# What is the probability that player A and B play against each other?

Players with same skill take part in a competition. The probability of winning each game is 0.5. At first, we divide a group of $2^n$ people to random pairs that play against each other. Then we will do the same for $2^{n-1}$ winners and this continues until there is only one winner. What is the probability that player A and B play against each other?

I know we should calculate the sum of probabilities that two players play against each other in the first round, then for second round and etc. So how can i calculate the probability that two players play at round K?

• Did you try a small example to see what would have to happen for A and B to meet in the second round, or the third round, and then see how it may generalize from that? – Bram28 Oct 20 '17 at 16:17

• It is a knockout competition with $m=2^n$ players so $m-1$ knockouts or matches or parings are needed to produce a winner

• There are ${m \choose 2}=\frac{m(m-1)}{2}$ equally-likely possible parings of the $m$ players, so the probability any particular match involves both players A and B is $\frac{2}{m(m-1)}$, and so the expected number of the $m-1$ matches which involve both players A and B is $(m-1) \times \dfrac{2}{m(m-1)}=\dfrac{2}{m}$

• Since players A and B can only meet $0$ or $1$ times, the probability they meet is $\dfrac{2}{m} = \dfrac{1}{2^{n-1}}$

• Elegantly simple. – true blue anil Oct 21 '17 at 2:14
• $P(A , B \text{ play at round } 1) = \frac{1}{2^{n-1}}$

When we have $2^{n}$ people, then we have at most $\log^{2^{n}}_{2}$ rounds. For example, assume we have only $2$ players, then $P(A , B \text{ play at round } 1) = \frac{1}{2^{\log^{2}_{2}-1}} = 1$

• $P(A , B \text{ play at round } 2) = P(A , B \text{ did not play at round } 1 \text{ and } A , B \text{ both win at the first round}) = (1-\frac{1}{2^{n-1}})(\frac{1}{2})^{2}$

• $P(A , B \text{ play at round } 3) = P(A , B \text{ did not play at round } 1,2 \text{ and } A , B \text{ both win at the first, second rounds}) = (1-\frac{1}{2^{n-1}})(1-\frac{1}{2^{n-2}})(\frac{1}{2})^{4}$

$$...$$

• $\forall k<n: \quad P(A , B \text{ play at round } k) = (\frac{1}{2})^{2(k-1)}\times \Pi_{i=1}^{k}(1-\frac{1}{2^{n-i}})$

$$...$$

• $\bbox[5px,border:2px solid #C0A000]{P(A , B \text{ play at round } n \text{ and } n>1) = (\frac{1}{2})^{2(n-1)}\times \Pi_{i=1}^{n-1}(1-\frac{1}{2^{n-i}})}$
• I think you're missing minus ones. Assume there are only two players. Is the probability that they will play $\frac12$? Also each round you have to multiply with that stages probability that they will play. – karakfa Oct 20 '17 at 16:32
• @karakfa, your first comment is not correct. I explain it by an example. – Hasan Heydari Oct 20 '17 at 16:42
• yes, my mistake $2^n$ people means for two people n=1. – karakfa Oct 20 '17 at 16:48

Hmm...I don't see any real difference between this and an ordinary single-elimination tournament where the seeding is set from the start. Maybe I'm missing something.

Anyway, assuming that there isn't any difference, we observe that Player A has $2^n-1$ different possible opponents in the first round. Of those, $2^{n-1}$ are in the opposite half of the draw, and both they and Player A would have to win $n-1$ games to meet; this happens with probability $\frac{1}{4^{n-1}}$. $2^{n-2}$ are in the same half, but opposite quarters, and both they and Player A would have to win $n-2$ games to meet; this happens with probability $\frac{1}{4^{n-2}}$. And so on, until we get to the $2^{n-n} = 2^0 = 1$ single player who meets Player A in the first round. Altogether, the probability of Player A and a given Player B meeting eventually is

\begin{align} \frac{1}{2^n-1} \sum_{k=1}^n 2^{n-k}\frac{1}{4^{n-k}} & = \frac{1}{2^n-1} \sum_{k=1}^n \frac{1}{2^{n-k}} \\ & = \frac{1}{2^n-1} \sum_{k=0}^{n-1} \frac{1}{2^k} \\ & = \frac{1}{2^n-1} \left(2-\frac{1}{2^{n-1}}\right) \\ & = \frac{1}{2^n-1} \times \frac{2^n-1}{2^{n-1}} \\ & = \frac{1}{2^{n-1}} \end{align}

I'll come back to edit this if I think (or someone points out) that there's a substantive difference between random reseeding and not reseeding.

There's also a pretty simple proof by induction: For $n = 1$ (two players), the probability is clearly $\frac{1}{2^n-1} = 1$. For $n > 1$, the probability that they meet in the first round is $\frac{1}{2^n-1}$. If they don't meet in the first round (with probability $\frac{2^n-2}{2^n-1}$), then they must both win their first games (with probability $\frac14$) to get to the next round. With reseeding, this is clearly the case of $n-1$, so with the premise that the probability of them meeting at that point is $\frac{1}{2^{n-1-1}} = \frac{1}{2^{n-2}}$, the overall probability for case $n$ is

\begin{align} \frac{1}{2^n-1}+\frac{2^n-2}{2^n-1} \times \frac14 \times \frac{1}{2^{n-2}} & = \frac{1}{2^n-1} \left(1+\frac{2^n-2}{2^n}\right) \\ & = \frac{1}{2^n-1} \times \frac{2^{n+1}-2}{2^n} \\ & = \frac{1}{2^n-1} \times \frac{2^n-1}{2^{n-1}} \\ & = \frac{1}{2^{n-1}} \end{align}


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# To find the units digit of a large number?

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To find the units digit of a large number? [#permalink]  03 Jul 2013, 21:58
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So if we are given a question what is the units digit of $$777^{777}$$

we find the pattern for 7 (7,9,3,1)

then we divide $$\frac{777}{4}$$ and the remainder is 1 so the units digit is $$7^1$$ which is 7?

Is this correct?
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Re: To find the units digit of a large number? [#permalink]  03 Jul 2013, 22:08
fozzzy wrote:
So if we are given a question what is the units digit of $$777^{777}$$

we find the pattern for 7 (7,9,3,1)

then we divide $$\frac{777}{4}$$ and the remainder is 1 so the units digit is $$7^1$$ which is 7?

Is this correct?

