Math Help - 2, 5, 8, 11, 14...

1. 2, 5, 8, 11, 14...

How can I prove that no square number can be in the sequence?

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...

Is it to do with mod 5? And if so how would I do it

2. Originally Posted by Natasha
How can I prove that no square number can be in the sequence?

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...

Is it to do with mod 5? And if so how would I do it
First you need find the general term of your sequence. Now it looks to me
as though the sequence is:

$a_n=2+3n, n=0,1,2,..$

Which suggests you consider the squares modulo 3. Now any natural number
may be written:

$N=3k+\rho$,

for some natural number $k$, and $\rho= 0, 1\ \mbox{or}\ 2$. So

$N^2=9k^2+6k \rho + \rho^2$.

Therefore $N^2(mod\ 3)$ is $\rho^2(mod\ 3)$, but

$
\rho^2(mod\ 3)=0, 1, \mbox{or}\ 1(mod\ 3) \mbox{as}\ \rho=0,1 \mbox{or}\ 2
$
.

So as $a_n, n=0,1,2,..$ is congruent to $2(mod\ 3)$ it cannot be a square for any
natural number $n$

RonL

3. Originally Posted by CaptainBlack
First you need find the general term of your sequence. Now it looks to me
as though the sequence is:

$a_n=2+3n, n=0,1,2,..$

Which suggests you consider the squares modulo 3. Now any natural number
may be written:

$N=3k+\rho$,

for some natural number $k$, and $\rho= 0, 1\ \mbox{or}\ 2$. So

$N^2=9k^2+6k \rho + \rho^2$.

Therefore $N^2(mod\ 3)$ is $\rho^2(mod\ 3)$, but

$
\rho^2(mod\ 3)=0, 1, \mbox{or}\ 1(mod\ 3) \mbox{as}\ \rho=0,1 \mbox{or}\ 2
$
.

So as $a_n, n=1,2,..$ is congruent to $2(mod\ 3)$ it cannot be a square for any
natural number $n$

RonL
Captain is it the same to say that the general form 3n - 1 is the same as 3n + 2??

4. Originally Posted by Natasha
Captain is it the same to say that the general form 3n - 1 is the same as 3n + 2??
Yes, but the +2 form is slightly more convienient when dealing with modular
arithmetic.

RonL

5. Originally Posted by CaptainBlack
Yes, but the +2 form is slightly more convienient when dealing with modular
arithmetic.

RonL
I see. Thanks!


# Find the Ratio in Which the Point P(X, 2) Divides the Line Segment Joining the Points A(12, 5) and B(4, −3). Also, Find the Value of X. - Mathematics

Find the ratio in which the point P(x, 2) divides the line segment joining the points A(12, 5) and B(4, −3). Also, find the value of x.

#### Solution 1

Let the point P (x, 2) divide the line segment joining the points A (12, 5) and B (4, −3) in the ratio k:1.

Then, the coordinates of P are ((4k+12)/(k+1),(-3k+5)/(k+1))

Now, the coordinates of P are (x, 2).

therefore (4k+12)/(k+1)=x and (-3k+5)/(k+1)=2

(-3k+5)/(k+1)=2

-3k+5=2k+2

5k=3

k=3/5

Substituting k=3/5 " in"  (4k+12)/(k+1)=x

we get

x=(4xx3/5+12)/(3/5+1)

x=(12+60)/(3+5)

x=72/8

x=9

Thus, the value of x is 9.

Also, the point P divides the line segment joining the points A(12, 5) and (4, −3) in the ratio 3/5:1  i.e. 3:5.

#### Solution 2

Let k be the ratio in which the point P(x,2)  divides the line joining the points

A(x_1 =12, y_1=5) and B(x_2 = 4, y_2 = -3 ) . Then

x= (kxx4+12)/(k+1) and 2 = (kxx (-3)+5) /(k+1)

Now,

 2 = (kxx (-3)+5)/(k+1) ⇒ 2k+2 = -3k +5 ⇒ k=3/5

Hence, the required ratio is3:5 .

Concept: Section Formula
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.3 | Q 20 | Page 29
RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 5


# How to evaluate this definite double integral?

$$\int\int_D 6x\sqrt{y^2-x^2}dA, D=\{(x,y)|0\leq y\leq 2, 0 \leq x \leq y\}$$

I tried: $$\int_0^2 \int_0^y 6x\sqrt{y^2-x^2}dxdy$$

But that is incorrect.

• Why do you say that's incorrect? Looks fine to me... – Eli Berkowitz Dec 14 '15 at 3:32
• Your integral looks perfect! Now just evaluate it, try a u-substitution for the inside integral. – Alex Dec 14 '15 at 3:37
• I got -8, I must have made an arithmetic error or something! – d0rmLife Dec 14 '15 at 3:39

Setting $y^2-x^2 = t$, we obtain $-2xdx = dt$. Hence, we obtain $$I = \int_0^2 \int_0^y 6x\sqrt{y^2-x^2}dxdy = \int_0^2 \int_{y^2}^0(-3\sqrt{t}dt)dy = \int_0^2 \int_0^{y^2} 3\sqrt{t}dtdy = \int_0^22y^3 dy = 8$$
• Why does the range of the inner integral change to $y^2,0$ instead of $y^2,y^2-y$ - I thought you put the original range of integration into the u substitution equation? – d0rmLife Dec 14 '15 at 3:47
• @d0rmLife because the first inner integral goes from $x=0$ to $x=y$, and when replacing $y^2-x^2=t$ in the first case you get $y^2-0^2=y^2=t$ and in the second $y^2-y^2=0=t$. When you u-substitute you must change the range accordingly or else it is "wrong": you can get away with it if you evaluate the result in terms of x instead of u in the end, but still it is best to change it. – Fede Poncio Dec 14 '15 at 4:13
Notice, $$\int_{0}^2\int_0^y 6x\sqrt{y^2-x^2} \ dxdy=\int_{0}^2\left(\int_0^y 6x\sqrt{y^2-x^2} \ dx\right)dy$$ $$=\int_{0}^2\left(-3\int_0^y (y^2-x^2)^{1/2} \ d(y^2-x^2)\right)dy$$ $$=\int_{0}^2dy\left(-3\frac{2(y^2-x^2)^{3/2}}{3}\right)_{0}^{y}$$ $$=\int_{0}^2\left(2y^3\right)\ dy$$ $$=\left(2\frac{y^4}{4}\right)_{0}^{2}=\left(\frac{2^4}{2}-0\right)=\color{red}{8}$$


# Recognizing the sequence 1/16, 1/8, 3/16, 1/4, 5/16, …

What is the missing number? $$\frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}, \ \ \ [?]$$

$$A. \frac{5}{4}\quad B. \frac{3}{4}\quad C. \frac{5}{8}\quad D. \frac{3}{8}$$

Spoiler: Answer is $D$, but I don't know why.

Thanks

-
OMG, of course. Thank you for the heads up. –  user1495024 Aug 2 at 1:22
Am I the only one who thought $\frac{1}{2}$ before reading the possible options? :) –  Thomas Aug 2 at 4:15
No, the next two numbers should definitely be 1/2 and 7/16 :-) –  CompuChip Aug 2 at 8:23
@Thomas I did consider it, but then thought it wasn't just right. First of all, I'd be generalizing from just two data points. Secondly I'd be ignoring the other three data points, which supposedly were there for a reason. –  kasperd Aug 2 at 9:37

$$\frac{1}{16}, \frac{1}{8}=\frac{2}{16}, \frac{3}{16}, \frac{1}{4}=\frac{4}{16}, \frac{5}{16}$$

So the $i$th term is of the form $$\frac{i}{16}$$ Therefore, the next term is $$\frac{6}{16}=\frac{3}{8}$$

-

$$\frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}$$ The above is the same as $\displaystyle\frac1{16},\frac2{16},\frac3{16},\frac4{16},\frac5{16}$.

-

Another sequence-recognizing technique is to look at the difference between consecutive terms.

In this case, $\frac{1}{8}-\frac{1}{16} = \frac{1}{16}$, $\frac{3}{16}-\frac{1}{8} = \frac{1}{16}$, $\frac{1}{4}-\frac{3}{16} = \frac{1}{16}$, and $\frac{5}{16}-\frac{1}{4} = \frac{1}{16}$.

Since the difference between consecutive terms is $\frac{1}{16}$, the next term should be $\frac{5}{16}+\frac{1}{16} =\frac{6}{16} =\frac{3}{8}$.

-


1. ## relations help (2)

hey guys, question:

Determine whether the following relation is a reflexive, symmetric or transitive. If any is an equivalence relation, describe its equivalence classes.

xRy iff x has the same integer part as y, on <----- I need help on what this relation is asking? thanks

2. Originally Posted by jvignacio
hey guys, question:

Determine whether the following relation is a reflexive, symmetric or transitive. If any is an equivalence relation, describe its equivalence classes.

xRy iff x has the same integer part as y, on <----- I need help on what this relation is asking? thanks
The integer part of x is the integer n such that $n\le x < n+1$. (There is exactly one such integer for each real x.) It is usually denoted by [x].

For example:
[3.14]=3
[0.52]=0
[1]=1
[-1.26]=-2

So 2 and 2.5 are in relation, since [2]=2=[2.5], but 1.5 and 2.5 are not.

Hope this helps.

3. Originally Posted by kompik
the integer part of x is the integer n such that $n\le x < n+1$. (there is exactly one such integer for each real x.) it is usually denoted by [x].

For example:
[3.14]=3
[0.52]=0
[1]=1
[-1.26]=-2

so 2 and 2.5 are in relation, since [2]=2=[2.5], but 1.5 and 2.5 are not.

Hope this helps.
Is that like saying the next full integer down?

4. Originally Posted by jvignacio
Is that like saying the next full integer down?
Exactly.

5. Originally Posted by kompik
Exactly.
So in this case,

- It IS reflexive since x has the same integer part as x.
- It IS symmetric since if x has the same integer part as y then y has the same integer part as x.
- It IS transitive since if x has the same integer part as y and if y has the same integer part as z then x has the same integer part as z.

correct?

6. Originally Posted by jvignacio
So in this case,

- It IS reflexive since x has the same integer part as x.
- It IS symmetric since if x has the same integer part as y then y has the same integer part as x.
- It IS transitive since if x has the same integer part as y and if y has the same integer part as z then x has the same integer part as z.

correct?
Yes, it is an equivalence relation.

Another way to see this is to notice that you're in fact given a decomposition of R and every decomposition gives you an equivalence relation. (But if you haven't heard much about the correspondence between equivalences and decompositions at your lessons, you should perhaps ignore this comment.)

