text:
# Distribution of point estimator

I am given the model $Y_i=\alpha_0+\beta_0 X_i+\epsilon_0$, where $i=1,2,...,n$, $X_i$ are fixed numbers and $\epsilon \sim N(0, \sigma^2).$ I am also given that $\sigma^2$ and the parameters $(\alpha_0,\beta_0)$ for $E(Y_i)=\alpha_0+\beta_0 X_i$ are unknown.

$(\alpha^*,\beta^*)$ estimate $(\alpha_0,\beta_0)$ and are found by minimizing $\sum_{i=1}^n (Y_i-\alpha-\beta X_i)^2.$

We get that $\alpha^*=\widehat{Y}-\beta^* \widehat{X}$ and $\beta^*=\frac{\sum_{i=1}^n(X_i-\widehat{X})Y_i}{\sum_{i=1}^n(X_i-\widehat{X})^2}$, where $\widehat{Y}=\frac{Y_1+Y_2+...+Y_n}{n}$ and $\widehat{X}=\frac{X_1+X_2+...+X_n}{n}$.

I have to find the distribution for $\beta^*$. I know that it has normal distribution, thus $\beta^* \sim N(E[\beta^*], var(\beta^*))$.

I think $E[\beta^*]$ is equal to $\beta_0$, but I am not entirely sure. For $var(\beta^*)$, I think it has something to do with $\epsilon \sim N(0, \sigma^2)$. Since $var(\epsilon)=\sigma^2=\sum\frac{(X_i-\widehat{X})^2}{n-1}$, I was thinking that maybe I had to substitute this into $\beta^*=\frac{\sum_{i=1}^n(X_i-\widehat{X})Y_i}{\sum_{i=1}^n(X_i-\widehat{X})^2}$ to get $var(\beta^*)$ with respect to $\sigma^2$, but I am not sure how.

An explanation would be greatly appreciated. I am having a really hard time understanding even basic concepts of this topic (and class).

You are right that $E(\beta^*) = \beta_0.$ You can show this by computing $$E(\beta^*) = \frac{\sum_i(X_i-\widehat X) E(Y_i)}{\sum_i(X_i-\widehat X)^2}$$ and plugging in $E(Y_i) = \alpha_0 + \beta_0 X_i + E(\epsilon_i) = \alpha_0 + \beta_0 X_i$ and summing it up. (As a hint, note that $\sum_i (X_i-\widehat X) = 0$).
For the variance you have made the mistake of conflating the parameter $\sigma^2$ with its estimator $\frac{1}{n-1}\sum_i (\epsilon_i-\widehat \epsilon)^2.$ These are not equal (one is random and one isn't). (Furthermore you wrote $\sum_i (X_i -\widehat X)^2$, with $X$'s rather than with $\epsilon$'s and the sample variance of $X$ has nothing to do with $\sigma^2.)$
To get the variance of $\beta^*,$ you can use the formula $$Var(\beta^*) = E((\beta^*-E(\beta^*))^2).$$
You can use the formula for $\beta^*$ to write this as $$Var(\beta^*) = E\left[\left(\frac{\sum_i (X_i-\widehat X)(Y_i-E(Y_i))}{\sum_i(X_i-\widehat X)^2}\right)^2\right]=E\left[\left(\frac{\sum_i (X_i-\widehat X)\epsilon_i}{\sum_i(X_i-\widehat X)^2}\right)^2\right]$$ and it's just a matter of doing this sum using the fact that $E(\epsilon_i^2) = \sigma^2$ and $E(\epsilon_i\epsilon_j) = 0$ for $i\ne j,$
• Thank you for your very helpful response. For $E[\beta^*]$, if $\sum_i (X_i-\widehat X) = 0$, why wouldn't the whole summation be equal to 0 or have division by 0? – Silvia Rossi Nov 12 '17 at 2:50
• Cause that does not imply that $\sum_i (X_i-\widehat X)X_i = 0.$ In fact, it implies that $\sum_i (X_i-\widehat X)X_i = \sum_i (X_i-\widehat X)^2.$ – spaceisdarkgreen Nov 12 '17 at 3:03


text:
## Stream: general

### Topic: timeout when working with ideal of polynomial ring

#### Casper Putz (Aug 14 2019 at 14:17):

I was working with the following setup:

variables (α : Type*) [discrete_field α]
variables (I : ideal (polynomial α))

For some reason I get a deterministic timeout whenever I do anything with I. For example: this gives a timeout:

#check (I : submodule (polynomial α) (polynomial α))

But also writing down I.carrier or a membership statement p ∈ I gives a timeout. Working over an integral_domain α or a field α also causes the timeout. However, in the following setup I do not get a timeout:

variables (α : Type*) [comm_ring α] [division_ring α] [decidable_eq α]
variables (I : ideal (polynomial α))
#check I.carrier

(While a field is just a comm_ring and division_ring, so there must be some weird instance kicking in or something)

The problem is similar to the problem posted here: https://leanprover.zulipchat.com/#narrow/stream/113488-general/topic/deterministic.20timeout.20with.20ideal.2Equotient.2Emk.
I tried increasing the class instance depth, and tried to find out what was happening using trace.class_instances but I could not manage to find the problem. Anybody any idea of why this may happen (and how one could fix it)?