Absolutely Correct.
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Re: To find the units digit of a large number? [#permalink]  03 Jul 2013, 22:10
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fozzzy wrote:
So if we are given a question what is the units digit of $$777^{777}$$

we find the pattern for 7 (7,9,3,1)

then we divide $$\frac{777}{4}$$ and the remainder is 1 so the units digit is $$7^1$$ which is 7?

Is this correct?

Yes.

The units digit of 777^777 = the units digit of 7^777.

7^1 has the units digit of 7;
7^2 has the units digit of 9;
7^3 has the units digit of 3;
7^4 has the units digit of 1.
7^5 has the units digit of 7 AGAIN.

The units digit repeats in blocks of 4: {7, 9, 3, 1}...

The remainder of 777/4 is 1, thus the units digit would be the first number from the pattern, so 7.

Hope it's clear.
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Re: To find the units digit of a large number? [#permalink]  03 Jul 2013, 22:23
One final question here is another example

If we have the find the units digit of $$344^{328}$$

$$4^1$$ is 4
$$4^2$$ is 16
$$4^3$$ is 4

so the repeating block over here {4,6}

In this case the remainder is 0 so the units digit of this expression is 6?

so if there are 4 repeating blocks and the remainder is 0 we raise it to the 4th power ( some examples would be 3,7 etc) in this current example its the 2nd power?
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Re: To find the units digit of a large number? [#permalink]  03 Jul 2013, 22:26
1
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fozzzy wrote:
One final question here is another example

If we have the find the units digit of $$344^{328}$$

$$4^1$$ is 4
$$4^2$$ is 16
$$4^3$$ is 4

so the repeating block over here {4,6}

In this case the remainder is 0 so the units digit of this expression is 6?

so if there are 4 repeating blocks and the remainder is 0 we raise it to the 4th power ( some examples would be 3,7 etc) in this case its the 2nd power?

Yes, if the remainder is 0, then take the last digit from the block.
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Re: To find the units digit of a large number? [#permalink]  07 Jul 2013, 20:00
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fozzy,

you are correct in both cases. here is how i like to think about it:
7^{777} example:
If we divide 777 by 4, the quotient is 194 and the remainder is 1.
This means that we will have 194 of {7, 9, 3, 1} repeating blocks, and we will have 1 more term left, and the units digit of 7^{777} will be 7, the first term in the repeating block.

When we look at the case of 344^{328}, the units digit of powers of 4 cycle as {4, 6}, when we divide 328 by 2, the remainder is 0, this means that there will be exactly 164 blocks consisting of {4,6} without any remainder and the units digit of 344^{328} will be 6.

here are problems using the same concept(some hard):
#1) if-n-is-a-positive-integer-what-is-the-remainder-when-82380.html
#2) what-is-the-remainder-when-7-345-7-11-2-is-divided-by-26794.html
#3) remainder-when-7-4n-3-6-n-104848.html
#4) if-3-4n-1-is-divided-by-10-can-the-remainder-be-0-a-4662.html
#5) what-s-the-remainder-of-2-x-divided-by-10-1-x-is-an-even-9651.html

cheers,
dabral
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Re: To find the units digit of a large number? [#permalink]  16 Mar 2015, 15:27
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Re: To find the units digit of a large number? [#permalink]  01 Apr 2015, 23:35
Finding the UNIT’S DIGIT - please share some more useful tricks if its there.
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Re: To find the units digit of a large number?   [#permalink] 01 Apr 2015, 23:35
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# Find all the integer values of k for x^3 - kx + (k + 11) = 0 has at least one positive integer solution

#### Taran

##### New member
Hi, this question was in a year 11 extension maths textbook in the enrichment section. I have the answer as k>17 and k<-11 because I graphed it on GeoGebra. The Graph can be found here: https://ggbm.at/xpegwwtq. While I know the answers I would like to know how to work it out using algebra.

Here is the Question:
Consider the cubic equation x^3 - kx + (k + 11) = 0, find all the integer values of k for which the equation has at least one positive integer solution for x

Thanks, Taran

#### Opalg

##### MHB Oldtimer
Staff member
Hi, this question was in a year 11 extension maths textbook in the enrichment section. I have the answer as k>17 and k<-11 because I graphed it on GeoGebra. The Graph can be found here: https://ggbm.at/xpegwwtq. While I know the answers I would like to know how to work it out using algebra.

Here is the Question:
Consider the cubic equation x^3 - kx + (k + 11) = 0, find all the integer values of k for which the equation has at least one positive integer solution for x

Thanks, Taran
Hi Taran , and welcome to MHB.

If $x=n$ is a positive integer solution of the equation, then $n^3 - kn + k + 11 = 0$, so that $$k = \frac{n^3+11}{n-1} = \frac{(n-1)(n^2+n+1) + 12}{n-1} = n^2+n+1 + \frac{12}{n-1}.$$ For that to be an integer, $n-1$ must be a factor of $12$. You can then tabulate the possible values of $n$ and $k = \frac{n^3+11}{n-1}$, as follows: $$\begin{array}{c|cccccc} n-1&1&2&3&4&6&12 \\ n&2&3&4&5&7&13 \\ k&19&19&25&34&59&184 \end{array}.$$ So the only possible values for $k$ are $19,\ 25,\ 34,\ 59,\ 184$ (which all agree with your condition that $k>17$).

Last edited:

#### MarkFL

##### Pessimist Singularitarian
Staff member
Hi Taran , and welcome to MHB.

If $x=n$ is a positive integer solution of the equation, then $n^3 - kn + k + 11 = 0$, so that $$k = \frac{n^3+11}{n-1} = \frac{(n-1)(n^2+n+1) + 12}{n-1} = n^2+n+1 + \frac{12}{n-1}.$$ For that to be an integer, $n-1$ must be a factor of $12$. You can then tabulate the possible values of $n$ and $k = \frac{n^3+11}{n-1}$, as follows: $$\begin{array}{c|cccccc} n-1&1&2&3&4&6&12 \\ n&2&3&4&5&7&13 \\ k&19&19&25&34&59&184 \end{array}.$$ So the only possible values for $k$ are $19,\ 25,\ 34,\ 59,\ 184$ (which all agree with your condition that $k>17$).
Hello, Chris!

This question was posted on another site, and I found your reply so insightful, I took the liberty of posting it there, for the benefit of several there trying to solve it.

#### Taran

##### New member
Hi, Thank you so much!!! This question had my class stumped. That answer makes so much sense. It's been bugging me for a while and I'm very thankful for your help.