7. Originally Posted by kompik
Yes, it is an equivalence relation.

Another way to see this is to notice that you're in fact given a decomposition of R and every decomposition gives you an equivalence relation. (But if you haven't heard much about the correspondence between equivalences and decompositions at your lessons, you should perhaps ignore this comment.)
Yeah ive never herd of that... Not yet anyway. What would the equivalence class be in this case now that its an equivalence relation? All numbers?

8. Originally Posted by jvignacio
Yeah ive never herd of that... Not yet anyway. What would the equivalence class be in this case now that its an equivalence relation? All numbers?
No.
Take, for instance, the equivalent class of 3.14. It contains all numbers such that x R 3.14.
This is equivalent to
[x]=[3.14]=3.
And what are the numbers such that [x]=3? Precisely the numbers from the interval $\langle 3,4)$, right?
Can you find the remaining classes?

9. Originally Posted by kompik
No.
Take, for instance, the equivalent class of 3.14. It contains all numbers such that x R 3.14.
This is equivalent to
[x]=[3.14]=3.
And what are the numbers such that [x]=3? Precisely the numbers from the interval $\lange 3,4)$, right?
Can you find the remaining classes?
Sorry whats in the interval for [x]=3? theres a latex error..

10. Originally Posted by jvignacio
Sorry whats in the interval for [x]=3? theres a latex error..
Sorry, did not notice that. I've edited my post.

11. Originally Posted by kompik
No.
Take, for instance, the equivalent class of 3.14. It contains all numbers such that x R 3.14.
This is equivalent to
[x]=[3.14]=3.
And what are the numbers such that [x]=3? Precisely the numbers from the interval $\langle 3,4)$, right?
Can you find the remaining classes?
Ok I understand the equivalent class of 3.14 and the numbers such that [x]=3 are all numbers between 3 and 4 but Which remaining classes are you referring too? Since my question is a x and y question, how can I write this in a general form. If you get what I mean...

12. Originally Posted by jvignacio
Ok I understand the equivalent class of 3.14 and the numbers such that [x]=3 are all numbers between 3 and 4 but Which remaining classes are you referring too? Since my question is a x and y question, how can I write this in a general form. If you get what I mean...
If I were your teacher, I would expect answer to be something like:
The equivalent classes are intervals .... (fill in the dots) for $n\in\mathbb{N}$.
(I guess the example with 3.14 might help you to see what to fill in.)

13. Originally Posted by kompik
If I were your teacher, I would expect answer to be something like:
The equivalent classes are intervals .... (fill in the dots) for $n\in\mathbb{N}$.
(I guess the example with 3.14 might help you to see what to fill in.)
intervals $\langle n,n+1)$ , $n \in \mathbb{N}$ ?

14. Originally Posted by jvignacio
intervals $\langle n,n+1)$ , $n \in \mathbb{N}$ ?
Exactly.

15. Originally Posted by kompik
Exactly.
mate thanks alot for the help. Really appreciate your time.

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# Use proof by induction to prove $\frac{1}{n!}<\frac{1}{2^n-1}$ for all $n\geq 4$

Use proof by induction to prove that that $\frac{1}{n!}<\frac{1}{2^n-1}$ for all $n\geq 4$, .\Base case: $$\frac{1}{4}=\frac{1}{24}\leq \frac{1}{2^4-1}$$ Inductive hypothesis: Assume there exists $k\in \mathbb{N}$ s.t.

$$\frac{1}{k!}\leq\frac{1}{2^k-1}$$ Inductive step: Show that:$$\frac{1}{(k+1)!}\leq\frac{1}{2^{k+1}-1}$$ Now, $$\frac{1}{(k+1)!}=\frac{1}{k!}\cdot\frac{1}{k+1}$$ Using the hypothesis $$\frac{1}{(k+1)!}\leq\frac{1}{2^k-1}\cdot \frac{1}{k+1}$$ Because $n\geq4, \frac{1}{k+1}<\frac{1}{2}$ $$\frac{1}{(k+1)!}\leq\frac{1}{2^k-1}\cdot \frac{1}{2}=\frac{1}{2^{k+1}-2}\leq\frac{1}{2^{k+1}-1}$$ Hence by mathematical induction we have proved that $\frac{1}{n!}<\frac{1}{2^n-1}$ for all $n\geq 4$

Firstly I need to know if the proof is correct, secondly it has to be as concise as possible hence I would like to know if there are any lines I can change/delete

And lastly can anyone explain to me why every sentence starts with "\" It looks perfectly fine in www.sharelatex.com ;(

• You must start with $n=4$. – User3101 Feb 18 '15 at 16:11
• I see and I am lost know because it is a question copied from the coursework given by a professor, I just blindly followed so lets assume $n\geq4$ – Scavenger23 Feb 18 '15 at 16:12
• So I should say Assume that $\frac{1}{k!}\leq\frac{1}{2^k-1}$ holds for some $k\geq4$ and then proceed ? – Scavenger23 Feb 18 '15 at 16:21
• One error I notice in your logic - in the last statement you state that $\frac 1{2^{k+1}-2} \leq \frac 1{2^{k+1}-1}$. This is incorrect. The latter (-1) fraction has a larger denominator than the -2 fraction, which means it's a smaller fraction. – Duncan Feb 18 '15 at 16:34

All in all it looks like you have the idea down on how induction works. It looks like you are correctly making the induction step to get your result. I have a couple things to point out, however.

$(1)$ During your induction step you introduce the "$\geq$" symbol, when initially you are trying to write a proof about a strict inequality (which should only use the "$>$" symbol). If the strict inequality holds, it is not incorrect to use "$\geq$", but for the sake of consistency you should just use one.

$(2)$ Your induction hypothesis is not stated quite right. It is not enough to assume there exists $k \in \Bbb{N}$ such that the inequality holds. You want to make the stronger assumption that you can find this $k$, and the inequality holds for all $n \leq k$. It is in doing this that you will be allowed to conclude at the end of the proof that the inequality holds for all $n \geq 4$, instead of just $n=4$, and one arbitrary $k$.

$(3)$. If you are looking for a more concise proof, I would instead prove the equivalent statement that $2^n-1<n!$ for all $n \geq 4$. It is clear for $n \geq 4$ that $$2^n-1<2^n = 2 \cdot 2\cdot 2\cdot 2\cdot \ldots < 1 \cdot 2\cdot 3 \cdot 4 \cdot \ldots = n!$$ This is easier than dealing with fractions. But, upon proving this result one need merely flip the inequality around and put the quantities on each side under a numerator of $1$.

• Never mind (4). in the preview mode every sentence had "\" in front of it while in here as you see Everything is fine. I see the mistake $< and \leq$ thank you for pointing it out. Thank you. I started induction only 2 weeks ago and it is a difficult topic for me. – Scavenger23 Feb 18 '15 at 16:37
• You see I didn't have the "flipping the fractions" tool in the toolbox so I didn't even think about what you have done in (3). The grammar unfortunately will remain an issue, since English is not my native language. – Scavenger23 Feb 18 '15 at 16:42
• You know what now I realised the whole mistake. I can't copy correctly. The question is not asking to prove $\frac{1}{n!}<\frac{1}{2^n-1}$ but $\frac{1}{n!}<\frac{1}{2^{n-1}}$ – Scavenger23 Feb 18 '15 at 16:50
• I don't want to use you but if I finish the right question this time, would you be able to have a one last look on it? I will stop spamming now since the website tells me to avoid extended discussions in comments. – Scavenger23 Feb 18 '15 at 16:54
• Use proof by induction to prove that that $\frac{1}{n!}<\frac{1}{2^n-1}$ for all $n\geq 3$\\I will use the fact that $k!>2^{k-1}$ is an equivelant statement to $\frac{1}{k!}<\frac{1}{2^k-1}$ .\\Base case $3!>2^{2}$ Inductive step: Assume $k!>2^{k-1}$ holds for all $n\leq k$ for some $k \in \bbb{N}$\\\ Inductive hypothesis\\Show that $(k+1)!>2^k$\\It is clear that when $k\geq 3$\\$2^{k-1}<2^k=2*2*2*2...<1*2*3*4...=k!<(k+1)!$ – Scavenger23 Feb 18 '15 at 17:18

It's much easier proving that $2^n<1+n!$, for $n\ge4$, which is completely equivalent to your assignment. The base step is obvious. Suppose it holds for $n$; then $$2^{n+1}=2\cdot2^n<2\cdot(1+n!)=2+2\cdot n!<1+(n-1)\cdot n!+2\cdot n!=1+(n+1)!$$ because $(n-1)n!>1$.

You can, if you want, transform this into a proof of your assigned inequality, but it's not necessary.

Rewrite as$$n!\ge2^n.$$ Then $$4!\ge2^4$$ and $$n!\ge2^n\land n+1\ge2\implies(n+1)!\ge2^{n+1}.$$


# How do I prove that $\lim_{x→0} x⋅\ln x=0$

How do I prove (without L'Hôpital's rule) that $$\lim_{x→0} x⋅\ln x = 0$$.

I'm trying to get some intuitive sense for this, but it's quite hard. It's like trying to prove that $x$ goes faster to $\ 0 \$ then $\ln x$ goes to $-∞$, right ?

I tried this, for $x∈(0,1)$ $$x \ln x=x⋅-∫_x^1 \frac{1}{t}dt>x⋅\frac{(1-x)}{-x}=-1+x$$

So when $x→0$, I conclude that $x⋅\ln x≥-1$.

-
I don't know how to implement this, but how about squeeze theorem? It was the first thing that came to mind. –  000 Dec 13 '12 at 22:35

Since $\ln(\frac{1}{x})=-\ln x$, $$\lim_{x\to 0^{+}} x \cdot \ln x = \lim_{x\to \infty} \frac{1}{x} \cdot\ln \frac{1}{x} = - \lim_{x\to \infty} \frac{\ln x}{x}$$

But $\ln x$ is "slower" than $x$ (since $e^x$ is faster than $x$), so this limit is zero.

(Rigorously: for positive $x$, $2 e^x > x^2$ (by comparing power series of both sides). Plug $\sqrt{x}$ instead of $x$ and take logarithm of both sides: $\ln x < \ln 2 + \sqrt{x} =o(x)$.)