#### Rob Lewis (Aug 14 2019 at 20:20):

This is yet another type class issue. The temporary fix: instance mpp : module (polynomial α) (polynomial α) := by apply_instance

#### Kevin Buzzard (Aug 14 2019 at 20:36):

So what is happening here? Making mpp an instance makes type class inference behave differently? Basically it finds it earlier? Could one use priorities to somehow fix this?

#### Rob Lewis (Aug 14 2019 at 20:46):

The logs are migraine inducing, so I don't have an exact answer. But basically yeah, if mpp is the last instance added it's the first one tried, and the resolution path avoids some very costly search path. It's possible a local priority change could fix this instance, and I don't think we ever systematically adopted this idea from Floris (which we should): https://leanprover.zulipchat.com/#narrow/stream/113488-general/topic/more.20type.20class.20inference.20issues/near/167026105

#### Casper Putz (Aug 14 2019 at 21:06):

Thanks Rob for the quick fix!

#### Casper Putz (Aug 14 2019 at 21:11):

Ah that makes sense. I agree about the migraine haha. I saw it doing some weird stuff like using discrete_field.to_local_ring.to_comm_ring instead of directly using that a field is a commutative ring

#### Scott Morrison (Aug 14 2019 at 21:29):

Hi Casper, if you're working on polynomials, and struggling with timeouts, you might be interested in completing the work I started on the decidable_eq branch, stripping out all the decidability conditions from polynomials and multivariable polynomials. Unfortunately I haven't had time to finish, but I hope it's mostly downhill from where I got up to --- just fix the remaining errors!

#### Reid Barton (Aug 14 2019 at 21:55):

I didn't realize you had already started this, I was going to propose it as a possible fix as well.

#### Casper Putz (Aug 15 2019 at 08:38):

Hi Scott, I may actually do that! I think I will give it a look this weekend.

#### Scott Morrison (Sep 03 2019 at 05:08):

I made quite a bit more progress on this now, but still have problems in power_series.lean and free_comm_ring.lean. Any help (feel free to commit direct to the decidable_eq branch) much appreciated.

#### Johan Commelin (Sep 03 2019 at 05:11):

I can probably help with power_series, since I wrote it not too long ago.

#### Johan Commelin (Sep 03 2019 at 05:11):

What kind of problems are you seeing?

#### Johan Commelin (Sep 03 2019 at 07:10):

@Scott Morrison I already have errors for mv_polynomial.C_add on L52 of mv_polynomial.lean

#### Johan Commelin (Sep 03 2019 at 07:11):

For some reason Lean doesn't agree that C a is single 0 a, even though it's true by definition. If I then hit it with another refl it shuts up and moves on.

Stupid Lean...

#### Scott Morrison (Sep 03 2019 at 07:58):

Weird, that's working for me.

#### Johan Commelin (Sep 03 2019 at 08:08):

@Scott Morrison Where is the first error that you hit in power series?

L581?

#### Casper Putz (Sep 03 2019 at 08:10):

I will try to fix free_comm_ring.lean. Did a bit already

#### Scott Morrison (Sep 03 2019 at 10:00):

@Johan Commelin, the first error I see is in /src/ring_theory/power_series.lean at line 560:

/Users/scott/projects/lean/mathlib/src/ring_theory/power_series.lean:560:36: error: type mismatch at application
mul_inv_of_unit (1 - φ) ?m_1 h
term
h
has type
1 - coeff 0 φ = ↑u
but is expected to have type
coeff 0 (1 - φ) = ↑?m_1

#### Casper Putz (Sep 03 2019 at 10:31):

I fixed the errors in free_comm_ring.lean

#### Johan Commelin (Sep 03 2019 at 12:29):

The biggest source of errors in power_series.lean seems to come from the fact that single unit.star n is no longer defeq to n, for n : ℕ...

#### Johan Commelin (Sep 03 2019 at 12:30):

I'm not sure how that happened. Because this is a statement about unit and nat, which shouldn't be affected by polynomials going classical.

#### Mario Carneiro (Sep 03 2019 at 12:34):

basically you can't evaluate single anymore by defeq

But why not?

#### Mario Carneiro (Sep 03 2019 at 12:34):

there is a proof of n = n by classical in there

This is stupid

#### Mario Carneiro (Sep 03 2019 at 12:35):

it's something like single unit.star n := if n = n then 1 else 0

#### Mario Carneiro (Sep 03 2019 at 12:35):

where the if is no longer decidable because classical

#### Mario Carneiro (Sep 03 2019 at 12:35):

solution: stop using defeq

#### Mario Carneiro (Sep 03 2019 at 12:36):

use simp

#### Johan Commelin (Sep 03 2019 at 12:36):

You might as well say, "stop using definitions"

not at all

#### Johan Commelin (Sep 03 2019 at 12:36):

And simp is slow... we learnt that.