Thanks again, Taran


# Prove that $\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$

In a related question the following integral was evaluated $$\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{\mathrm{d}x/2}{1 + \cos x \sin x} =\int_0^{2\pi} \frac{\mathrm{d}x/2}{2 + \sin x} \,\mathrm{d}x =\int_{-\infty}^\infty \frac{\mathrm{d}x/2}{1+x+x^2}$$ I noticed something interesting, namely that \begin{align*} \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x & = \int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \\ & = \int_0^{\pi} \frac{(\cos x)^2}{1 - \cos x \sin x} \,\mathrm{d}x = \int_0^{\pi} \frac{(\sin x)^2}{1 - \cos x \sin x} \,\mathrm{d}x \end{align*} The same trivially holds if the upper limits are changed to $\pi/2$ as well ($x \mapsto \pi/2 -u$). But I had problems proving the first equality. Does anyone have some quick hints?

• consider evaluating $$\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} - \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x = \int_0^{\pi} \frac{\cos (2x)}{1 +\frac{1}{2}\sin(2 x)}\,\mathrm{d}x$$ – ganeshie8 Oct 2 '14 at 10:27
• Are you not satisfied yet with the answers below? – Anastasiya-Romanova 秀 Oct 7 '14 at 12:07
• @N3buchadnezzar Oh sorry, I didn't know that. Get well soon ᕙ( ^ₒ^ c) – Anastasiya-Romanova 秀 Oct 8 '14 at 8:21
• Someday ;)${}{}{}$ – N3buchadnezzar Oct 8 '14 at 10:12
• math.stackexchange.com/questions/61605/… If you want further discussion ping me in chat. – N3buchadnezzar Oct 8 '14 at 10:32

Split the integral into two terms with limit $\left[0,\frac{\pi}{2}\right]$ and $\left[\frac{\pi}{2},\pi\right]$ \begin{align} \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi/2} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x +\int_{\pi/2}^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \\ \end{align} Using identity \begin{align} \int_a^{b} f(x) \,\mathrm{d}x = \int_a^{b} f(a+b-x) \,\mathrm{d}x \end{align} Also the facts that $\cos\left(\frac{\pi}{2}-x\right)=\sin x$, $\sin\left(\frac{\pi}{2}-x\right)=\cos x$, $\cos\left(\frac{3\pi}{2}-x\right)=-\sin x$, and $\sin\left(\frac{3\pi}{2}-x\right)=-\cos x$, we get \begin{align} \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x &=\int_0^{\pi/2} \frac{(\sin x)^2}{1 + \sin x \cos x} \,\mathrm{d}x +\int_{\pi/2}^{\pi} \frac{(-\sin x)^2}{1 + \sin x \cos x} \,\mathrm{d}x \\ &=\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x\tag{1} \end{align}

Let \begin{align} I=\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \end{align} Since \begin{align} \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \end{align} then \begin{align} 2I&=\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x +\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x\\ &=\int_0^{\pi} \frac{(\cos x)^2 + (\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x\\ I&=\frac{1}{2}\int_0^{\pi} \frac{1}{1 + \cos x \sin x} \,\mathrm{d}x\tag{2} \end{align}

Using $(2)$, we get \begin{align} I&=\int_0^{\pi} \frac{1}{2 + 2\cos x \sin x} \,\mathrm{d}x\\ &=\int_0^{\pi} \frac{1}{2 + \sin (2x)} \,\mathrm{d}x\qquad\Rightarrow\qquad x\mapsto2x\\ &=\frac{1}{2}\int_0^{2\pi} \frac{1}{2 + \sin x} \,\mathrm{d}x\tag{3} \end{align}

Subtract the two integrals in question and get

$$\int_0^{\pi} dx \frac{\cos{2 x}}{1+\frac12 \sin{2 x}} = \frac12 \int_0^{2 \pi} du \frac{\cos{u}}{1+\frac12 \sin{u}}$$

This may be shown to be equal to the complex integral

$$-\frac{i}{4} \oint_{|z|=1} \frac{dz}{z} \frac{z+z^{-1}}{1+\frac{1}{4 i} (z-z^{-1})} = \oint_{|z|=1}\frac{dz}{z} \frac{z^2+1}{z^2+i 4 z-1}$$

The poles of the integrand within the unit circle are at $z=0$ and $z=-(2-\sqrt{3}) i$; their respective residues are $-1$ and $1$. By the residue theorem, therefore, the integral is zero and the two original integrals are equal.

A quick hint is noticing the symmetry. A rigorous proof is that $$\int\limits_a^b {\frac{{f(\cos x)}}{{g(\sin x\cos x)}}dx} = \int\limits_a^b {\frac{{f\left( {\sin \left( {\frac{\pi }{2} - x} \right)} \right)}}{{g\left( {\cos \left( {\frac{\pi }{2} - x} \right)\sin \left( {\frac{\pi }{2} - x} \right)} \right)}}dx} = \int\limits_{\frac{\pi }{2} - b}^{\frac{\pi }{2} - a} {\frac{{f\left( {\sin u} \right)}}{{g\left( {\cos u\sin u} \right)}}du}$$Hope it helps ;)

$$\frac{\cos(\pi/2-x)^2}{1+\cos(\pi/2-x)\sin(\pi/2-x)} = \frac{\sin(x)^2}{1+\sin(x)\cos(x)}$$

The integrands are both periodic with period $\pi$, so it suffices to verify the identity over any interval of length $\pi$, including $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Then, since both integrands are even functions, it suffices to check that the identity holds when integrating over $\left[0, \frac{\pi}{2}\right]$, that is, that $$\int_0^{\frac{\pi}{2}} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\frac{\pi}{2}} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x.$$

But one can show this simply by substituting $x = \frac{\pi}{2} - u$ on either side.

• Your answer was the first I got, andthe first one I used. However for later readers who also stumble uppon this integral I feel the answer given by Anastasiya is clearer. – N3buchadnezzar Oct 7 '14 at 17:30
• The important thing is that you've learned something from posting here, which it seems you have. – Travis Oct 8 '14 at 3:23

$$I=\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$ $$J=\int_0^{\pi/2} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$