-
Is it always true that you can rewrite $\lim_{x→0^+}f(x) = \lim_{x→∞}f(1/x)$? –  Kasper Dec 13 '12 at 22:48
@Kasper, that's a great question. My intuition is no, and the reason is that there are tricky situations that arise when moving from a one-sided limit to a two-sided limit. I would say, however, that it may be true that $$\lim_{x \to 0}f(x)=\lim_{x \to \infty}f\left(\frac{1}{x}\right).$$ However, I am uncertain. –  000 Dec 13 '12 at 22:55
@Kasper - The answer is "Yes". Try to prove it using the definition of one-sided limit at 0 and limit at infinity. Show that if one limit exists, the other must equal it. –  Ofir Dec 13 '12 at 23:03
@Limitless - There's no problem because the limit at infinity is not two-sided. Actually, your proposed equality is false, because in the LHS you evaluate $f$ on both negative and positive values, and at the RHS you evaluate $f$ only on positive values. Take $f(x)=\frac{1}{x}$ to reach a contradiction. –  Ofir Dec 13 '12 at 23:05
@Ofir Let $s_n>0$ and $\lim_{n→∞} s_n = 0$. By definition $$\lim_{x→0^+}f(x)=\lim_{n→∞}f(s_n)$$. Let $t_n=1/s_n$. Then $\lim_{n→∞}t_n=+∞$. So $$\lim_{x→0^+}f(x)=\lim_{n→∞}f(s_n)=\lim_{n→∞}f(1/t_n)=\lim_{x→∞}f(1/x)$$ Do you mean something like this ? –  Kasper Dec 13 '12 at 23:21

$$x \ln x=-x⋅\int_x^1 \frac{1}{t}dt$$

Now, let $0<a$ be arbitrary. Then, for all $x \in (0,1)$ we have

$$\frac{1}{t^{1-a}}\leq \frac{1}{t} \leq \frac{1}{t^{1+a}}$$

Thus

$$-x⋅\int_x^1 \frac{1}{t^{1+a}}dt \leq -x \ln(x) \leq -x⋅\int_x^1 \frac{1}{t^{1-a}}dt$$ $$-x⋅\frac{x^a-1}{a} \leq -x \ln(x) \leq -x⋅\frac{x^{-a}-1}{-a}⋅$$

now let $x \to 0$.

P.S. The argument works directly with any $0<a<1$, so picking $a=\frac{1}{2}$ makes the proof much cleaner....

-

Make the change $x = e^{-y}$: $$\lim_{x \to 0} x \ln x = - \lim_{y \to \infty} y e^{-y}$$

Show that for $y > 0$: $$e^y > \frac{y^2}{2!}$$

And use the squeeze theorem.

-


# What am I doing wrong with this second order ODE?

I have an ODE: $$\ddot v=-kt^{-1}\dot v+kt^{-2}v$$ I want to solve it by reducing it to a first order ODE, by defining $u(t)=t^{m}v(t)$ If I rewrite the ODE in terms of $u$, this gives me: $$\ddot u =(-k+2m)t^{-1}\dot u +(km+k-1-m^2)t^{-2}u$$ Setting the coefficient of $u$ to zero gives $2m=k+\sqrt{k^2-4+4k}$, which I will write as $k+\alpha$.

This gives me the single order ODE $$\ddot u=\alpha t^{-1}\dot u$$ Solving it gives $$\dot u=2c\alpha t$$ $$u(t)=c\alpha t^2$$ Plugging this back into $v$, gives $$v(t)=c\alpha t^{2-m}$$ $$\dot v = (2-m)c\alpha t^{1-m}$$ $$\ddot v = (1-m)(2-m)c\alpha t ^{-m}$$

These equations are supposed to hold for any $k$. If we plug this into the original ODE, however, this reduces to the equation $$3k=4+\sqrt{k^2-4+4k}$$

Which obviously does not hold for all $k$.

What am I doing wrong?

• Hint: Since this is a Euler-Cauchy type, let $v = t^m$. Find $v''$ and $v'$ and substitute in, solve for $m$ and solve.
– Moo
Mar 30 '17 at 17:39
• most likely your solution changes between (over)damped and oscillatory depending on $k$ Mar 30 '17 at 18:00
• Do you need to reduce it to a first order ODE? Or are you fine solving it with any method? Mar 30 '17 at 18:06
• I want to learn as much as possible about these things, so I was working on reducing it to a first order ODE. So I'd like to learn about other methods as well, but also I'd like to make sure I can do this method Mar 30 '17 at 18:09
• Please check again all signs and coefficients, the one before $u$ should reduce to $-(m+k)(m-1)$. Mar 30 '17 at 18:53

What you should realize is that you don't need to do a change of variables here but recognize that your solution is a Euler-Cauchy problem and therefore can be solved by:

Let $v=t^m$ then subbing the values into your DE you get:

$$m(m-1)t^{m-2} = -kmt^{m-2} + kt^{m-2} \\ \implies m(m-1) +km -k =0 \\ \implies m=1,m=-k \\$$

therefore giving:

$$v(t) = At^1 + Bt^{-k} \implies v(t) = At + Bt^{-k}$$

Which does hold $\forall k \in \mathbb{R}$ but you should also note that it does hold $\forall t \in (-\infty,0) \cup (0,+\infty)$ because $t=0$ is a case basis in that if $k>0$ then $t \neq 0$ but if $k<0$ then $t\in (-\infty,\infty)$

Since you did say you want to learn about the reduction to first order just note then when you do that you typically let u be something like $u= t^m \dot{v}$ so then $\dot{u}$ would be equal to a second derivative of v.

In your first integration, you should get from $$\ddot u=αt^{-1}\dot u$$ to $$\ln|\dot u|=α\ln|t|+c$$ or $$\dot u = C·t^α$$ leading to $$u=\frac{C}{α+1}·t^{α+1}+D$$


# Proving ⊢ A∨(A→B) using Hilbert system

I'm self-studying mathematical logic from "Introduction to Mathematical Logic" by Detlovs and Podnieks (available free here under CC license). Unfortunately, it doesn't come with any solutions. I'm stuck trying to prove ⊢ A∨(A→B), from Exercise 1.4.2 (i) on page 43. The text appears to be using the Hilbert-style deduction method.

Relevant axioms (classical logic)

• $$L_1: B \to (C \to B)$$
• $$L_2: (B \to (C \to D)) \to ((B \to C) \to (B \to D))$$
• $$L_3: B \to B \lor C$$
• $$L_4: C \to B \lor C$$
• $$L_8: (B \to D) \to ((C \to D) \to (B \lor C \to D))$$
• $$L_9: (B \to C) \to ((B \to \lnot C) \to ¬B)$$
• $$L_{10}: \lnot B \to (B \lor C)$$
• $$L_{11}: B \lor \lnot B$$

Relevant inference rules

• Modus Ponens

Attempt

1. $$(A \to A \lor (A \to B)) \to ((\lnot A \to A \lor (A \to B)) \to (A \lor \lnot A \to A \lor (A \to B)))$$ --- $$L_8$$
2. $$A \to A \lor (A \to B)$$ --- $$L_6$$
3. $$A \lor \lnot A$$ --- $$L_{11}$$
4. $$(\lnot A \to A \lor (A \to B)) \to (A \lor \lnot A \to A \lor (A \to B))$$ --- from (1) and (2) by MP
5. $$(\lnot A \to A \lor (A \to B)) \to A \lor (A \to B)$$ --- from (3) and (4) by MP
6. $$(\lnot A \to (A \to A \lor (A \to B))) \to ((\lnot A \to A) \to (\lnot A \to A \lor (A \to B)))$$ --- $$L_2$$
7. $$(\lnot A \to (A \to A \lor (A \to B)))$$ --- $$L_{10}$$
8. $$(\lnot A \to A) \to (\lnot A \to A \lor (A \to B))$$ --- from (6) and (7) by MP

I'm stuck at this point because $$\lnot A \to A$$ appears to be a contradiction. I'm also not sure whether this is the right approach. It looks alright at formula 5, but I'm not sure how to prove $$\lnot A \to A \lor (A \to B)$$, since it requires proving that $$A \lor (A \to B)$$ is always true, which is what we're trying to prove in the first place.

The text states that it can be solved using 14 formulas, 13 being the shortest yet.

• The "proof strategy" is to use L11: $A \lor \lnot A$ and derive $A \lor (A \to B)$ from $A$ with L6 and from $\lnot A$ with L10. Then use L8 – Mauro ALLEGRANZA Jul 5 '20 at 16:48
• @MauroALLEGRANZA I seem to have done the first three steps in my attempt (if I understood correctly); substituting B=A, C=¬A for L8 seems to yield (A→D)→((¬A→D)→D) after simplifying. The only straightforward substitution for D that I haven't tried seems to be D=(¬A→A∨(A→B)) but that leads to a double negation case with L10, which hasn't been proved yet. – user383527 Jul 5 '20 at 20:36

## 1 Answer

I assume that you are forced not tu use the Deduction Theorem.

But you can use the so-called Law of Syllogism (transitivity of $$\to$$).

If so, here is a sketch of a derivation:

1. $$\vdash A \to (A \lor (A \to B))$$ --- L6

2. $$\vdash \lnot A \to (A \to B)$$ --- L10

3. $$\vdash (A \to B) \to (A \lor (A \to B))$$ --- L7

4. $$\vdash \lnot A \to (A \lor (A \to B))$$ --- from 2. and 3. by Syllogism.

Now we can "cook them" together using L8:

1. $$\vdash (A \to (A \lor (A \to B))) \to [(\lnot A \to (A \lor (A \to B))) \to ((A \lor \lnot A) \to (A \lor (A \to B)))]$$

Now, from 5., 1. and 4. by MP twice:

1. $$\vdash (A \lor \lnot A) \to (A \lor (A \to B))$$

Finally, using L11: $$\vdash A \lor \lnot A$$, by MP:

$$\vdash A \lor (A \to B)$$.

• Smart use of the law of syllogism! The text has proved it as a theorem, so it is straightforward to show it from (2) and (3). In total, this gives 13 formulas (verbosely written), which the text says is the shortest proof. – user383527 Jul 6 '20 at 22:35


# Sums of rising factorial powers

Doodling in wolfram, I found that $$\sum^{k}_{n=1}1=k$$ The formula is pretty obvious, but then you get $$\sum^{k}_{n=1}n=\frac{k(k+1)}{2}$$ That is a very well known formula, but then it gets interesting when you calculate $$\sum^{k}_{n=1}n(n+1)=\frac{k(k+1)(k+2)}{3}\\ \sum^{k}_{n=1}n(n+1)(n+2)=\frac{k(k+1)(k+2)(k+3)}{4}$$ And so on. There is an obvious pattern that I really doubt is a coincidence, but I have no idea how to prove it in the general case. Any ideas?