#### Mario Carneiro (Sep 03 2019 at 12:36):

use definitional equations

#### Mario Carneiro (Sep 03 2019 at 12:36):

use rw [single]

#### Johan Commelin (Sep 03 2019 at 12:37):

Yep, but this single is hidden behind 4 definitions...

#### Johan Commelin (Sep 03 2019 at 12:37):

And rfl used to blast through them

#### Mario Carneiro (Sep 03 2019 at 12:37):

that's not a good thing

#### Johan Commelin (Sep 03 2019 at 12:37):

And now I need to delta 5 things, and than I can rw finsupp.single_zero or something silly like that.

#### Mario Carneiro (Sep 03 2019 at 12:37):

Why are you unfolding 4 definitions?

#### Mario Carneiro (Sep 03 2019 at 12:38):

you should only ever need to unfold one at most at a time

#### Mario Carneiro (Sep 03 2019 at 12:38):

maybe two, but that's bad practice

4 is abuse

#### Johan Commelin (Sep 03 2019 at 12:38):

Because power_series.C is defined as mv_power_series.C which is defined as mv_power_series.monomial 0, and power_series.monomial is defined as mv_power_series.monomial, and I these definitions to commute by rfl.

#### Mario Carneiro (Sep 03 2019 at 12:39):

What are you proving?

#### Johan Commelin (Sep 03 2019 at 12:39):

@[simp] lemma coeff_C_zero (a : α) : coeff 0 (C a) = a :=

#### Mario Carneiro (Sep 03 2019 at 12:40):

the proof should be mv_power_series.coeff_C_zero

#### Johan Commelin (Sep 03 2019 at 12:40):

by { convert coeff_monomial' 0 a, delta C mv_power_series.C monomial, rw finsupp.single_zero }

#### Mario Carneiro (Sep 03 2019 at 12:40):

that's one definition unfold

#### Johan Commelin (Sep 03 2019 at 12:41):

Because single unit.star 0 is not defeq to 0

#### Mario Carneiro (Sep 03 2019 at 12:41):

I'm not 100% on the exact definitions of everything here, but you should be able to unfold things until it's a statement about mv_polynomial, and then apply the appropriate lemma

#### Johan Commelin (Sep 03 2019 at 12:42):

Nope, even that doesn't work

#### Johan Commelin (Sep 03 2019 at 12:43):

This definition might be relevant:

/-- The nth coefficient of a formal power series.-/
def coeff (n : ℕ) : power_series α → α := mv_power_series.coeff (single () n)

#### Johan Commelin (Sep 03 2019 at 12:43):

This used to be harmless, but now it is not

#### Johan Commelin (Sep 03 2019 at 12:43):

Because of that single () n tagging at the end

#### Johan Commelin (Sep 03 2019 at 12:44):

I can't understand why we can't have classical coefficients living together with a constructively decidable unit and nat

#### Mario Carneiro (Sep 03 2019 at 12:44):

I see this in the file now:

@[simp] lemma coeff_monomial' (n : σ →₀ ℕ) (a : α) :
coeff n (monomial n a) = a := if_pos rfl

@[simp] lemma coeff_C_zero (a : α) : coeff 0 (C a : mv_power_series σ α) = a :=
coeff_monomial' 0 a

are you saying that the two coeffs are different here?

#### Johan Commelin (Sep 03 2019 at 12:45):

No, those coeffs are the same

#### Mario Carneiro (Sep 03 2019 at 12:45):

I can't understand why we can't have classical coefficients living together with a constructively decidable unit and nat

That's called accepting a decidable_eq instance, and we tried that

#### Mario Carneiro (Sep 03 2019 at 12:45):

then what's the problem? they look defeq to me

#### Mario Carneiro (Sep 03 2019 at 12:45):

def C (a : α) : mv_power_series σ α := monomial 0 a

Yes...

#### Johan Commelin (Sep 03 2019 at 12:46):

But power_series.C is no longer defeq to power_series.monomial 0

#### Mario Carneiro (Sep 03 2019 at 12:46):

why? Did the definition change?

No

#### Mario Carneiro (Sep 03 2019 at 12:46):

It says that it is defined as exactly that

#### Johan Commelin (Sep 03 2019 at 12:46):

Someone removed a [decidable_eq α]. Em tasol.

#### Johan Commelin (Sep 03 2019 at 12:47):

Over here it says

/-- The constant formal power series.-/
def C : α → power_series α := mv_power_series.C

#### Mario Carneiro (Sep 03 2019 at 12:47):

oh, I was looking at the wrong C

#### Johan Commelin (Sep 03 2019 at 12:48):

I mean... I can fix these errors, but the barrier between mv_power_series unit and power_series just got a lot bigger. Because we don't have good definitional reductions anymore.

#### Mario Carneiro (Sep 03 2019 at 12:49):

I see, so the problem is that there is no theorem power_series.C a = monomial 0 a

#### Mario Carneiro (Sep 03 2019 at 12:49):

if you had that it would be a single rw

#### Johan Commelin (Sep 03 2019 at 12:50):

Aha... I guess that theorem was missing, because it used to be defeq...