Use $\int_a^b f(x)dx=\int_a^nf(a+b-x)dx$ and $\int_0^{2a}f(x)dx=\int_0^a f(x)dx+f(2a-x)dx$

$$I=\int_0^{\pi/2} \frac{(\cos (\pi-x))^2}{1 + \cos (\pi-x) \sin (\pi-x)} \,\mathrm{d}x+\frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$ $$I=\int_0^{\pi/2} \frac{(\cos x)^2}{1 - \cos x \sin x} \,\mathrm{d}x+\frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$ $$I=\int_0^{\pi/2} \frac{(\cos (\pi/2-x))^2}{1 - \cos (\pi/2-x) \sin (\pi/2-x)} \,\mathrm{d}x+\frac{(\cos (\pi/2-x))^2}{1 + \cos (\pi/2-x) \sin (\pi/2-x)} \,\mathrm{d}x$$ $$I=\int_0^{\pi/2} \frac{(\sin x)^2}{1 - \sin x \cos x} \,\mathrm{d}x+\frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$ $$I=\int_0^{\pi/2} \frac{(\sin x)^2}{1 - \sin x \cos x} \,\mathrm{d}x+\frac{(\sin (\pi-x))^2}{1 - \cos (\pi-x) \sin(\pi-x)} \,\mathrm{d}x$$ $$I=\int_0^{\pi} \frac{(\sin x)^2}{1 - \sin x \cos x} \,\mathrm{d}x$$

Hope you can now show that: $$J=\int_0^{\pi/2} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x=\int_0^{\pi/2} \frac{(\cos (\pi/2-x))^2}{1 + \cos (\pi/2-x) \sin (\pi/2-x)} \,\mathrm{d}x=\int_0^{\pi/2} \frac{(\sin x)^2}{1 + \sin x \cos x} \,\mathrm{d}x$$


# How to minimize residual sum of squares of an exponential fit?

I have the following data and would like to fit a negative exponential growth model to it:

Days <- c( 1,5,12,16,22,27,36,43)
Emissions <- c( 936.76, 1458.68, 1787.23, 1840.04, 1928.97, 1963.63, 1965.37, 1985.71)
plot(Days, Emissions)
fit <- nls(Emissions ~ a* (1-exp(-b*Days)), start = list(a = 2000, b = 0.55))
curve((y = 1882 * (1 - exp(-0.5108*x))), from = 0, to =45, add = T, col = "green", lwd = 4)

The code is working and a fitting line is plotted. However, the fit is visually not ideal, and the residual sum of squares seems to be quite huge (147073).

How can we improve our fit? Does the data allow a better fit at all?

We could not find a solution to this challenge on the net. Any direct help or linkage to other websites/posts is greatly appreciated.

-
In this case, if you consider a regression model $\text{Emissions}_i=f(\text{Days}_i,a,b)+\epsilon_i$, where $\epsilon_i\sim N(0,\sigma)$, then you obtain similar estimators. By plotting the confidence regions, one can observe how these values are contained in the confindence regions. You cannot expect a perfect fit unless you interpolate the points or use a more flexible nonlinear model. – user10525 Jul 23 '12 at 13:12
I changed the title because "negative exponential model" means something different than described in the question. – whuber Jul 23 '12 at 13:15
Thanks for making the question clearer (@whuber) and thanks for your answer (@Procrastinator). How can I calculate and plot the confidence regions. And, what would be a more flexible nonlinear model? – Strohmi Jul 23 '12 at 13:18
You need an additional parameter. See what happens with fit <- nls(Emissions ~ a* (1- u*exp(-b*Days)), start = list(a = 2000, b = 0.1, u=.5)); beta <- coefficients(fit); curve((y = beta["a"] * (1 - beta["u"] * exp(-beta["b"]*x))), add = T). – whuber Jul 23 '12 at 13:20
@whuber - maybe you should post that as an answer? – jbowman Jul 23 '12 at 13:25

A (negative) exponential law takes the form $y=-\exp(-x)$. When you allow for changes of units in the $x$ and $y$ values, though, say to $y = \alpha y' + \beta$ and $x = \gamma x' + \delta$, then the law will be expressed as

$$\alpha y' + \beta = y = -\exp(-x) = -\exp(-\gamma x' - \delta),$$

which algebraically is equivalent to

$$y' = \frac{-1}{\alpha} \exp(-\gamma x' - \delta) - \beta = a\left(1 - u\exp(-b x')\right)$$

using three parameters $a = -\beta/\alpha$, $u = 1/(\beta\exp(\delta))$, and $b = \gamma$. We can recognize $a$ as a scale parameter for $y$, $b$ as a scale parameter for $x$, and $u$ as deriving from a location parameter for $x$.

As a rule of thumb, these parameters can be identified at a glance from the plot:

• The parameter $a$ is the value of the horizontal asymptote, a little less than $2000$.

• The parameter $u$ is the relative amount the curve rises from the origin to its horizontal asymptote. Here, the the rise therefore is a little less than $2000 - 937$; relatively, that's about $0.55$ of the asymptote.

• Because $\exp(-3) \approx 0.05$, when $x$ equals three times the value of $1/b$ the curve should have risen to about $1-0.05$ or $95\%$ of its total. $95\%$ of the rise from $937$ to almost $2000$ places us around $1950$; scanning across the plot indicates this took $20$ to $25$ days. Let's call it $24$ for simplicity, whence $b \approx 3/24 = 0.125$. (This $95\%$ method to eyeball an exponential scale is standard in some fields that use exponential plots a lot.)

Let's see what this looks like:

plot(Days, Emissions)
curve((y = 2000 * (1 - 0.56 * exp(-0.125*x))), add = T)

Not bad for a start! (Even despite typing 0.56 in place of 0.55, which was a crude approximation anyway.) We can polish it with nls:

fit <- nls(Emissions ~ a * (1- u * exp(-b*Days)), start=list(a=2000, b=1/8, u=0.55))
beta <- coefficients(fit)
plot(Days, Emissions)
curve((y = beta["a"] * (1 - beta["u"] * exp(-beta["b"]*x))), add = T, col="Green", lwd=2)

The output of nls contains extensive information about parameter uncertainty. E.g., a simple summary provides standard errors of estimates:

> summary(fit)

Parameters:
Estimate Std. Error t value Pr(>|t|)
a 1.969e+03  1.317e+01  149.51 2.54e-10 ***
b 1.603e-01  1.022e-02   15.69 1.91e-05 ***
u 6.091e-01  1.613e-02   37.75 2.46e-07 ***

We can read and work with the entire covariance matrix of the estimates, which is useful for estimating simultaneous confidence intervals (at least for large datasets):

> vcov(fit)
a             b             u
a 173.38613624 -8.720531e-02 -2.602935e-02
b  -0.08720531  1.044004e-04  9.442374e-05
u  -0.02602935  9.442374e-05  2.603217e-04

nls supports profile plots for the parameters, giving more detailed information about their uncertainty:

> plot(profile(fit))

Here is one of the three output plots showing variation in $a$:

E.g., a t-value of $2$ corresponds roughly to a 95% two-sided confidence interval; this plot places its endpoints around $1945$ and $1995$.