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Pascal's triangle comes to mind here –  Omnomnomnom Sep 4 '13 at 3:47
@Omnomnomnom I thought about it but I was unable to introduce binomials there. However, you can generalize the hypothesis using factorials, but usually working with factorials is harder than not doing so(At least I would have a very hard time trying to do so). –  chubakueno Sep 4 '13 at 3:51
These are essentially the sums for the general $d$ dimensional simplex. –  Jaycob Coleman Sep 4 '13 at 4:02
It's overkill, but you could expand and apply en.wikipedia.org/wiki/Faulhaber%27s_formula –  dls Sep 4 '13 at 4:11
@dls I think that in that case the interesting part would go backwards: That expanding and a applying Faulhaber´s yields such a regular and simple result :) –  chubakueno Sep 4 '13 at 4:22

You can argue any given case by induction. I will take your last,$$\sum^{k}_{n=1}n(n+1)(n+2)=\frac{k(k+1)(k+2)(k+3)}{4}$$ for the example, but I think it is easy to see how it gets carried forward. The base case is simply $1\cdot 2\cdot 3=1\cdot 2\cdot 3\cdot \frac 44$ If it is true up to $k$, then $$\sum^{k+1}_{n=1}n(n+1)(n+2)\\=\sum^{k}_{n=1}n(n+1)(n+2)+(k+1)(k+2)(k+3)\\=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)\frac {k+4-k}4\\=\frac{(k+1)(k+2)(k+3)(k+4)}{4}$$

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Thanks! Just changing the $2$, $3$ and $4$ by $x$,$x+1$ and $x+2$ does the work of generalizing it very nicely. –  chubakueno Sep 4 '13 at 3:59

The easy way to deal, for example, with $\sum_{i=1}^n i(i+1)(i+2)(i+3)$ is to let $F(i)=i(i+1)(i+2)(i+3)(i+4)$. We calculate $F(i)-F(i-1)$. We get $$i(i+1)(i+2)(i+3)(i+4)-(i-1)(i)(i+1)(i+2)(i+3).$$ There is a common factor of $i(i+1)(i+2)(i+3)$. When we "take it out" we are left with $(i+4)-(i-1)=5$.

Let $G(i)=\frac{F(i)}{5}$. Then by our calculation $i(i+1)(i+2)(i+3)=G(i)-G(i-1)$.

Now consider the sum $\sum_{i=1}^n i(i+1)(i+2)(i+3)$. This is $$(G(1)-G(0))+(G(2)-G(1))+G(3)-G(2)) +\cdots+(G(n)-G(n-1)).$$ Observe the telescoping. Since $G(0)=0$, the above sum is equal to $G(n)$. Thus $$\sum_{i=1}^n i(i+1)(i+2)(i+3)=G(n)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}.$$

Exactly the same idea works in general.

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The hypothesized equality can be written as follows: for any $m$, we conjecture $$\sum^{k}_{n=1}\frac{(n+m)!}{(n-1)!}=\frac{(k+m+1)!}{(m+2)(k-1)!}$$ Dividing both sides by $(m+1)!$, we have $$\sum^{k}_{n=1}\frac{(n+m)!}{(n-1)!(m+1)!}=\frac{(k+m+1)!}{(m+2)!(k-1)!}$$ Or, in other words $$\sum^{k}_{n=1}\binom{n+m}{m+1}=\binom{k+m+1}{m+2}$$ I'm not sure how to prove this (yet), but it seems very likely that there's a neat trick for all this.

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Partial sums of sequences $a_n$ that can be expressed as polynomials in $n$ are easily found using discrete calculus.

We start with the discrete version of the Fundamental Theorem of Integral Calculus:

\begin{align*} \sum\limits_{n=1}^k a_n &= \sum\limits_{n=1}^k (\Delta b)_n \\ &= (b_2 - b_1)+(b_3 - b_2)+ \ldots + (b_k - b_{k-1}) + (b_{k+1} - b_k) \\ &= b_{k+1} - b_1 \end{align*}

where $(\Delta b)_n = b_{n+1} - b_n$ is the forward difference. Finding the partial sum has now been reduced to finding a sequence $b_n$ such that $(\Delta b)_n = a_n$.

We will find $b$, the antiderivative of $a$, using falling powers, which are defined by

$$n^{\underline{k}} = n(n-1)(n-2)\ldots (n-k+1)$$

where $k$ is an integer and, by a second definition, $n^{\underline{0}}=1$. For example

$$n^{\underline{3}} = n(n-1)(n-2).$$

We now need one more result: the discrete derivative of $n^{\underline{k}}$ is given by

\begin{align*} \Delta n^{\underline{k}} &= (n+1)^{\underline{k}} - n^{\underline{k}} \\ &= (n+1)n^{\underline{k-1}} - n^{\underline{k-1}}(n-k+1) \\ &= kn^{\underline{k-1}} \end{align*}

Let's now find the partial sum for a particular case:

\begin{align} \sum^{k}_{n=1}n(n+1)(n+2) &= \sum^{k}_{n=1} (n+2)^{\underline{3}} \\ &= \sum^{k}_{n=1} \Delta \left[\frac{1}{4} (n+2)^{\underline{4}}\right] \\ &= \frac{(k+3)(k+2)(k+1)k}{4} - \require{cancel}\cancelto{0}{\frac{(1+2)(1+1)(1-0)(1-1)}{4}} \end{align}

The general case:

\begin{align} \sum^{k}_{n=1} (n+p)^{\underline{p+1}} &= \sum^{k}_{n=1} \Delta \left[\frac{1}{p+2} (n+p)^{\underline{p+2}}\right] \\ &= \frac{(k+1+p)(k+p)\ldots [(k+1+p)-(p+2)+1)]}{p+2} \\ &= \frac{(k+1+p)(k+p)\ldots k}{p+2} \end{align}

where $p>0$ is an integer.

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[I haven't finished yet. Will be gradually improved]

If we do it this way, then probably, it will become more insightful.

$$\sum^{k}_{n=1}1=k$$

then we preserve everything from the right side exactly as it is. $$\sum^{k}_{n=1}n=\frac{k(k+1)}{1 \cdot 2}$$ [there will be a picture of the right triangle]

Now again, preserve everything from the right side: $$\sum^{k}_{n=1}\frac{n(n+1)}{1 \cdot 2}=\frac{k(k+1)(k+2)}{1 \cdot 2 \cdot 3}$$

[there will be a picture of the 6 pyramids composed into rectangular parallelepiped]

$$\sum^{k}_{n=1}\frac{k(k+1)(k+2)}{1 \cdot 2 \cdot 3}=\frac{k(k+1)(k+2)(k + 3)}{1 \cdot 2 \cdot 3 \cdot 4}$$

[There should be a description of the connection between simple combinations and combinations with repetition]

$${n \choose k} = \frac{n \cdot (n - 1) \cdot \ldots \cdot (n - k + 1)}{k \cdot (k - 1) \cdot \ldots \cdot 1} = \frac{n^{\underline{k}}}{k!}$$ $$\left(\!\middle(\genfrac{}{}{0pt}{}{n}{k}\middle)\!\right) = \frac{n \cdot (n + 1) \cdot \ldots \cdot (n + k - 1)}{k \cdot (k - 1) \cdot \ldots \cdot 1} = \frac{n^{\overline{k}}}{k!}$$

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You should use the sandbox if you are part way through an answer. –  Michael Albanese 2 days ago
@Michael: Sorry, I don't understand how to use it. Probably, I just don't have the rights to use that post or I need to do something beforehand, but I don't see [Add answer] button there and I cannot edit them neither. –  Pixar 2 days ago
Ah, you might not have enough reputation yet. Oh well, keep it in mind for future use. –  Michael Albanese 2 days ago


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( l A simple and rough upper bound for the sum of binomial coefficients can be obtained using the binomial theorem: which is valid by for all integers = Rekursive Darstellung und Pascalsches Dreieck, Der Binomialkoeffizient in der Kombinatorik, Summen mit alternierenden Binomialkoeffizienten, Summen von Binomialkoeffizienten mit geraden bzw. {\displaystyle {\tbinom {n}{k}}=0} k setzt. ) 0 Um unnötigen Rechenaufwand zu vermeiden, berechnet man im Fall Then it follows from our earlier definition for the sum of binomial coefficients that. ( 1\quad 4 \quad 6 \quad 4 \quad 1\\ k -Binomialsymbolen: ( , ) ⋅ Addition obiger Gleichungen ) n 1 n und {\displaystyle \operatorname {Re} z>0} . M Die wörtliche Übersetzung von „ Suppose that the customer is a meat lover. The following is one specific path that the person might take. The reasoning is based on the multiplication principle (see here). , 1 Some properties make use of symmetry, some deal with expansion, but they all can be proved rather intuitively. \binom MN = \frac{M!}{N!(M-N)!} n For each k, the polynomial {\displaystyle z} {\displaystyle {\tbinom {t}{k}}} ( The Chu–Vandermonde identity, which holds for any complex-values m and n and any non-negative integer k, is, and can be found by examination of the coefficient of )^3}[/math], $\sum_{k=-a}^a(-1)^k{a+b\choose a+k} {b+c\choose b+k}{c+a\choose c+k} = \frac{(a+b+c)!}{a!\,b!\,c! 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(5−3)!3!=10.\binom{5}{3} = \frac{5!}{(5-3)! − | {\displaystyle M} , , [/math], that is clear since the RHS is a term of the exponential series [math] e^k=\sum_{j=0}^\infty k^j/j! of binomial coefficients. ∞ Möglichkeiten der Wahl des ersten Tupel-Elements. ∞ In ordering a pizza, a customer can choose from a list of 10 toppings: mushroom, onion, olive, bell pepper, pineapple, spinach, extra cheese, sausage, ham, and pepperoni. {\displaystyle -z,-s\notin \mathbb {N} } − = 5040. {\displaystyle m\leq n} }\\ , 5 k This formula is used in the analysis of the German tank problem. ) 1 {\displaystyle \sum _{k=0}^{[{\frac {n-1}{2}}]}{\binom {n}{2k+1}}=2^{n-1}} Assume that all the paths from any point to any point in the above diagram are available for walking. This gives. roten und 2 terms in this product is 0 {\color{blue}1 \qquad 3 \qquad 3 \qquad 1} \\ is a natural number for any natural numbers n and k. There are many other combinatorial interpretations of binomial coefficients (counting problems for which the answer is given by a binomial coefficient expression), for instance the number of words formed of n bits (digits 0 or 1) whose sum is k is given by + This article incorporates material from the following PlanetMath articles, which are licensed under the Creative Commons Attribution/Share-Alike License: Binomial Coefficient, Upper and lower bounds to binomial coefficient, Binomial coefficient is an integer, Generalized binomial coefficients. The point 1 is known to be in each -subset of the first type and the other points must be chosen from objects. ) = {\displaystyle n} A combinatorial interpretation of this formula is as follows: when forming a subset of elements (from a set of size ), it is equivalent to consider the number of ways you can pick elements and the number of ways you can exclude elements. Now, if you know your stuff about triangular numbers, you can say that. □\displaystyle { y }^{ 3 }{ a \choose 3 } = { 3 }^{ 3 }{ 10 \choose 3 } = 3240.\ _\squarey3(3a)=33(310)=3240. {\displaystyle k=m=n} und This is the statement that each row of Pascal's triangle is symmetric in either one of two ways. { (x+y) }^{ a }=\sum _{ i=0 }^{ a }{ a \choose i } { x }^{ a-i }{ y }^{ i }.(x+y)a=i=0∑a(ia)xa−iyi. n


# How many bit strings?

How many bit strings of length $8$ have either exactly two $1$-bit among the first $4$ bits or exactly two $1$-bit among the last $4$ bits?