#### Mario Carneiro (Sep 03 2019 at 12:51):

And the proof boils down to single () 0 = 0, which should be in finsupp

#### Mario Carneiro (Sep 03 2019 at 12:52):

single_zero

#### Johan Commelin (Sep 03 2019 at 12:55):

so why does decidable_eq cause problem afterall?

format: plain


text:
Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 06 Feb 2016, 18:34

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Victor's job requires him to complete a series of identical

Author Message
TAGS:
Senior Manager
Joined: 29 Jan 2011
Posts: 365
Followers: 0

Kudos [?]: 120 [3] , given: 87

Victor's job requires him to complete a series of identical [#permalink]  11 Sep 2011, 13:17
3
KUDOS
6
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

66% (04:19) correct 34% (03:06) wrong based on 225 sessions
Victor's job requires him to complete a series of identical jobs. If Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. If Victor works for 144 days and is supervised for half the time, he will finish a total of 36 jobs. How long would it take Victor to complete 10 jobs without any supervision?

A. 34
B. 52
C. 60
D. 70
E. 92
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Mar 2013, 01:55, edited 2 times in total.
Renamed the topic and edited the question.
Director
Joined: 01 Feb 2011
Posts: 758
Followers: 14

Kudos [?]: 85 [0], given: 42

Re: Victor's job [#permalink]  11 Sep 2011, 13:33
supervised not supervised

time per job t-3 t

In 144 days when victor was supervised for half of the time , he finished 36 jobs.

=> During 72 days of supervision he finished 36 jobs.

=> 36(t-3) = 72

=> t = 5

=> victor took 5 days to finish a each unsupervised job.

time taken by victor to finish 10 jobs with out supervision = 10t = 50
Senior Manager
Joined: 29 Jan 2011
Posts: 365
Followers: 0

Kudos [?]: 120 [0], given: 87

Re: Victor's job [#permalink]  11 Sep 2011, 13:39
Spidy001 wrote:
supervised not supervised

time per job t-3 t

In 144 days when victor was supervised for half of the time , he finished 36 jobs.

=> During 72 days of supervision he finished 36 jobs.

=> 36(t-3) = 72

=> t = 5

=> victor took 5 days to finish a each unsupervised job.

time taken by victor to finish 10 jobs with out supervision = 10t = 50

OA is 60 but.... there is some mistake with your solution
Director
Joined: 01 Feb 2011
Posts: 758
Followers: 14

Kudos [?]: 85 [1] , given: 42

Re: Victor's job [#permalink]  11 Sep 2011, 13:57
1
KUDOS
my bad. ignore the previous post. i thought the question said 36 jobs in 72 days of supervision.

Where as Questions says it took 36 jobs in 144 days out of which he worked halftime supervised.

Supervised not supervised
time per job t-3 days t days

=> jobs in 72 days of supervision + jobs in 72 days when not supervised = 36

=> 72/(t-3) + 72/t = 36

solving we get t = 1 or 6

=> t=6 ( t = 1 is not possible as t-3 would be negative)

=> time taken by victor each unsupervised job = 6 days.

=> time taken by victor to finish 10 unsupervised jobs = 10t = 60 days.
Manager
Status: Prepping for the last time....
Joined: 28 May 2010
Posts: 200
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 630 Q47 V29
GPA: 3.2
Followers: 0

Kudos [?]: 21 [1] , given: 21

Re: Victor's job [#permalink]  12 Sep 2011, 01:34
1
KUDOS
60 days it is

72/ (x-3) + 72/x = 36

=> x^2 -7x+6 =0 => x= 1 or 6

X = 1 is omitted coz x-3 = -2 if x=1

so x=6 and time taken for 10 jobs under no supervision = 10x6 = 60
_________________

Two great challenges: 1. Guts to Fail and 2. Fear to Succeed

Senior Manager
Joined: 03 Aug 2011
Posts: 279
GMAT 1: 640 Q44 V34
GMAT 2: 700 Q42 V44
GMAT 3: 680 Q44 V39
GMAT 4: 740 Q49 V41
GPA: 3.7
WE: Project Management (Energy and Utilities)
Followers: 11

Kudos [?]: 56 [0], given: 912

Re: Victor's job [#permalink]  15 Nov 2012, 23:21
Guys,

I've made th same error as Spidy001. This advanced RTW problem is otherwise not that hard, but I wonder what strategies you apply in order to fully grasp the right meaning of the problem. At first, I also thought the 5 days per job scenario was right. Based on this problem, do you have any tips on that?

_________________

Thank you very much for reading this post till the end! Kudos?

Intern
Joined: 15 Nov 2012
Posts: 1
Concentration: Technology, Strategy
GPA: 2.6
Followers: 0

Kudos [?]: 1 [1] , given: 0

Re: Victor's job [#permalink]  15 Nov 2012, 23:57
1
KUDOS
bgpower wrote:
Guys,

I've made th same error as Spidy001. This advanced RTW problem is otherwise not that hard, but I wonder what strategies you apply in order to fully grasp the right meaning of the problem. At first, I also thought the 5 days per job scenario was right. Based on this problem, do you have any tips on that?

bgpower,

I can imagine why Spidy001 made that mistake. The question says that he was supervised for half that time, but it doesn't say that Victor did not work during the time he was unsupervised.
So, what the question intended to say was, Victor worked under supervision for 72 days and without supervision for the remaining 72. What Spidy001 has done is, he hasn't taken into account the 72 days that Victor worked without supervision.