-
Oh, I almost forgot: res <- residuals(fit); res %*% res tells us that introducing the third parameter $u$ reduces the sum of squares to $2724$ (compared to $147073$ as stated in the question). – whuber Jul 23 '12 at 14:23
All well and good whuber. But maybe the OP had some reason to pick the exponential model (or maybe it is just because it is well known). I think first the residuals should be looked at for the exponential model. Plot them against potential covariates to see if there is structure there and not just large random noise. Before jumping into more sophisticated models try to see if a fancier model could possibly help. – Michael Chernick Jul 23 '12 at 14:25
Why don't you look at the original plot, Michael? It will make it abundantly obvious why at least one additional parameter is needed. Please note, too, that in a comment to the question the OP has asked, "what would be a more flexible nonlinear model?" One implication of the initial analysis offered in this answer is that it should be considered out of the ordinary to fit an exponential with fewer than three parameters: there must be some inherent constraint operating in such cases (such as an intrinsically determined unit of measurement or an intrinsic location for $x$). – whuber Jul 23 '12 at 14:32
I was not criticizing your answer! I did not see any residual plots. All I was suggesting is that plots of residuals vs potential covariates should be the first step in finding a better model. If I thought I had an answer to put up there I would have given an answer rather than raised my point as a constant. I thought you gave a great response and i was among the ones who gave you a +1. – Michael Chernick Jul 23 '12 at 14:58


Is there a general threshhold for which a large condition number becomes “problematic?”

I think the answer is probably no in the linear algebra community, but I'd say anything above $$10^8$$ and you're starting to lose too much precision, making the situation problematic. In some statistics settings, where they use OLS, I've seen it stated that even values over $$20$$ are problematic, but that seems pretty extreme. How do you guys assess whether a condition number is "problematic?"

• I would consider anything above $1/\sqrt{\epsilon_{\text{mach}}}$ problematic, so it would be floating point system dependent. But it also depends on the size of the linear system. You can create small linear systems (3x3) with a condition number at O(1000) and the solution a computer finds could be out of acceptable tolerance. In general, I tell my students if they are using double precision the danger zone starts at 1000, and they need to be particularly careful if it is above $1/\sqrt{\epsilon_{\text{mach}}}$ – Abdullah Ali Sivas May 7 at 13:30
• @AbdullahAliSivas Is it also dependent on what method you're using to solve the linear system? If it's a direct method, I imagine a large condition number is (relatively) less problematic than an iterative method? Also, what is the logic behind $\frac{1}{\sqrt{\epsilon_{mach}}}$. I know what machine epsilon is, but I think I'm having trouble seeing the direct connection. – David May 7 at 13:35
• The condition number does not have much to do with stability of the method you are using. For example, say you are using the conjugate gradient method. The error coming from the matrix is at the mat-vec step. So the issue there would be caused by the condition number of the mat-vec operation not the condition number of the matrix. So both a direct method and CG would struggle similarly in that sense. But that doesn't have anything to do with the fact that an ill-conditioned problem is hard to solve independent of however stable the algorithm is. – Abdullah Ali Sivas May 7 at 14:08
• $1/\sqrt{\epsilon_{\text{mach}}}$ is some kind of heuristic for me; that is just where I notice things start to break. Like a small perturbation of $\epsilon_{\text{mach}}$ to the matrix (or the right-hand side vector) may cause upto an error of $\sqrt{\epsilon_{\text{mach}}}$ even if you assume that you have a perfect algorithm which does not introduce and propagate errors and is not affected by the floating point arithmetic. And my error tolerance is usually at $10^{-8}$ level which is close to $1/\sqrt{\epsilon_{\text{mach}}}$ in double precision :). – Abdullah Ali Sivas May 7 at 14:26
• @AbdullahAliSivas Why not make all of this into an answer? :-) – Wolfgang Bangerth May 7 at 16:10

This answer is my comments compiled and edited together. I apologize for the repetition.

I would consider anything above $$1/\sqrt{\epsilon_{\text{mach}}}$$ problematic, so it would be floating point system dependent. But it also depends on the size of the linear system. You can create small linear systems (3x3) with a condition number at O(1000) and the solution a computer finds could be out of acceptable tolerance. In general, I tell my students if they are using double precision the danger zone starts at 1000, and they need to be particularly careful if it is above $$1/\sqrt{\epsilon_{\text{mach}}}$$.

I want to note that the condition number does not have much to do with stability of the method you are using. Even when it is connected, for example, if you are using the conjugate gradient method, the error coming from the matrix is at the mat-vec step. So the issue there would be caused by the condition number of the mat-vec operation not the condition number of the matrix.

As Federico Poloni pointed out in the comments, the condition number of the matrix-vector multiplication problem is bounded from above by the condition number of the matrix $$A$$, e.g. $$\|A\|\|A^{-1}\|$$. However, I think we should rather consider the slightly more general bound for the condition number of mat-vec $$Ax$$: $$\|A\|\|x\|/\|Ax\|$$. While we can immediately take the condition number of the problem equal to the condition number of the matrix, I usually see it written as $$\alpha\kappa(A)$$ with $$\alpha=(\|x\|/\|Ax\|)/(\|A^{-1}\|)$$. The idea, I believe, is that mat-vec and msolve are inverses of each other, and if one of them is well-conditioned, the other one will be ill-conditioned, similar to multiplication and division (some exceptions apply: $$1\times x$$ and $$x/1$$ are both well-conditioned). So, for matrices with high condition number msolves, it still may be the case that $$(\|x\|/\|Ax\|)\ll \|A^{-1}\|$$. Hence, the condition number of the mat-vec operation may be much lower than the condition number of the matrix. As an extreme example, if $$A$$ is singular $$Ax$$ is still meaningful and may be well-conditioned, but $$\kappa(A)=\infty$$ and $$A^{-1}x$$ is an ill-posed problem.

So both a direct method and CG would struggle similarly if the elementary steps of the algorithm have issues related to input errors. But this is not the same thing as the conventional wisdom that an ill-conditioned problem is hard to solve independent of how stable the algorithm you employ is.