My solution:

A bit only contains $0$ and $1$, so $2$ different numbers, i.e., $0$ and $1$. For the first part we have $2^6=64$ ways. Similar for the other way. Hence there exists $2^4=16$ bit strings. Is my answer true?

Update: I mean $2^6+2^6-2^4=112$ bit strings

I lost your logic. By symmetry, amount of strings with exactly 2 ones in the first four (call this group $F$) is identical to the ones with exactly 2 ones in the last four (call this group $L$).

Then your desired amount is $|F| + |L| - |F \cap L|$.

To compute $F$, note that you have exactly $\binom{4}{2} = 6$ ways to pick the location of the ones in the first four, which fixes your picks to be ones and the other two of the first four to be zeros. In other words, this fixes the first four bits, and the rest can be manipulated in $2^4=16$ ways.

Can you finish computing $|F|$? We already said $|L|=|F|$ and can you compute $|F \cap L|$ by a similar technique?

• I still belive that it should be $2^6+2^6-2^4=112$ ways.
– user530832
Feb 14, 2018 at 16:40
• @M.Rasmussen your mistake is not automatically setting the other 2 bits to zero. The way you compute, consider, for example, the first group, with exactly 2 ones in the first four bits. $2^6$ you are getting counts the number of bit strings with 6 places, but (a) you did not count that there are multiple ways in spreading the ones among the first four and (b) you are including the string 11111111 as legal, whereas you must have exactly 2 ones in the first four, and this string has 4. Feb 14, 2018 at 17:25
• Ohh okay. I see, so all other "empty" spaces must be 0 if you know what I mean.
– user530832
Feb 14, 2018 at 17:49
• Update: I get 156 now. But to make sure I understand it, what if I change the problem to this: "How many bit strings of length $12$ have either exactly four $1$-bit among the first $6$ bits or exactly four $1$-bit among the last $6$ bits?" Then I get the answer: 122655 is that correct? I did the same proces like the original question.
– user530832
Feb 14, 2018 at 17:57
• @M.Rasmussen Then both elementary groups are $|F| = |L| = \binom{6}{4} \cdot 2^6$ and their intersection is $|F \cap L| = \binom{6}{4} \binom{6}{4}$ so you end up with $$|F| + |L| - |F\cap L| = 2 \cdot \binom{6}{4} \cdot 2^6 - \binom{6}{4}^2 = 960 - 225 = 715.$$ Feb 14, 2018 at 21:45

Let $A$ be the set of bit strings with exactly two $1$-bit among the first $4$ bits, and $B$ be the set of bit strings with exactly two $1$-bit among the last $4$ bits.

\begin{align} \#A &= \binom{4}{2} 2^4 = 6\cdot2^4 \\ \#B &= 2^4 \binom{4}{2} = 6\cdot2^4 \\ \#A\cap B &= \binom{4}{2}^2 = 6^2 \\ \#A\cup B &= \#A + \#B - \# A \cap B \\ &= 6 (2^4 \cdot 2 - 6) \\ &= 6 \cdot 26 = 156 \end{align}

Among the first $4$ bits, choose $2$ to set them to one and the other two would be set to $0$ and there are $4$ of them with absolute freedom.

$$\binom{4}{2}\cdot 2^4$$

Similar when we focus on the last $4$ bits.

When we focus on intersection. We would pick $2$ from the first $4$ and pick $2$ from the last $4$.

So my overall answer would be

$$2\binom42 \cdot 2^4 - \binom42^2$$

Remark: I think your mistake is thinking that you can set arbitary $6$ bits to anything.

• I see many of you get 156, but I just think on Inclusion–exclusion principle and the way I did it here was drawing the problem: i.stack.imgur.com/jYOXA.png
– user530832
Feb 14, 2018 at 16:38
• The keyword is exactly $2$ ones in the first $4$ bits, which means exactly $2$ ones and exactly $2$ zeros in the first $4$ bits. Feb 14, 2018 at 16:59
• Update: I get 156 now. But to make sure I understand it, what if I change the problem to this: "How many bit strings of length 12 have either exactly four 1-bit among the first 6 bits or exactly four 1-bit among the last 6 bits?" Then I get the answer: 122655 is that correct? I did the same proces like the original question
– user530832
Feb 14, 2018 at 17:58
• @M.Rasmussen: There are not even that many 12 bit strings period. There are only 4096 strings of length 12 so how could there possibly be 122655 of them that have some property? You need to get in the habit of checking your answers for reasonableness. Feb 14, 2018 at 20:55
• @M.Rasmussen would you like to show the working of how you obtain that number? Feb 14, 2018 at 21:58

How many bit strings of length 8 have either exactly two 1-bit among the first 4 bits or exactly two 1-bit among the last 4 bits?

The simplest solution is the best. There are only 256 possibilities to check so just list them!

00000011 00000101 00000110 00001001 00001010 00001100 00010011 00010101
00010110 00011001 00011010 00011100 00100011 00100101 00100110 00101001
00101010 00101100 00110000 00110001 00110010 00110011 00110100 00110101
00110110 00110111 00111000 00111001 00111010 00111011 00111100 00111101
00111110 00111111 01000011 01000101 01000110 01001001 01001010 01001100
01010000 01010001 01010010 01010011 01010100 01010101 01010110 01010111
01011000 01011001 01011010 01011011 01011100 01011101 01011110 01011111
01100000 01100001 01100010 01100011 01100100 01100101 01100110 01100111
01101000 01101001 01101010 01101011 01101100 01101101 01101110 01101111
01110011 01110101 01110110 01111001 01111010 01111100 10000011 10000101
10000110 10001001 10001010 10001100 10010000 10010001 10010010 10010011
10010100 10010101 10010110 10010111 10011000 10011001 10011010 10011011
10011100 10011101 10011110 10011111 10100000 10100001 10100010 10100011
10100100 10100101 10100110 10100111 10101000 10101001 10101010 10101011
10101100 10101101 10101110 10101111 10110011 10110101 10110110 10111001
10111010 10111100 11000000 11000001 11000010 11000011 11000100 11000101
11000110 11000111 11001000 11001001 11001010 11001011 11001100 11001101
11001110 11001111 11010011 11010101 11010110 11011001 11011010 11011100
11100011 11100101 11100110 11101001 11101010 11101100 11110011 11110101
11110110 11111001 11111010 11111100

There are 156 such strings.

Do you still believe there are only 112? If so, then which ones did I either list twice, or list in error?

• I am convinced, thank you
– user530832
Feb 15, 2018 at 13:13

As a programmer (and a relatively naive mathematician) I immediately thought of reducing the 8-bit string to a hexadecimal string. Two digits, each made of four bits exactly as the problem is divided.

And, out 16 digits (0-F), 6 of them are valid (exactly 2 1-bits) digits.

So, we have 16×6 valid numbers (the first digit can be anything) and out of the the remaining 16×10 numbers, 6×10 of those will also be valid.

Now we have 16×6 + 6×10 = 156


# GCF of 28 and 32

The gcf of 28 and 32 is the largest positive integer that divides the numbers 28 and 32 without a remainder. Spelled out, it is the greatest common factor of 28 and 32. Here you can find the gcf of 28 and 32, along with a total of three methods for computing it. In addition, we have a calculator you should check out. Not only can it determine the gcf of 28 and 32, but also that of three or more integers including twenty-eight and thirty-two for example. Keep reading to learn everything about the gcf (28,32) and the terms related to it.

## What is the GCF of 28 and 32

If you just want to know what is the greatest common factor of 28 and 32, it is 4. Usually, this is written as

gcf(28,32) = 4

The gcf of 28 and 32 can be obtained like this:

• The factors of 28 are 28, 14, 7, 4, 2, 1.
• The factors of 32 are 32, 16, 8, 4, 2, 1.
• The common factors of 28 and 32 are 4, 2, 1, intersecting the two sets above.
• In the intersection factors of 28 ∩ factors of 32 the greatest element is 4.
• Therefore, the greatest common factor of 28 and 32 is 4.

Taking the above into account you also know how to find all the common factors of 28 and 32, not just the greatest. In the next section we show you how to calculate the gcf of twenty-eight and thirty-two by means of two more methods.

## How to find the GCF of 28 and 32

The greatest common factor of 28 and 32 can be computed by using the least common multiple aka lcm of 28 and 32. This is the easiest approach:

gcf (28,32) = $\frac{28 \times 32}{lcm(28,32)} = \frac{896}{224}$ = 4

Alternatively, the gcf of 28 and 32 can be found using the prime factorization of 28 and 32:

• The prime factorization of 28 is: 2 x 2 x 7
• The prime factorization of 32 is: 2 x 2 x 2 x 2 x 2
• The prime factors and multiplicities 28 and 32 have in common are: 2 x 2
• 2 x 2 is the gcf of 28 and 32
• gcf(28,32) = 4

In any case, the easiest way to compute the gcf of two numbers like 28 and 32 is by using our calculator below. Note that it can also compute the gcf of more than two numbers, separated by a comma. For example, enter 28,32. The calculation is conducted automatically.

The gcf is...
Frequently searched terms on our site also include:

## Use of GCF of 28 and 32

What is the greatest common factor of 28 and 32 used for? Answer: It is helpful for reducing fractions like 28 / 32. Just divide the nominator as well as the denominator by the gcf (28,32) to reduce the fraction to lowest terms.

$\frac{28}{32} = \frac{\frac{28}{4}}{\frac{32}{4}} = \frac{7}{8}$.

## Properties of GCF of 28 and 32

The most important properties of the gcf(28,32) are:

• Commutative property: gcf(28,32) = gcf(32,28)
• Associative property: gcf(28,32,n) = gcf(gcf(32,28),n) $\hspace{10px}n\hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$

The associativity is particularly useful to get the gcf of three or more numbers; our calculator makes use of it.

To sum up, the gcf of 28 and 32 is 4. In common notation: gcf (28,32) = 4.

If you have been searching for gcf 28 and 32 or gcf 28 32 then you have come to the correct page, too. The same is the true if you typed gcf for 28 and 32 in your favorite search engine.

Note that you can find the greatest common factor of many integer pairs including twenty-eight / thirty-two by using the the search form in the sidebar of this page.

