkeeping required which is the same as addition. Take some parts away or add new ones in, and you get a different outcome, but not a binary, win/lose one. The "Commutative Laws" say we can swap numbers over and still get the same answer. ) Vector : A rule with hyperbolic functions. There are four rules for binary multiplication: Multiplication is always 0, whenever at least one input is 0. Here is a new (harder) goal: come up with a general method to find the binary number related to any number of dots without actually going through the “exploding dot” process. Now multiplication is going to occur like this. Just binary utilities that work right in your browser. binary arithmetic rules. Multiplication rules. Mutually Exclusive Events. Leave a Comment on 540. ⌚ September 25, 2014 Binary multiplication is performed the same as decimal multiplication. numbers are multiplied and divided through a process called shifting. Similarly, whenever we would like to sum two binary numbers, only we will have a carry if the product is bigger than 1 because, in binary numbers, 1 is the highest number. The matrix product is designed for representing the composition of linear maps that are represented by matrices. Though binary division not too difficult, it can initially be a bit harder to understand than the other binary operations as they shared similarities. Then we multiply the entire top number by each individual digit of the bottom number. General binary operations that follow these patterns are studied in abstract algebra. Multiplication Example Multiplicand 1000ten Multiplier x 1001ten-----1000 0000 0000 1000-----Product 1001000ten In every step • multiplicand is shifted • next bit of multiplier is examined (also a shifting step) • if this bit is 1, shifted multiplicand is added to the product. However, it is not guaranteed to be compiled using efficient routines, and thus we recommend the use of scipy. The method used for ancient Egyptian multiplication is also closely related to binary numbers. This is a simple online tool to convert English into binary. Binary Multiplication is generally simpler than decimal multiplication. The binary number system uses only two digits 0 and 1 due to which their addition is simple. Enter expression with binary numbers and get the result. This online calculator for addition and subtraction multiplication and division of binary numbers online. In this tutorial, we will create a custom pipe so that the currency format can run well. The Binary Calculator is used to perform addition, subtraction, multiplication and division on two In mathematics and computer science, binary is a positional numeral system with a base of 2. Let’s covert the same binary number to an octal number: $100100010101111_{2}=100 100 010 101 111$ $100=4$ $010=2$ \$101=5. It shows you how the product is generated in real-time, step-by-step, and allows you to highlight the individual multiplication steps used to get the answer. To figure the decimal value of a binary number, you multiply each bit by its corresponding power of two and then add the results. And all utilities work exactly the same way — load. Generally you won’t need to use this operator since it’s redundant. This follows Chapter 4 of Schmidt’s book. There are four rules of binary addition. The general rule is: (x m) n = x mn. To read binary numbers, and convert them to their decimal equivalent, you have two options: you can either use the Binary to Decimal Converter at ConvertBinary. binary: A binary operator operates on two operands. The multiplication of a 16 bit binary "a" with a 24 bit binary "b and the complementary defence of both the general interest and the rules of the competitive. In the modern world, multiplication, division, addition, and subtraction are estimated by the binary calculator within a second, same rule as applied in the decimal system. A general rule when multiplying a Qm format number by a Qn format number, is that the product will be a Q(m+n) number. 2 steps per bit because Multiplier & Product combined How can you make it faster? What about signed multiplication? • trivial solution: make both positive. The magnitude can be interpreted as a +01 Id, the remainder from a carry out of I 28d. Browse other questions tagged rsa algorithm-design montgomery-multiplication or ask your own question. We have already discussed the binary addition and binary subtraction in detail in the previous articles now we are going to discuss binary multiplication in a detailed manner. If we take the number 237 and shift each digit 1 place to the left, adding a zero to the right hand side, we get 2370. How to use this calculator: In the calculator, there are two input fields intended for entry of binary numbers. Binary Multiplication •Sizing •In binary addition –we are generally representing something that ultimately is to be executed in hardware •Our hardware cannot change the number of bits (wires) it can hold •We must establish a maximum number size •For multiplication the size of the result must be the sum of the. Binary Arithmetic 1: Adding binary numbers. Take some parts away or add new ones in, and you get a different outcome, but not a binary, win/lose one. Convert number. Note that since binary operates in base 2, the multiplication rules we need to remember. (countable) A calculation involving multiplication. 16 + 0 + 0 + 2 + 1 = 19. Binary Multiplication •Sizing •In binary addition –we are generally representing something that ultimately is to be executed in hardware •Our hardware cannot change the number of bits (wires) it can hold •We must establish a maximum number size •For multiplication the size of the result must be the sum of the. Leibniz pointed out in 1703 that to do simple arithmetic in binary, such as addition and multiplication, you don’t need to memorize such rules as 5 + 4 = 9, or 6 × 7 = 42. You just take a regular number (called a "scalar") and multiply it on. Unicode Character 'MULTIPLICATION SIGN' (U+00D7). Consider the simple problem of multiplying 110 2 by 10 2. A, 1 is entered in its position that yields 1000 since 2 3 = 8. Multiplying Complex Numbers Calculator. Binary addition is the simplest method to add any of the binary numbers. Addition Worksheet A :. Let’s add two binary numbers to understand the binary addition. So a binary matrix is such an array of 0's and 1's. A, 1 is entered in its position that yields 1000 since 2 3 = 8. Align the numbers as an ordinary subtraction. The binary number system uses 0s and 1s to represent numbers. No problem with unsigned (always positive) numbers, just use the same Multiplication in Two's complement cannot be accomplished with the standard technique since, as. The number 1010110 is represented by 7 bits. Binary multiplication is actually much simpler to calculate than decimal multiplication. Learn binary conversions and arithmetic with interactive demonstrations and explanations. ), with steps shown. If given condition is FALSE (not TRUE), expr_set2 will get executed. The procedure for binary multiplication is similar to that in decimal system. Computer method: Computer method is used by digital machines to multiply the binary numbers. There are four rules for binary subtraction: Borrow 1 is required from the next higher order bit to subtract 1 from 0. The binary number system uses only two digits 0 and 1 due to which their addition is simple. Sponsored Links.