Secondly, as data given to us was in the form of (days/job) and not (jobs/day), you should divide 72 by (t-3) and (t) rather than multiplying it.

This sort of an error can be avoided by simply taking a step back to understand what the question says and writing the facts down, rather than trying to hold them in your head.
Current Student
Joined: 03 Aug 2012
Posts: 915
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Followers: 18

Kudos [?]: 477 [0], given: 322

Re: Victor's job requires him to complete a series of identical [#permalink]  23 Mar 2013, 22:00
siddhans wrote:
Victor's job requires him to complete a series of identical jobs. If Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. If Victor works for 144 days and is supervised for half the time, he will finish a total of 36 jobs. How long would it take Victor to complete 10 jobs without any supervision?

I have a query

R T D
S (1/t) 72 (72/t)
U (1/(t+3) 72 (72/(t+3))

Total jobs = 36

(72/t) + [(72/(t+3)] = 36

=> 2/t + 2/(t+3) = 1
=> 2t + 6 + 2t = t^2 + 3t
=> t^2 - t - 6 = 0

t = 3 or t =-2 (can't be negative)
Hence , 3

Plz advice where I am wrong as I took the other way around by taking (t+3)

Rgds,
TGC
_________________

Rgds,
TGC!
_____________________________________________________________________
I Assisted You => KUDOS Please
_____________________________________________________________________________

Math Expert
Joined: 02 Sep 2009
Posts: 31228
Followers: 5342

Kudos [?]: 62057 [0], given: 9427

Re: Victor's job requires him to complete a series of identical [#permalink]  24 Mar 2013, 02:07
Expert's post
targetgmatchotu wrote:
siddhans wrote:
Victor's job requires him to complete a series of identical jobs. If Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. If Victor works for 144 days and is supervised for half the time, he will finish a total of 36 jobs. How long would it take Victor to complete 10 jobs without any supervision?

I have a query

R T D
S (1/t) 72 (72/t)
U (1/(t+3) 72 (72/(t+3))

Total jobs = 36

(72/t) + [(72/(t+3)] = 36

=> 2/t + 2/(t+3) = 1
=> 2t + 6 + 2t = t^2 + 3t
=> t^2 - t - 6 = 0

t = 3 or t =-2 (can't be negative)
Hence , 3

Plz advice where I am wrong as I took the other way around by taking (t+3)

Rgds,
TGC

We are told that if Victor is supervised at work, he finishes each job three days faster than if he is unsupervised.

Supervised = t days;
Unsupervised = (t + 3) days.

Victor to complete 10 jobs without any supervision will need 10(t + 3) = 60.

Hope it helps.
_________________
Current Student
Joined: 03 Aug 2012
Posts: 915
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Followers: 18

Kudos [?]: 477 [0], given: 322

Re: Victor's job requires him to complete a series of identical [#permalink]  24 Mar 2013, 02:25
Bunuel wrote:
targetgmatchotu wrote:
siddhans wrote:
Victor's job requires him to complete a series of identical jobs. If Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. If Victor works for 144 days and is supervised for half the time, he will finish a total of 36 jobs. How long would it take Victor to complete 10 jobs without any supervision?

I have a query

R T D
S (1/t) 72 (72/t)
U (1/(t+3) 72 (72/(t+3))

Total jobs = 36

(72/t) + [(72/(t+3)] = 36

=> 2/t + 2/(t+3) = 1
=> 2t + 6 + 2t = t^2 + 3t
=> t^2 - t - 6 = 0

t = 3 or t =-2 (can't be negative)
Hence , 3

Plz advice where I am wrong as I took the other way around by taking (t+3)

Rgds,
TGC

We are told that if Victor is supervised at work, he finishes each job three days faster than if he is unsupervised.

Supervised = t days;
Unsupervised = (t + 3) days.

Victor to complete 10 jobs without any supervision will need 10(t + 3) = 60.

Hope it helps.

Hi Bunuel

As I would follow the same method of RTW chart for any problem in TNW, it would be really helpful if you can QUOTE the error in my method .

Rgds,
TGC
_________________

Rgds,
TGC!
_____________________________________________________________________
I Assisted You => KUDOS Please
_____________________________________________________________________________

Math Expert
Joined: 02 Sep 2009
Posts: 31228
Followers: 5342

Kudos [?]: 62057 [0], given: 9427

Re: Victor's job requires him to complete a series of identical [#permalink]  24 Mar 2013, 02:30
Expert's post
targetgmatchotu wrote:
Bunuel wrote:
targetgmatchotu wrote:
I have a query

R T D
S (1/t) 72 (72/t)
U (1/(t+3) 72 (72/(t+3))

Total jobs = 36

(72/t) + [(72/(t+3)] = 36

=> 2/t + 2/(t+3) = 1
=> 2t + 6 + 2t = t^2 + 3t
=> t^2 - t - 6 = 0

t = 3 or t =-2 (can't be negative)
Hence , 3

Plz advice where I am wrong as I took the other way around by taking (t+3)

Rgds,
TGC

We are told that if Victor is supervised at work, he finishes each job three days faster than if he is unsupervised.