Questions and comments related to the gcf of 28 and 32 are really appreciated. Use the form below or send us a mail to get in touch.

Please hit the sharing buttons if our article about the greatest common factor of 28 and 32 has been useful to you, and make sure to bookmark our site.

Posted in Greatest Common Factor
###### 2 comments on “GCF of 28 and 32”
1. POOFA says:

WOW! THIS IS A GREAT ANSWER! THANK YOU! 🙂

2. POOFA says:

WOW! THIS IS A GREAT ANSWER! THANK YOU! 🙂

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# Seeking methods to solve $\int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx$

As part of going through a set of definite integrals that are solvable using the Feynman Trick, I am now solving the following:

$$\int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx$$

I'm seeking methods using the Feynman Trick (or any method for that matter) that can be used to solve this definite integral.

If one wishes to use "Feynman's Trick," then begin by defining a function $$I(a)$$, $$a>1$$ as given by

$$I(a)=\int_0^{\pi/2}\log(a+\tan^2(x))\,dx \tag1$$

Differentiation of $$(1)$$ reveals

\begin{align} I'(a)&=\int_0^{\pi/2} \frac{1}{a+\tan^2(x)}\,dx\\\\ &=\frac{\pi/2}{a-1}-\frac{\pi/2}{\sqrt a (a-1)}\tag2 \end{align}

Integration of $$(2)$$ yields

\begin{align} I(a)&=\frac\pi2\left(\log(a-1)+\log\left(\frac{\sqrt a+1}{\sqrt{a}-1}\right) \right)\\\\ &=\pi \log(\sqrt a+1)\tag3 \end{align}

Finally, setting $$a=2$$ in $$(3)$$, we obtain the coveted result

$$\int_0^{\pi/2}\log(2+\tan^2(x))\,dx=\pi \log(\sqrt 2+1)$$

• How did you resolve the constant of Integration in Step (3)? – user150203 Nov 21 '18 at 4:38
• @davidg Apology for the sign error. Note $I(0)=0$. – Mark Viola Nov 21 '18 at 5:08
• Hi Mark ! Long time no speak. Nice solution $\to +1$. Cheers. – Claude Leibovici Nov 21 '18 at 6:22
• @MarkViola do you know of any other ‘tricks that would work? – user150203 Nov 21 '18 at 6:51
• Hi David. You could try writing the logarithm as $\log(1+\cos^2(x))-\log(\sin^2(x))$ and expanding the first term as $\sum_{n=1}^\infty \frac{(-1)^{n-1}\cos^{2n}(x)}{n}$ and proceeding. I haven't tried this, but it might be worth pursuing. – Mark Viola Nov 21 '18 at 15:47

My approach

Let

\begin{align} \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx &= \int_{0}^{\frac{\pi}{2}} \ln\left|1 + \left(1 + \tan^2(x)\right) \right| \:dx \\ &= \int_{0}^{\frac{\pi}{2}} \ln\left|1 + \sec^2(x) \right| \:dx \\ &= \int_{0}^{\frac{\pi}{2}} \ln\left|\frac{\cos^2(x) + 1}{\cos^2(x)} \right| \:dx \\ &= \int_{0}^{\frac{\pi}{2}} \left[ \ln\left|\cos^2(x) + 1 \right| - \ln\left|\cos^2(x)\right| \right]\:dx \\ &= \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx - \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx \end{align}

Now

$$\int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx = 2\int_{0}^{\frac{\pi}{2}} \ln\left|\cos(x)\right|\:dx = 2\cdot-\frac{\pi}{2}\ln(2) = -\pi \ln(2)$$

For detail on this definite integral, see guidance here

We now need to solve

$$\int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx$$

Here, Let

$$I(t) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + t \right|\:dx$$

Thus,

$$\frac{dI}{dt} = \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos^2(x) + t}\:dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{\cos(2x) + 1}{2} + t}\:dx = 2\int_{0}^{\frac{\pi}{2}} \frac{1}{\cos(2x) + 2t + 1}\:dx$$

Employ a change of variable $$u = 2x$$:

$$\frac{dI}{dt} = \int_{0}^{\pi} \frac{1}{\cos(u) + 2t + 1}\:du$$

Employ the Weierstrass substitution $$\omega = \tan\left(\frac{u}{2} \right)$$:

\begin{align} \frac{dI}{dt} &= \int_{0}^{\infty} \frac{1}{\frac{1 - \omega^2}{1 + \omega^2} + 2t + 1}\:\frac{2}{1 + \omega^2}\cdot d\omega \\ &= \int_{0}^{\infty} \frac{1}{t\omega^2 + t + 1} \:d\omega \\ &= \frac{1}{t}\int_{0}^{\infty} \frac{1}{\omega^2 + \frac{t + 1}{t}} \:d\omega \\ &= \frac{1}{t}\left[\frac{1}{\sqrt{\frac{t+1}{t}}}\arctan\left( \frac{\omega}{\sqrt{\frac{t+1}{t}}}\right)\right]_{0}^{\infty} \\ &= \frac{1}{t}\frac{1}{\sqrt{\frac{t+1}{t}}}\frac{\pi}{2} \\ &= \frac{1}{\sqrt{t}\sqrt{t + 1}}\frac{\pi}{2} \end{align}

And so,

$$I(t) = \int \frac{1}{\sqrt{t}\sqrt{t + 1}}\frac{\pi}{2}\:dt = \pi\ln\left| \sqrt{t} + \sqrt{t + 1}\right| + C$$

Now

$$I(0) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 0 \right|\:dx = -\pi \ln(2) = \pi\ln\left|\sqrt{0} + \sqrt{0 + 1} \right| + C \rightarrow C = -\pi \ln(2)$$

And so,

$$I(t) = \pi\ln\left| \sqrt{t} + \sqrt{t + 1}\right| -\pi \ln(2) = \pi\ln\left|\frac{\sqrt{t} + \sqrt{t + 1}}{2} \right|$$

Thus,

$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx = \pi\ln\left|\frac{\sqrt{1} + \sqrt{1 + 1}}{2} \right| = \pi\ln\left|\frac{1 + \sqrt{2}}{2} \right|$$

And Finally

\begin{align} \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx &= \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx - \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx \\ &= \pi\ln\left|\frac{1 + \sqrt{2}}{2} \right| - \left(-\pi \ln(2)\right) \\ &= \pi\ln\left|1 + \sqrt{2} \right| \end{align}

• Your question and your answer have a gap of just 1 minute. If you did solve the question earlier then you must've included your approach in the question itself. What is the need to post it as an answer? – Rohan Shinde Nov 21 '18 at 3:56
• @Digamma - Sorry, although I've been a member of Math StackExchange for 4 years, I've only become regularly active recently. When I asked a friend who is a member as to what is the best way of presenting a solution whilst asking for other solutions I was told this is the approach taken within the community. If this violate any of the rules, please advise and I will reposition accordingly. Also, I never assign my solution as 'the solution' to any post of this nature. – user150203 Nov 21 '18 at 4:00
• @Digamma The site encourages people to share the questions with answers. – Tianlalu Nov 21 '18 at 4:03
• @Tianlalu - Sorry, just to be clear - Have I employed the correct process here? – user150203 Nov 21 '18 at 4:10
• Thanks @mick - much appreciated :-) – user150203 Dec 3 '18 at 3:42


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# If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II)

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If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink]

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If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)?

(A) 3
(B) 4
(C) 5
(D) 21/4
(E) 7

(Still high on mods! Next week, will make questions on some other topic.)
[Reveal] Spoiler: OA

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Last edited by Bunuel on 06 Jul 2013, 02:33, edited 1 time in total. Added the OA. CEO Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2783 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Re: Question of the Day - II [#permalink] ### Show Tags 29 Oct 2010, 18:51 7 This post received KUDOS VeritasPrepKarishma wrote: Q. If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)? (A) 3 (B) 4 (C) 5 (D) 21/4 (E) 7 (Still high on mods! Next week, will make questions on some other topic.) put x=0 we get f(x) = 5..rule out D and E put x=1/4 we get f(x) = 0 + 11/4 + 5/4 = 16/4 = 4.. rule out C D and E Now the answer is either 3 or 4. Reason for above checking of values: for every value of x> 3 the f(x) is quite big because of 4x-1 for every value of x < -1 the f(x) if bigger than 3 and 4. for x >1/4 and x< 3 f(x) is bigger than 4. because f(x) = 4x-1 + 3-x + x+1 = 4x+3 > 3 => only 4 is the probable answer. Thus we only need to check x>=-1 and x<= 1/4 f(x) in this domain is = 1-4x + 3-x + x+1 = 5-4x => for f(x) to be minimum the x should be +ve => for x = 1/4 , f(x) = 4. This can be solved using graph as well by plotting the f(x) in different domains. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html Retired Moderator Joined: 02 Sep 2010 Posts: 803 Location: London Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 01:59 10 This post received KUDOS 1 This post was BOOKMARKED Let g(x) = |4x - 1| & h(x)=|x-3| + |x + 1| ... So f(x)=g(x)+h(x) Now h(x) is equal to 4 between -1 and 3 and is higher everywhere else, so it minimizes between -1 & 3 g(x) is minimum at x=(1/4) where it is equal to 0. Since -1<(1/4)<3 & f(x)=g(x)+h(x) ... It is straight forward to imply that f(x) will be minimum at 1/4, since both g(x) & h(x) are minimum at that point f(1/4)=g(1/4)+h(1/4)=0+4 Hence answer is 4 _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 08:53 45 This post received KUDOS Expert's post 13 This post was BOOKMARKED That's right! I can count on you guys to always give the correct answer and some innovative ways of solving... Let me give you the method I would use to solve it too... Where mods are concerned, the fact that mod signifies the distance from 0 on the number line is my best friend... So |x-3| is the distance from 3, |x + 1| is the distance from -1 and |4x - 1| is 4 times the distance from 1/4 So |4x - 1| + |x-3| + |x + 1| is the sum of distances from 3, -1 and four times the distance from 1/4 e.g. if x takes the value at point A, f(x) will be sum of length of red line, green line and blue line. the question here is, what is the minimum such sum possible? Attachment: Ques.jpg [ 4.56 KiB | Viewed 14954 times ] Can I say that minimum total distance will be covered from point 1/4? Attachment: Ques1.jpg [ 3.43 KiB | Viewed 14937 times ] The logic being that the distance from 3 to -1, which is 4 units has to be covered. Why to cover any distance from 1/4 at all? _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Question of the Day - II [#permalink]

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30 Oct 2010, 08:57
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In other words, as my mentor says, assume there is one guy on point 3, one on point -1 and 4 guys on point 1/4. If they have to meet up, but cover minimum distance, they should meet at point 1/4. What happens when there are 4 such points? Lets add another term (2x - 3) to f(x)...
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 11 Jul 2010 Posts: 224 Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 09:29 would the answer still remain 1/4? based on your 'distance' method? The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 16:17 3 This post received KUDOS Expert's post 3 This post was BOOKMARKED gmat1011 wrote: would the answer still remain 1/4? based on your 'distance' method? The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance Yes, that is right! The answer still remains 1/4. Since |2x - 3| = 2|x - 3/2| it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at -1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4. Attachment: Ques.jpg [ 5.2 KiB | Viewed 14965 times ] Try some other combinations. e.g. f(x) = |x - 1| + |x-3| + |x + 1| + |x + 6| f(x) = |2x - 3| + |4x + 7| etc _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Question of the Day - II [#permalink]

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09 Nov 2010, 23:23
Great Method Karishma !!