Note that my arbitrary choice of $$1/\sqrt{\epsilon_{\text{mach}}}$$ is basically a heuristic for me and, in my experience, it is just where things start to break. As an example, a small perturbation of the relative magnitude $$\epsilon_{\text{mach}}$$ to the matrix (or the right-hand side vector) may cause up to an error of $$\sqrt{\epsilon_{\text{mach}}}$$, even if you assume that you have a perfect algorithm which does not introduce and propagate errors and is not affected by the floating point arithmetic. And my error tolerance is usually at $$10^{−8}$$ level which is close to $$\sqrt{\epsilon_{\text{mach}}}$$ in double precision :)

• Thanks for the expanded answer! I still disagree on some points though: (1) if we are speaking about the usual relative condition number, then scalar matrix multiplication and division have both condition number 1, for each $x$. (2) I view matrix multiplication and solving linear systems as perfectly symmetrical operations, when seen as abstract maps: multiplying by $A$ isn't any different than multiplying by $A^{-1}$. The same vector-dependent modification to the condition number can be done for solving linear systems, by multiplying by $\alpha = ||b|| / ||A^{-1}b|| ||A||$. – Federico Poloni May 8 at 7:04
• I was thinking more in terms of this answer: math.stackexchange.com/a/3031746 . They are definitely symmetrical when thought as maps, and probably there are matrices for which msolve is well-conditioned but not mat-vec, I just haven't heard of them. – Abdullah Ali Sivas May 8 at 7:43


# what is the cube root of 125

The cube root of -8 is written as $$\sqrt[3]{-8} = -2$$. ... ∛125: 5 ∛216: 6 ∛1,000: 10 ∛1,000,000: 100 ∛1,000,000,000: 1000: The calculations were performed using this cube root calculator. When did organ music become associated with baseball? 5³ = 5 * 5 * 5 = 25 * 5 = 125 So the cube root of 125 is 5. So, in this case the cube root of 125 is 5. Here is the answer to questions like: What is the cube root of 125 or what is the cube root of 125? Someone help me with this ASAP ANSWER QUICK! The nearest previous perfect cube is … Thus, each edge of the cube is 5 cm long. That number is 5. Calculating n th roots can be done using a similar method, with modifications to deal with n. While computing square roots entirely by hand is tedious. Thus, each edge of the cube is 5 cm long. The cube root of 125 is what number cubes is 125. 80% of questions are answered in under 10 minutes Answers come with explanations, so that … See next answers. So, in this case the cube root of 125 is 5. Step 1) Set up 125 in pairs of two digits from right to left and attach one set of 00 because we want one decimal: … Learn more with Brainly! What is plot of the story Sinigang by Marby Villaceran? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. cube root of 125/512 = 5/8. Estimating higher n th roots, even if using a calculator for intermediary steps, is … The length of a side (edge) of a cube is equal to the cube root of the volume. Does the calculator support fractions? What is the birthday of carmelita divinagracia? While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information. The length of a side (edge) of a cube is equal to the cube root of the volume. Learn more with Brainly! 125 can be written as 125(e^0), 125(e^((2pi)i)), 125(e^((4pi)i)) where e is euler’s number, i is the imaginary unit, i^2=-1, and e^((theta)i)=cos(theta)+(i)(sin(theta)) where theta is an angle measured in radians. First we will find all factors under the cube root: 125 has the cube factor of 125. All Rights Reserved. Get free help! Find the interest rate … that Sally earned from the bank. Having trouble with your homework? Guess: 5.125 27 ÷ 5.125 = 5.268 (5.125 + 5.268)/2 = 5.197 27 ÷ 5.197 = 5.195 (5.195 + 5.197)/2 = 5.196 27 ÷ 5.196 = 5.196 Estimating an n th Root. Is evaporated milk the same thing as condensed milk? The cube root of a number answers the question "what number can I multiply by itself twice to get this number?". 5³ = 5 * 5 * 5 = 25 * 5 = 125 So the cube root of 125 is 5 Just right click on the above image, then choose copy link address, then past it in your HTML. The length of a side (edge) of a cube is equal to the cube root of the volume. For example, 5 is the cube root of 125 because 53 = 5•5•5 = 125, -5 is cube root of -125 because (-5)3 = (-5)•(-5)•(-5) = -125. Why don't libraries smell like bookstores? Yes, simply enter the fraction as a decimal floating point number and you will get the corresponding cube root. Who is the longest reigning WWE Champion of all time? Not sure about the answer? See also our cube root table from 1 to 1000. How to find the square root of 125 by long division method Here we will show you how to calculate the square root of 125 using the long division method with one decimal place accuracy. New questions in Mathematics. Therefore the cube roots of 125 are 5(e^0),5(e^((2pi)i/3)),5(e^((4pi)i/3)). 'CUBE ROOT OF 125' is a 13 letter phrase starting with C and ending with 5 Crossword clues for 'CUBE ROOT OF 125' Synonyms, crossword answers and other related words for CUBE ROOT OF 125 [five] We hope that the following list of synonyms for the word five will help you to finish your crossword today. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. How long does it take to cook a 23 pound turkey in an oven? 5(e^0)=5(1)=0 How long will the footprints on the moon last? For … As you can see the radicals are not in their simplest form. Let's check this with ∛125*1=∛125. Now extract and take out the cube root ∛125 * ∛1. Inter state form of sales tax income tax? The cube root of -64 is written as $$\sqrt[3]{-64} = -4$$. The cube root of 125 is 5 so therefore each edge is 5 cm which is about 2 inches A cube root of a number a is a number x such that x 3 = a, in other words, a number x whose cube is a. Please link to this page! Thus, each edge of the cube is 5 cm long. All information in this site is provided “as is”, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information. 125 is said to be a perfect cube because 5 x 5 x 5 is equal to 125. Cube of ∛125=5 which results into 5∛1; All radicals are now simplified. For example, 5 is the cube root of 125 because 5 3 = 5•5•5 = 125, -5 is cube root of -125 because (-5) 3 = (-5)• (-5)• (-5) = -125. Therefore, the real cube root of 125 is 5. Copyright © 2020 Multiply Media, LLC. How will understanding of attitudes and predisposition enhance teaching? What is the conflict of the story of sinigang? So, in this case the cube root of 125 is 5. This is the lost art of how they calculated the square root of 125 by hand before modern technology was invented. That number is 5. The cube root of 125 is what number cubes is 125. Cube Root of 125. Volume to (Weight) Mass Converter for Recipes, Weight (Mass) to Volume to Converter for Recipes. Since 125 is a whole number, it is a perfect cube. Who of the proclaimers was married to a little person? ( cube root of 125 is 5 and cube root of 512 is 8) hope it helps you... i want full solution The Brain; Helper; Not sure about the answer? A cube root of a number a is a number x such that x3 = a, in other words, a number x whose cube is a. After 42 months, Sally earned $238 in simple interest. Sally deposited$850 into her bank account for 42 months. WHO WANNA JOIN MY … How long will it take to cook a 12 pound turkey? What is the contribution of candido bartolome to gymnastics? Cube roots (for integer results 1 through 10) Cube root of 1 is 1; Cube root of 8 is 2; Cube root of 27 is 3; Cube root of 64 is 4; Cube root of 125 is 5; Cube root of 216 is 6; Cube root of 343 is 7; Cube root of 512 is 8; Cube root of 729 is 9 What details make Lochinvar an attractive and romantic figure?