Supervised = t days;
Unsupervised = (t + 3) days.

Victor to complete 10 jobs without any supervision will need 10(t + 3) = 60.

Hope it helps.

Hi Bunuel

As I would follow the same method of RTW chart for any problem in TNW, it would be really helpful if you can QUOTE the error in my method .

Rgds,
TGC

You calculated the time needed to complete 10 jobs WITH supervision (10t, less time) but we are asked to find the time needed to complete 10 jobs WITHOUT any supervision (10(t+3), more time).

Hope it's clear.
_________________
Current Student
Joined: 03 Aug 2012
Posts: 915
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Followers: 18

Kudos [?]: 477 [0], given: 322

Re: Victor's job requires him to complete a series of identical [#permalink]  27 Sep 2013, 22:31
1
This post was
BOOKMARKED
Nice Question !

Took me a good amount of time !

Rate(Supervised) = 1/(t-3)
Rate(Unsupervised) = 1/(t)

72/(t-3) + 72 /(t) = 36

t=6 or t=1

Cannot be 1 as 't-3' would be negative then => t=6

Rate (UN) = 1/6

Rate * Time = work

1/6 * time = 10

Time = 10*6 = 60
_________________

Rgds,
TGC!
_____________________________________________________________________
I Assisted You => KUDOS Please
_____________________________________________________________________________

Manager
Status: Please do not forget to give kudos if you like my post
Joined: 19 Sep 2008
Posts: 128
Location: United States (CA)
Followers: 0

Kudos [?]: 61 [0], given: 257

Re: Victor's job requires him to complete a series of identical [#permalink]  13 Oct 2013, 11:10
siddhans wrote:
Victor's job requires him to complete a series of identical jobs. If Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. If Victor works for 144 days and is supervised for half the time, he will finish a total of 36 jobs. How long would it take Victor to complete 10 jobs without any supervision?

A. 34
B. 52
C. 60
D. 70
E. 92

My approach:
Since we are asked Unsupervised time, i make that as t (variable in question). This way i do not have to worry about other stuff.

S => 1/(t-3)
U => 1/t

72days Job S + 72days job U = 36job
72/t + 72/(t-3) = 36

now pick a number to that LHS = RHS
t=6 => 72\6 + 72\3 = 36

now asked is 10t=6*10 = 60.
_________________

[Reveal] Spoiler:

Current Student
Joined: 26 Sep 2013
Posts: 221
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41
Followers: 3

Kudos [?]: 94 [0], given: 40

Re: Victor's job [#permalink]  17 Oct 2013, 11:14
raghavakumar85 wrote:
60 days it is

72/ (x-3) + 72/x = 36

=> x^2 -7x+6 =0 => x= 1 or 6

X = 1 is omitted coz x-3 = -2 if x=1

so x=6 and time taken for 10 jobs under no supervision = 10x6 = 60

edit: nevermind, had a mistake in my work
Senior Manager
Joined: 10 Mar 2013
Posts: 285
GMAT 1: 620 Q44 V31
GMAT 2: 690 Q47 V37
GMAT 3: 610 Q47 V28
GMAT 4: 700 Q50 V34
GMAT 5: 700 Q49 V36
GMAT 6: 690 Q48 V35
GMAT 7: 750 Q49 V42
GMAT 8: 730 Q50 V39
Followers: 1

Kudos [?]: 42 [0], given: 2403

Re: Victor's job requires him to complete a series of identical [#permalink]  23 Oct 2013, 18:47
Why multiply the following by 72 instead of 144 seeing that Victor worked for 144 days?
Unsupervised rate + Supervised rate = Combined Rate
1/u + 1/(u-3) = 1/c
144*(1/c)=36
c = 4
c = (u^2-3u)/(2u-3) = 4

Isn't this the same logic that was done here: "http://gmatclub.com/forum/running-at-their-respective-constant-rates-machine-x-takes-98599.html"?
Math Expert
Joined: 02 Sep 2009
Posts: 31228
Followers: 5342

Kudos [?]: 62057 [1] , given: 9427

Re: Victor's job requires him to complete a series of identical [#permalink]  23 Oct 2013, 23:42
1
KUDOS
Expert's post
TooLong150 wrote:
Why multiply the following by 72 instead of 144 seeing that Victor worked for 144 days?
Unsupervised rate + Supervised rate = Combined Rate
1/u + 1/(u-3) = 1/c
144*(1/c)=36
c = 4
c = (u^2-3u)/(2u-3) = 4

Isn't this the same logic that was done here: "http://gmatclub.com/forum/running-at-their-respective-constant-rates-machine-x-takes-98599.html"?