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Re: Question of the Day - II [#permalink]

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14 Jun 2011, 02:14
for x = 0, f(x) = 5. POE options C,D and E.
for x = 1/4, f(x) = 4.
B it is.
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Re: Question of the Day - II [#permalink]

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14 Jun 2011, 03:47
Very good question thanks for posting
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Re: Question of the Day - II [#permalink]

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15 Mar 2012, 11:07
VeritasPrepKarishma wrote:
gmat1011 wrote:

The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance

Yes, that is right! The answer still remains 1/4.
Since |2x - 3| = 2|x - 3/2| it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at -1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4.
Attachment:
Ques.jpg

Try some other combinations. e.g. f(x) = |x - 1| + |x-3| + |x + 1| + |x + 6|
f(x) = |2x - 3| + |4x + 7| etc

Karishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ?
In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?

Please correct me if I am wrong.
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Karishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ?
In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?

Please correct me if I am wrong.

Forget these numbers. Think logically.

My house is 10 miles away from your house. If we have to meet up, how much distance do we need to cover together? In any case, we need to cover 10 miles together at least, right? Either you come down to my place (you cover 10 miles) or I come down to yours (I cover 10 miles) or we meet mid way (10 miles covered together) or we meet up at a nice coffee place 2 miles further down from my house in the opposite direction in which case we will need to cover more than 10 miles (i.e. we cover 2 + 12 = 14 miles)

Now say, another friend is at my place. In which case will people cover minimum distance together? If two of us come down to your place, we cover 10+10 = 20 miles together but if you come down to our place, you cover only 10 miles. If instead, we meet midway, we cover 5+5 and you cover 5 miles so in all 15 miles. So less number of people should travel the entire distance.
If there are 4 people at point A and 2 at point B, minimum distance will be covered if people at point B travel to point A. So people at 3/2 should come down to 1/4.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 15 Mar 2012, 11:33 5 This post received KUDOS Expert's post 2 This post was BOOKMARKED ficklehead wrote: In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ? Please correct me if I am wrong. x is that point on the number line whose sum of distances from -6, -1, 1 and 3 is minimum. So basically there is a person each at points -6, -1, 1 and 3. You need to make them all meet by covering minimum distance. Distance between -6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: -6, -1, 0, 1 or 3 etc they will cover a distance of 9 together. If -1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at -1 travels down to 1 and -6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet. If they instead meet at -1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between -1 and 1, the total distance covered will be 11. To check, put x = 1. you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 put x = -1, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 put x = 0, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Question of the Day - II [#permalink]

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15 Mar 2012, 14:48
VeritasPrepKarishma wrote:
In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?

Please correct me if I am wrong.

x is that point on the number line whose sum of distances from -6, -1, 1 and 3 is minimum. So basically there is a person each at points -6, -1, 1 and 3. You need to make them all meet by covering minimum distance.
Distance between -6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: -6, -1, 0, 1 or 3 etc they will cover a distance of 9 together.
If -1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at -1 travels down to 1 and -6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet.
If they instead meet at -1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between -1 and 1, the total distance covered will be 11.

To check, put x = 1. you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11
put x = -1, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11
put x = 0, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11

Thanks Karishma for this detailed explanation.
I got it now.
I was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11.

To seal the concept, for |2x-3|+|4x+7|, minimum distance be : 3/2+7/4=13/4 ?
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I was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11.

To seal the concept, for |2x-3|+|4x+7|, minimum distance be : 3/2+7/4=13/4 ?

Look at the diagram below. Make a number line in such questions.
Attachment:

Ques3.jpg [ 5.1 KiB | Viewed 13919 times ]

For x and y to meet, they have to cover a distance of 9 together. For p and q to meet, they have to cover a distance of 2 together. They can meet anywhere between -1 and 1 and they will cover a total distance of 11 only. So x can take any value -1 < x < 1 and the value of the expression will be 11.

|2x-3|+|4x+7| = 2|x-3/2| + 4|x+7/4|

Attachment:

Ques4.jpg [ 5.35 KiB | Viewed 13934 times ]

There are 4 people at -7/4 and 2 people at 3/2. Distance between the two points is 7/4 + 3/2 = 13/4
For these people to meet covering the minimum distance, the 2 people X and Y should travel to point -7/4. (Make minimum people travel). So minimum distance that needs to be covered = 2*13/4 = 13/2 (because 2 people travel 13/4 each) which is the minimum value of the expression. The value of x when the expression takes minimum value is -7/4.
Check by putting x = -7/4. You get |2x-3|+|4x+7| = 13/2

Also see that when you put x = 0 or 3/2 etc, the value of the expression is higher.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 24 Jul 2011 Posts: 13 Re: Question of the Day - II [#permalink] ### Show Tags 16 Mar 2012, 09:01 Thanks a lot, Karishma. Intern Joined: 24 Jul 2011 Posts: 13 Re: Question of the Day - II [#permalink] ### Show Tags 16 Mar 2012, 19:48 I am wondering how can this method be used in questions where there are negative between terms : Ex: minimum value of : |x+6|-|x-1| ? Manager Joined: 28 Jul 2011 Posts: 238 Re: Question of the Day - II [#permalink] ### Show Tags 20 Mar 2012, 14:33 I got C = 5 f(x) = |4x-1| + |x-3| + |x+1| =(4x-1) + (x-3) + (x+1) or = - (4x-1) - (x-3) - (x+1) so, x= 1/2 then i used the value of x=1/2 f(x) = |4x-1| + |x-3| + |x+1| f(1/2) = |4(1/2)-1| + |(1/2)-3| + |(1/2)+1| = |1| + |-5/2| + |3/2| = 1 + 5/2 + 3/2 = 5 Is that correct? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 20 Mar 2012, 21:07 kuttingchai wrote: I got C = 5 f(x) = |4x-1| + |x-3| + |x+1| =(4x-1) + (x-3) + (x+1) or = - (4x-1) - (x-3) - (x+1) so, x= 1/2 then i used the value of x=1/2 f(x) = |4x-1| + |x-3| + |x+1| f(1/2) = |4(1/2)-1| + |(1/2)-3| + |(1/2)+1| = |1| + |-5/2| + |3/2| = 1 + 5/2 + 3/2 = 5 Is that correct? Put x = 1/4 and the minimum value you will get is 4. How did you get x = 1/2? I would suggest you to check out one of the approaches mentioned above. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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I am wondering how can this method be used in questions where there are negative between terms :
Ex: minimum value of : |x+6|-|x-1| ?

You can do it with a negative sign too.
You want to find the minimum value of (distance from -6) - (distance from 1)

Make a number line with -6 and 1 on it.
(-6)..........................(1)

Think of a point in the center of -6 and 1. Its distance from -6 is equal to distance from 1 and hence (distance from -6) - (distance from 1) = 0 .

What if instead, the point x is at -6? Distance from -6 is 0 and distance from 1 is 7 so (distance from -6) - (distance from 1) = 0 - 7 = -7

If you keep moving to the left, (distance from -6) - (distance from 1) will remain -7 so the minimum value is -7.
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# Computing $\int\limits_0^\infty x \left \lfloor{\frac1x}\right \rfloor \, dx$

This is an integral I computed but can't find the result online or on wolfram. So here's a proof sketch, please indulge this sanity check:

$$\int_0^\infty x \left \lfloor{\frac1x}\right \rfloor \ dx = \int_0^1 x \left \lfloor{\frac1x}\right \rfloor \ dx$$ $$= \sum_{n=1}^\infty \int_{1/(n+1)}^{1/n} nx \ dx =\sum_{n=1}^\infty\frac n2 \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$$ $$= \sum_{n=1}^\infty\frac n2 \left(\frac{2n+1}{n^2(n+1)^2}\right)$$ $$= \sum_{n=1}^\infty\frac{1}{(n+1)^2} + \frac12 \sum_{n=1}^\infty \frac{1}{n(n+1)^2}$$ $$= \frac{\pi^2}{6} -1 + \frac12\left(\sum_{n=1}^\infty \frac1n - \frac{1}{n+1} - \frac{1}{(n+1)^2}\right)$$ $$=\frac{\pi^2}{6} -1 + \frac12\left(\sum_{n=1}^\infty \frac1n - \frac{1}{n+1}\right) -\frac12\left(\sum_{n=1}^\infty\frac{1}{(n+1)^2}\right)$$ $$= \left(\frac{\pi^2}{6} -1\right) + \left(\frac12\cdot 1\right) - \frac12\left(\frac{\pi^2}{6} -1\right)$$ $$= \frac{\pi^2}{12}.$$

Basically, I used the Basel sum several times, and the fifth line follows from a partial sum decomposition. The seventh follows from the known result for the Basel sum, as well as the fact that the first series in the 6th line telescopes.

I hope this is all correct.

• @XanderHenderson For $x>1$, floor of $\frac{1}{x}$ is $0$ – BallBoy Dec 28 '17 at 3:34
• @XanderHenderson What you wrote is certainly not true, because you forgot the $x$ term. However, what Y. Forman says is correct, and I should've been more explicit there. – David Bowman Dec 28 '17 at 3:37
• Oi... derp. Sorry for being dyslexic. I missed the $x$. – Xander Henderson Dec 28 '17 at 3:39
• $$\int_0^1 x \lfloor 1/x \rfloor dx = \int_1^\infty \frac{1}{t} \lfloor t \rfloor \frac{dt}{t^2}= \sum_{n=1}^\infty \int_n^\infty t^{-3}dt = \sum_{n=1}^\infty \frac{n^{-2}}{2} = \frac{\zeta(2)}{2}$$ – reuns Dec 28 '17 at 3:43
• @reuns Nice, better post it as answer! – samjoe Dec 28 '17 at 4:40

Use

$$n\left(\frac1{n^2}-\frac1{(n+1)^2}\right)=\frac n{n^2}-\frac{n+1-1}{(n+1)^2}=\frac1n-\frac1{n+1}+\frac1{(n+1)^2}.$$

The first two terms do telescope and the Basel series remains.