# Prove: $\log_2(x)+\log_x(y)+\log_y(8)\geq \sqrt[3]{81}$

Prove that for every $x$,$y$ greater than $1$: $$\log_2(x)+\log_x(y)+\log_y(8)\geq \sqrt[3]{81}$$

What I've tried has got me to:

$$\frac{\log_y(x)}{\log_y(2)}+\log_x(y)+3\log_y(2)\geq \sqrt[3]{81}$$

I didn't really get far.. I can't see where I can go from here, especially not what to do with $\sqrt[3]{81}$.

This is taken out of the maths entry tests for TAU, so this shouldn't be too hard.

• $\log_xy=\frac1{\log_yx}$ – MalayTheDynamo Jul 16 '18 at 19:04
• A neat thing about $\log_a b = \frac 1{\log_b a}$ and $\frac {\log_b c}{\log_b a} = \log_a c$.... It means $\log_a b* \log_b c = \log_a c$ and so $\log_a b*\log_b c* \log_c d*....... *\log_y z = \log_a z$...... – fleablood Jul 16 '18 at 19:40

We have $$\log_2x+\log_xy+\log_y8=\frac{\log x}{\log2}+\frac{\log y}{\log x}+\frac{3\log2}{\log y}$$

and since $2,x,y>1$, we can deduce that their logarithms are non-negative, and so will their quotients.

Now use AM-GM:

$$\sqrt[3]{\frac{\log x}{\log2}\cdot\frac{\log y}{\log x}\cdot\frac{3\log2}{\log y}}\le\frac{\frac{\log x}{\log2}+\frac{\log y}{\log x}+\frac{3\log2}{\log y}}3$$ giving $$\frac{\log x}{\log2}+\frac{\log y}{\log x}+\frac{3\log2}{\log y}\ge3\sqrt[3]3=\sqrt[3]{81}$$ as required.

• While I did all these logarithmic exercises, I noticed that the AM-GM is actually a very common technique for the solution. Is there any reason for that or is it just a coincudence? – L0wRider Jul 17 '18 at 12:41
• @Maxim AM-GM is usually the second inequality that people learn (after $x^2 \ge 0$). So it's a good choice to use when writing a problem for an entrance exam, because many people are familiar with it. – Ovi Jul 17 '18 at 17:37
• @Ovi That's embarrassing. I learned about AM/GM/HM just last week :) – TheSimpliFire Jul 17 '18 at 19:22
• @TheSimpliFire Don't feel embarrased, I'm not saying that most people know about and how to use AM-GM. I'm saying that of the people who know inequalities, most know AM-GM; so if they put an inequality question, AM-GM is a good problem. However, the question of weather they should put inequalities questions is separate. – Ovi Jul 17 '18 at 21:39

Let's convert everything to $$\log$$ base $$2$$ so we have a common something to work with:

$$\log_2 x + \dfrac {\log_2 y}{\log_2 x} + \dfrac {\log_2 8}{\log_2 y} \ge \sqrt [3]{81} = 3 \sqrt[3]{3}$$

Now let $$a = \log_2 x$$ and $$b = \log_2 y$$. The condition $$x, y > 1$$ implies that $$a, b > 0$$. So now we have to prove

$$a + \dfrac ba + \dfrac {3}{b} \ge3 \sqrt[3]{3}$$

The cube root in there especially may remind us of AM-GM with $$3$$ terms. And indeed, the inequality $$\text{AM}\left(a, \dfrac ba, \dfrac 3b\right) \ge \text{GM}\left(a, \dfrac ba, \dfrac 3b\right)$$ gives exactly what is desired.

Hint: Show that $a+\frac{b}{a}+\frac{3}{b}\geq \sqrt[3]{81}$ for all $a,b>0$, using AM-GM.

Two things:

$\log_a b = \frac 1{\log_b a}$ and $\frac {\log_b c}{\log_b a} = \log_a c$ so $\log_a b\log_b c = \frac {\log_b c}{\log_b a} = \log_a c$.

And AM-GM says $\frac {a + b+ c}3 \ge \sqrt[3]{abc}$.

So.....

$\frac {\log_2 x + \log_x y + \log_y 8}3 \ge \sqrt[3]{\log_2 x \log_x y \log_y 8} = \sqrt[3]{\log_2 8}$


# Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$

I got this question in my maths paper

Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$ and find the sum if it exists.

I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you.

Using Partial Fraction Decomposition, $$\frac1{n(n+1)(n+2)}=\frac An+\frac B{n+1}+\frac C{n+2}$$

$$\implies 1=A(n+1)(n+2)+Bn(n+2)+Cn(n+1)$$ $$\implies 1=n^2(A+B+C)+n(3A+2B+C)+2A$$

Comparing the coefficients of the different powers (namely, $0,1,2$) of $n,$ we get $A=\frac12,B=-1,C=\frac12$

$$\implies\frac1{n(n+1)(n+2)}=\frac12\cdot\frac1n-\frac1{n+1}+\frac12\cdot\frac1{n+2}$$ $$=-\frac12\left(\underbrace{\frac1{n+1}-\frac1n}\right)+\frac12\left(\underbrace{\frac1{n+2}-\frac1{n+1}}\right)$$

Can you recognize the two Telescoping series?

Hint

$$\frac{2}{n(n+2)}=\frac{1}{n}-\frac{1}{n+2}$$

Now multiply both sides by $\frac{1}{n+1}$.