Victor's job requires him to complete a series of identical jobs. If Victor is supervised at work, he finishes each job three days faster than if he is unsupervised. If Victor works for 144 days and is supervised for half the time, he will finish a total of 36 jobs. How long would it take Victor to complete 10 jobs without any supervision?
A. 34
B. 52
C. 60
D. 70
E. 92

Rate when supervised = (job)/(time) = 1/t.
Rate when unsupervised = (job)/(time) = 1/(t+3).

For 144/2=72 days he is supervised and for 144/2=72 days he is unsupervised and does 36 jobs:
72/t + 72/(t+3) = 36 --> t=3 days --> t+3 = 6 days.

Victor to complete 10 jobs without any supervision will need 10(t + 3) = 60 days.

Hope it's clear.
_________________
Senior Manager
Joined: 23 Jan 2013
Posts: 488
Schools: Cambridge'16
Followers: 2

Kudos [?]: 39 [0], given: 36

Re: Victor's job requires him to complete a series of identical [#permalink]  24 Oct 2013, 05:33
i tried to do with 72/x + 72/x+3=36 but it requires huge amount of time with use quadratic equation. Better to backsolve the answers with start from 60 (choice C). 60/10=6 and x=3, it wins
Manager
Joined: 25 Oct 2013
Posts: 173
Followers: 1

Kudos [?]: 37 [0], given: 56

Re: Victor's job requires him to complete a series of identical [#permalink]  13 Jan 2014, 08:16
Temurkhon wrote:
i tried to do with 72/x + 72/x+3=36 but it requires huge amount of time with use quadratic equation. Better to backsolve the answers with start from 60 (choice C). 60/10=6 and x=3, it wins

Simplifying $$\frac{72}{{x}} + \frac{72}{{x+3}}=36$$ to $$\frac{2}{{x}}+\frac{2}{{x+3}}=1$$ may make it easier to solve the resultant quadratic equation.

We then have
$$x^2+3x=4x+6$$
$$x^2-x-6=0$$
$$(x-3)(x+2)=0$$ => $$x=3$$ (Supervised time). so 6 is unsupervised time per job. Therefore 6*10 = 60 hrs
_________________

Click on Kudos if you liked the post!

Practice makes Perfect.

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 8156
Followers: 416

Kudos [?]: 111 [0], given: 0

Re: Victor's job requires him to complete a series of identical [#permalink]  06 Jun 2015, 22:30
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Manager
Joined: 21 Jun 2014
Posts: 150
Location: United States
Concentration: General Management, Strategy
GMAT 1: 630 Q45 V31
GPA: 3.4
WE: Engineering (Computer Software)
Followers: 0

Kudos [?]: 44 [0], given: 59

Re: Victor's job requires him to complete a series of identical [#permalink]  06 Jul 2015, 06:56
Also keep in mind the number of jobs =10 so answer has to be a multiple of 10 ,only C and D are the choices in this case .

Regards,
Manish Khare
_________________

Regards,
Manish Khare
"Every thing is fine at the end. If it is not fine ,then it is not the end "

Re: Victor's job requires him to complete a series of identical   [#permalink] 06 Jul 2015, 06:56
Similar topics Replies Last post
Similar
Topics:
10 Machine A takes 10 hours to complete a certain job and starts that job 6 23 Sep 2015, 21:32
23 At Supersonic Corporation, the time required for a machine to complete 10 06 Oct 2014, 03:17
3 Tom is on a certain diet that requires him to limit the 6 01 Mar 2014, 11:23
22 A carpenter worked alone for 1 day on a job that would take him 6 more 9 27 Jun 2011, 11:00
8 A can complete the job in 2 hours and B can complete the sam 6 17 Mar 2011, 04:21
Display posts from previous: Sort by


text:
Why is the dimension of the vector space which contains only the zero vector $0$?

That means the basis is an empty list. I can see how the empty list spans the space (Basically, linear combinations of nothing are equal to $0$) but I don't see what it means for the empty list to be linearly independent. This is the definition of linear independence I am familiar with:

If $a_1v_1+...+a_mv_m = 0$ then $a_1 = ... = a_m = 0$

That definition does not seem to be applicable for an empty list of vectors. I assume you solve that problem with a better definition of linear independence.

Bonus: Is there a name for the vector space that contains only the zero vector?

Your definition omitted some crucial fine print. A set $B$ in a vector space is linear independent if: "For every subset $\{v_{1}, \dots, v_{m}\} \subset B$, if $\sum_{i} a_{i} v_{i} = 0$ for some scalars $a_{i}$, then $a_{i} = 0$ for all $i$." If $B = \emptyset$, this condition is vacuous.
A vector space with one element is sometimes called a trivial vector space, but often one simply denotes it by "$0$" without giving it a name.


text:
## The Annals of Probability

### A system of coalescing heavy diffusion particles on the real line

Vitalii Konarovskyi

#### Abstract

We construct a modified Arratia flow with mass and energy conservation. We suppose that particles have a mass obeying the conservation law, and their diffusion is inversely proportional to the mass. Our main result asserts that such a system exists under the assumption of the uniform mass distribution on an interval at the starting moment. We introduce a stochastic integral with respect to such a flow and obtain the total local time as the density of the occupation measure for all particles.