• @stressedout: of course, but much simpler. And also simpler than yours, which is also essentially the OP's. – Yves Daoust Sep 5 '18 at 6:28
• @stressedout: your solution isn't different, just longer. A one-liner is simpler. – Yves Daoust Sep 5 '18 at 6:38

Alternatively, one may follow the same line of thought and use summation by parts formula to calculate the infinite sum. Similarly, $$\int_0^{+\infty} x\lfloor \frac{1}{x}\rfloor dx =\sum_{n=1}^{\infty}\frac{n}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)=-\frac{1}{2}\sum_{n=1}^{\infty}f_n(g_{n+1}-g_n)$$

where $f_n = n$ and $g_n = 1/n^2$. Now, summation by parts for the last expression gives:

$$-\frac{1}{2}\sum_{n=1}^{\infty}f_n(g_{n+1}-g_n) = -\frac{1}{2}\left(\ \lim_{n\to\infty}\frac{n}{(n+1)^2} -1 -\sum_{n=2}^{\infty}\frac{1}{n^2}\right)$$

But it is well-known that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$

Hence,

$$-\frac{1}{2}\sum_{n=1}^{\infty}f_n(g_{n+1}-g_n) = -\frac{1}{2}\left(\ \lim_{n\to\infty}\frac{n}{(n+1)^2} -1 - (\frac{\pi^2}{6}-1)\right) = \frac{\pi^2}{12}$$


# Prove that each integer n ≥ 12 is a sum of 4's and 5's using strong induction

So I've been given the following problem: Prove that each integer n ≥ 12 is a sum of 4's and 5's What I have so far: (Basis):

n ≥ 12 Therefore,

12 ≤ 4(x) + 5(y)

x = 3 | y = 0

12 ≤ 4(3) + 5(0)

12 ≤ 12 = Correct

However, what I don't understand is how would I use the x & y variable in induction step. Where exactly would I place these? and how would I go on to solve this?

The trick to these sorts of problems is to realize that if we can find four consecutive integers ($$4$$ is the smallest of our numbers we're taking a combination of) that can be represented as a non-negative sum of the fours and fives.

For example $$12 = 4(3) + 5(0)$$

$$13 = 4(2) + 5(1)$$ $$14 = 4(1)+ 5(2)$$ $$15 = 4(0) + 5(3).$$

With this information is it clear how you could represent 16? Just take the solution from the 12 case, and add 4! (so increase $$x$$ by $$1$$.)

Here's the formal argument. Let $$P(n)$$ be the open sentence "$$n$$ can be written as a non-negative combination of $$4$$ and $$5$$". By what we've shown above, we know that $$P(k)$$ is true for $$k = 12,13,14,15.$$ We wish to prove that $$P(k+1)$$ is true. Our strong inductive hypothesis is to suppose that for some $$k \in \mathbb{Z}$$ that for every $$i$$ with $$12\leq i\leq k$$ that $$P(k)$$ is true, and we need to prove that $$P(k+1)$$ is true.

If $$k = 12,13$$ or $$14$$, we've already seen that $$p(k+1)$$ is also true, so suppose that $$k \geq 15$$. Then we know that $$12 \leq k-3\leq k$$ so then by our strong inductive hypothesis, there exists $$x,y \in \mathbb{Z}$$ such that $$k-3 = 4(x) + 5(y)$$. Then adding $$4$$ to both sides gives that $$k+1 = 4(x+1)+5(y)$$ so that $$P(k+1)$$ is true. Thus by the principle of mathematical induction, $$P(n)$$ must be true for all $$n \geq 12$$.

• Wow this is very detailed and helped me understand it more, thank you! – MathNoob Feb 15 '19 at 4:15
• No problem! I'm glad this helped! – JonHales Feb 15 '19 at 4:17

$$4x+5y=\underbrace{4(x-1)+5y}_{n-4}+4$$

So, if $$n-4$$ is expessible so will be $$n$$

$$12=4\cdot3$$

$$13=2\cdot4+5$$

$$14=4+2\cdot5$$

$$15=3\cdot5$$


# Finding dimension of a kernel from a linear transformation.

I'm trying to solve an exercise where I'm asked to find this linear transformation's kernel and its dimension:

$$f(x_1,x_2,x_3,x_4)=(x_1-x_2,2x_3+x_4)$$

So, from the definition of kernel and converting to parametric equations, I found $x_1=x_2=\lambda$, $x_3=-\frac{1}{2}\mu$ and $x_4=\mu$. Therefore, I assumed: $$\ker(f)=\{(\lambda,\lambda,-\frac{1}{2}\mu,\mu): \lambda,\mu \in \mathbb{R}\}$$

And since the kernel can be described with only one vector (with two parameters), I thought $\dim(\ker(f))=1$, but my textbook says it's actually equal to $2$. Where is my reasoning flawed?

• A basis for the kernel is $\{(1,1,0,0), (0,0,-\frac{1}{2}, 1) \}$ – leibnewtz Jan 9 '18 at 23:20
• @leibnewtz Hm, so that means I can only 'fix' one parameter at once? I thought it'd be possible to establish something like $\lambda=1$, $\mu=2$ so that a basis for the kernel is $\{(1,1,-1,2)\}$. Why isn't that possible? – Manuel Jan 9 '18 at 23:22
• Of course you can do what you say...but that way you'll only get one vector, and you already know you need two lin. ind. vectors to have a basis... – DonAntonio Jan 9 '18 at 23:30
• Since $f:\mathbb R^4\to\mathbb R^2$, its image is at most two-dimensional, so you know that its kernel must be at least two-dimensional. – amd Jan 9 '18 at 23:35
• @Manuel Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Feb 4 '18 at 0:09

It is all correct, the dimension of the ker(f) is 2 because you have 2 free parameters that define it $(\lambda,\mu)$.
• @Manuel They are equivalent concepts. As noted in the comments you can set $\lambda=1$ and $\mu=0$ and define a basis vector, then $\lambda=0$ and $\mu=1$ and define a second vector linearly independent from the first, thus the dimension is 2. This is true for any number of free parameter (EG a line or a plane in $\mathbb{R^3}$). – user Jan 9 '18 at 23:25
• @Manuel There are two vectors in your description of the kernel: $\left(\lambda,\lambda,-\frac12\mu,\mu\right)=\lambda(1,1,0,0)+\mu\left(0,0,-\frac12,1\right)$; every vector in the kernel is a linear combination of these two linearly-independent vectors. – amd Jan 9 '18 at 23:36


Is every group an ideal class group of a number field?

The inverse Galois problem asks whether every finite group appears as the Galois group of some finite extension of $\mathbb Q$. I was wondering to what extent the analogous problem for ideal class groups has been investigated. More precisely, consider the following question:

Is every finite abelian group the ideal class group of some number field (finite extension of $\mathbb Q$)?

I'd be interested to hear about any partial results, as I suppose this question is still open. I'd be also interested in any results about a weaker problem:

Is every positive integer the ideal class number of some number field?

Again, any reference, even to a partial result, will be appreciated.

• Not an answer to your question, but are you aware of Claborn's theorem that says every finite abelian group is the class group of a Dedekind domain (though not necessarily, as far as one can tell from Claborn's work, of a ring of integers)? – Steven Landsburg Oct 16 '16 at 19:52
• @StevenLandsburg I was not aware of that result, since, to be honest, I am not so interested in general Dedekind's domain. I will definitely take a look at this result though. – Wojowu Oct 16 '16 at 19:55
• duplicate: math.stackexchange.com/questions/10949/… – Franz Lemmermeyer Oct 16 '16 at 20:36
• @FranzLemmermeyer Thanks, I haven't seen that question when searching about the topic. I suppose at this point my question can be closed. – Wojowu Oct 16 '16 at 20:39
• The system does not allow closing a question on one site as a duplicate of a question on a different site. One option, Wojowu, is for you to post an answer here, summarizing what's over there, and linking to it, and then accept your answer. – Gerry Myerson Oct 16 '16 at 22:04

The recent paper by Homlin, Jones, Kurlberg, McLeman, and Petersen (Experimental Math., to appear) is devoted to these questions especially in the context of imaginary quadratic fields. One should expect that every natural number arises as the class number of an imaginary quadratic field. Refining an earlier conjecture of Soundararajan, in this paper a precise asymptotic is formulated for the number of imaginary quadratic fields with a given class number. They also formulate conjectures on what kind of groups can arise as class groups of imaginary quadratic fields -- for example, one expects that for any odd prime $p$, $({\Bbb Z}/p{\Bbb Z})^3$ is not the class group of any imaginary quadratic field. The paper gives much data on such questions together with many related references.

• I think $(\dfrac{\Bbb Z}{2\Bbb Z})^3$ can occur as a class group for $D=-420$, and maybe it is better to restrict $p$ to the odd primes, i.e: $p\neq2$. – Davood KHAJEHPOUR May 8 '18 at 5:51
• Thanks for the reference! I will look into it as soon as I can. – Wojowu May 8 '18 at 7:26

Just for the sake of completeness, here I am posting the answer of Pete Clark which was posted in the math.SE question. I am making this post CW so as not to create the unseemly impression that I am gaining any undeserved reputation from this. Here it is:

Virtually nothing is known about the question of which abelian groups can be the ideal class group of (the full ring of integers of) some number field. So far as I know, it is a plausible conjecture that all finite abelian groups (up to isomorphism, of course) occur in this way. Conjectures and heuristics in this vein have been made, but unfortunately for me I'm not so familiar with them.

The situation for imaginary quadratic fields is different. Here there is an absolute bound on the size of an integer $$k$$ such that the class group of an imaginary quadratic field can be isomorphic to $$(\mathbb{Z}/2\mathbb{Z})^k$$. Conditionally on the Generalized Riemann Hypothesis, the largest such $$k$$ is $$4$$. This has do to with idoneal numbers, of which the following paper provides a very fine survey:

http://www.mast.queensu.ca/~kani/papers/idoneal-f.pdf

Actually the truth is slightly stronger: let $$H_D$$ be the class group of the imaginary quadratic field $$\mathbb{Q}(\sqrt{-D})$$. Then, as $$D$$ tends to negative infinity through squarefree numbers, the size of $$2H_D$$ (the image of multiplication by $$2$$) tends to infinity. See for instance

http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.0358v2.pdf

for some recent explicit bounds on this.

For $S$-class groups: Perret, Marc On the ideal class group problem for global fields. Zbl 0933.11053 J. Number Theory 77, No. 1, 27-35 (1999). https://zbmath.org/?any=&au=perret&ti=On+the+Ideal+Class+Group+Problem+for+Global+Fields&so=&ab=&cc=&ut=&an=&la=&py=&rv=&sw=&dm=