• This is a good hint. – Mark Bennet Oct 28 '17 at 17:51

Hint. You may write $$\frac{1}{n(n+1)(n+2)} =\frac{1}{2}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\left(\frac{1}{(n+2)}-\frac{1}{(n+1)}\right)$$ giving two telescoping sums $$\sum_{n=1}^N\frac{1}{n(n+1)(n+2)} =\frac{1}{2}-\frac{1}{2(N+1)}+\frac{1}{2(N+2)}-\frac{1}{4}.$$

• That is a great clean answer!! +1 Not like that show off of Jack who always has to show his "ability" in complicating easy things. – Von Neumann Sep 20 '16 at 20:44
• @FourierTransform: the benefit of complicating things is that my approach also computes $$\sum_{k\geq 1}\frac{1}{k(k+1)(k+2)(k+3)(k+4)(k+5)}$$ in two steps. It is, in fact, equivalent to telescoping. – Jack D'Aurizio Sep 20 '16 at 20:47
• @DarioGutierrez Thanks. I just started with your identity:$\frac{1}{n(n+1)(n+2)} =\frac{1}{2}(\frac{1}{n}+\frac{1}{(n+2)})-\frac{1}{(n+1)}$, but $\frac{1}{2}(\frac{1}{n}+\frac{1}{(n+2)})-\frac{1}{(n+1)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\left(\frac{1}{(n+2)}-\frac{1}{(n+1)}\right)$. – Olivier Oloa Sep 20 '16 at 20:51
• @FourierTransform It's always nice to have different points of view. That was what $\texttt{Jack D'Aurizio}$ did it. Usually, I receive a lot of negative comments or/and down-votes in my own answers because I posted some weird overkill answer. What 'some' people don't understand is that I'm adding something different to the bag of answers. By the way, behind me I have a guy who behaves like a 'MSE-police officer' who down-vote, propose to close, review my $\LaTeX$, change my editions, etc... a kind of insane user. Also, I enjoy your answers too which are 'full creative'. Best Regards. – Felix Marin Sep 20 '16 at 21:43
• @FourierTransform We should have different answers and different ideas, else, what would be the point of answering in the first place? Also, different ideas lead to learning and building of intuition, which is one reason I dislike the school system to some level. I agree with Felix, we enjoy both of your answers, and we want both. Sometimes, answers aren't meant for the OP, but for the community. – Simply Beautiful Art Sep 20 '16 at 22:04

There is an alternate method and is as follows.

Notice that $$\frac{1}{n(n+1)(n+2)} = \frac{(n-1)!}{(n+2)!} = \frac{1}{2!} \, B(n,3)$$ where $B(x,y)$ is the Beta function. Using an integral form of the Beta function the summation becomes \begin{align} S &= \sum_{n=1}^{\infty} \frac{1}{n \, (n+1) \, (n+2)} \\ &= \frac{1}{2} \, \int_{0}^{1} \left( \sum_{n=1}^{\infty} x^{n-1} \right) \, (1-x)^{2} \, dx \\ &= \frac{1}{2} \, \int_{0}^{1} \frac{(1-x)^{2}}{1-x} \, dx = \frac{1}{2} \, \int_{0}^{1} (1-x) \, dx \\ &= \frac{1}{4} \end{align}

This leads to the known result \begin{align} \sum_{n=1}^{\infty} \frac{1}{n \, (n+1) \, (n+2)} = \frac{1}{4}. \end{align}

Using Euler's beta function, $$S=\sum_{k\geq 1}\frac{(k-1)!}{(k+2)!}=\sum_{k\geq 1}\frac{\Gamma(k)}{\Gamma(k+3)}=\frac{1}{2}\sum_{k\geq 1}B(k,3)$$ hence: $$S = \frac{1}{2}\int_{0}^{1}(1-x)^2\sum_{k\geq 1}x^{k-1}\,dx=\frac{1}{2}\int_{0}^{1}(1-x)\,dx = \color{red}{\frac{1}{4}}.$$ A straightforward generalization of this approach gives the identity: $$\sum_{n\geq 1}\frac{1}{n(n+1)\cdots(n+N)}=\color{red}{\frac{1}{N\cdot N!}}.$$

Writing

$$\frac{1}{n(n+1)(n+2)} = \frac{1}{n+1}\left(\frac{1}{2n} - \frac{1}{2(n+2)}\right) = \frac{1}{2n(n+1)} - \frac{1}{2(n+1)(n+2)},$$

we see that the sum telescopes to

$$\frac{1}{2(1)(2)} = \frac{1}{4}.$$

• This is the best form to use - simple and straightforward. – Mark Bennet Oct 28 '17 at 17:51

Alternatively, take $$\frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1},$$ and integrate three times with lower limit $0$, giving \begin{align*} -\log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n} \\ x + (1-x)\log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} \\ \frac{3}{4}x^2 - \frac{1}{2}x - \frac{1}{2} (1-x)^2 \log{(1-x)} &= \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)(n+2)}, \end{align*} and (as @Clement C reminds me) we then apply Abel's theorem to take the limit as $x \to 1$, which gives $1/4$ as the answer.

• Isn't there a bit more to the argument, then? The original power series has radius of convergence $1$, so the equality and all integration theorems only apply on $(-1,1)$. To apply the conclusion to $x=1$ at the end (outside the open disc of convergence of the original series, and where the final LHS is only defined by continuity), don't you need Abel's theorem? – Clement C. Mar 26 '15 at 14:12
• I suppose, since you also have to take the limit to get $0\log{0} "=" 0$. I'll put a note. – Chappers Mar 26 '15 at 14:17

You are on the right track. Fix $N \geq 1$. Then \begin{align} \sum_{n=1}^{N} \frac{1}{n \cdot (n+1) \cdot (n+2)} &= \sum_{n=1}^N \left(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4}\right) = \frac{1}{2}\sum_{n=1}^N \frac{1}{n}-\sum_{n=1}^{N}\frac{1}{n+1}+\frac{1}{2}\sum_{n=1}^{N} \frac{1}{n+2}\\ &= \frac{1}{2}\sum_{n=1}^N \frac{1}{n}-\sum_{n=2}^{N+1}\frac{1}{n}+\frac{1}{2}\sum_{n=3}^{N+2} \frac{1}{n} \\ &= \frac{1}{2}\left(1+\frac{1}{2}\right) + \frac{1}{2}\sum_{n=3}^N \frac{1}{n}-\left(\frac{1}{2}+\frac{1}{N+1}\right)-\sum_{n=3}^{N}\frac{1}{n}+\\ &\quad \left(\frac{1}{N+1}+\frac{1}{N+2}\right)+\frac{1}{2}\sum_{n=3}^{N} \frac{1}{n} \\ \end{align}

Can you continue from there? (the partial sums cancel out, and you only have a few remaining terms. Taking the limit $N\to\infty$ will give you the limit.)

• Interesting approach... Different from everything I learned.... Thanks! – Frank Mar 26 '15 at 14:09

Here's another way to do it:$$\sum_{n\ge 1}\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\sum_{n\ge 1}\bigg(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\bigg)=\frac{1}{2}\cdot\frac{1}{1\cdot 2}=\frac{1}{4}.$$One advantage is an easy generalisation to a problem for which the partial fractions decomposition would get thorny:$$\sum_{n\ge 1}\frac{1}{\sum_{j=0}^k(n+j)}=\frac{1}{k!\cdot k}.$$