#### Article information

Source
Ann. Probab., Volume 45, Number 5 (2017), 3293-3335.

Dates
Revised: July 2016
First available in Project Euclid: 23 September 2017

https://projecteuclid.org/euclid.aop/1506132039

Digital Object Identifier
doi:10.1214/16-AOP1137

Mathematical Reviews number (MathSciNet)
MR3706744

Zentralblatt MATH identifier
06812206

#### Citation

Konarovskyi, Vitalii. A system of coalescing heavy diffusion particles on the real line. Ann. Probab. 45 (2017), no. 5, 3293--3335. doi:10.1214/16-AOP1137. https://projecteuclid.org/euclid.aop/1506132039

#### References

• [1] Arratia, R. A. (1979). Brownian motion on the line. PhD dissertation, Univ. Wisconsin, Madison.
• [2] Arratia, R. A. (1981). Coalescing Brownian motion and the Voter model on $\mathbb{Z}$. Unpublished manuscript.
• [3] Billingsley, P. (1999). Convergence of Probability Measures, 2nd ed. Wiley, New York.
• [4] Chernega, P. P. (2012). Local time at zero for Arratia flow. Ukrainian Math. J. 64 616–633.
• [5] Dawson, D. A. (1993). Measure-valued Markov processes. In Ecole D’ete de Probabilites SaintFlour XXI. Lecture Notes in Math 154. Springer, Berlin.
• [6] Dawson, D. A., Li, Z. and Wang, H. (2001). Superprocesses with dependent spatial motion and general branching densities. Electron. J. Probab. 6 33 pp. (electronic).
• [7] Dawson, D. A., Li, Z. and Zhou, X. (2004). Superprocesses with coalescing Brownian spatial motion as large-scale limits. J. Theoret. Probab. 17 673–692.
• [8] Dorogovtsev, A. A. (2003). Stochastic flows with interaction and measure-valued processes. Int. J. Math. Math. Sci. 63 3963–3977.
• [9] Dorogovtsev, A. A. (2004). Measure-valued Markov processes and stochastic flows on abstract spaces. Stoch. Stoch. Rep. 76 395–407.
• [10] Dorogovtsev, A. A. (2004). One Brownian stochastic flow. Theory Stoch. Process. 10 21–25.
• [11] Dorogovtsev, A. A. (2005). Some remarks on a Wiener flow with coalescence. Ukrainian Math. J. 57 1550–1558.
• [12] Dorogovtsev, A. A. (2006). A stochastic integral with respect to the Arratia flow. Dokl. Akad. Nauk 410 156–157.
• [13] Dorogovtsev, A. A. (2007). Measure-Valued Processes and Stochastic Flows. Institute of Mathematics of NAS of Ukraine, Kiev.
• [14] Dorogovtsev, A. A., Gnedin, A. V. and Vovchanskii, M. B. (2012). Iterated logarithm law for sizes of clusters in Arratia flow. Theory Stoch. Process. 18 1–7.
• [15] Dorogovtsev, A. A. and Kotelenez, P. (1997). Smooth Stationary Solutions of Quasilinear Stochastic Differential Equations. Finite Mass. Preprint No. 97. Department of Mathematics Case Western Reserv Univ. Cleveland, Ohio.
• [16] E, Weinan, Rykov, Yu. G. and Sinai, Ya. G. (1996). Generalized variational principles, global weak solutions and behavior with random initial data for systems of conservation laws arising in adhesion particle dynamics. Comm. Math. Phys. 177 349–380.
• [17] Ethier, S. N. and Kurtz, T. G. (1986). Markov Processes: Characterization and Convergence. Wiley, New York.
• [18] Evans, S. N., Morris, B. and Sen, A. (2013). Coalescing systems of non-Brownian particles. Probab. Theory Related Fields 156 307–342.
• [19] Fleming, W. H. and Viot, M. (1979). Some measure-valued Markov processes in population genetics theory. Indiana Univ. Math. J. 28 817–843.
• [20] Gorostiza, L. G. and Roelly, S. (1991). Some properties of the multitype measure branching process. Stochastic Process. Appl. 37 259–274.
• [21] Gorostiza, L. G. and Wakolbinger, A. (1991). Persistence criteria for a class of critical branching particle systems in continuous time. Ann. Probab. 19 266–288.
• [22] Ikeda, N. and Watanabe, Sh. (1981). Stochastic Differential Equations and Diffusion Processes. North-Holland Mathematical Library, Amsterdam.
• [23] Jacod, J. and Shiryaev, A. N. (2003). Limit Theorems for Stochastic Processes, 2nd ed. Springer, Berlin.
• [24] Kallenberg, O. (2002). Foundations of Modern Probability, 2nd ed. Springer, New York.
• [25] Karlikova, M. P. (2005). On a weak solution of an equation for an evolutionary flow with interaction. Ukrainian Math. J. 57 1055–1065.
• [26] Konarovs’kyi, V. V. (2010). System of sticking diffusion particles of variable mass. Ukrainian Math. J. 62 97–113.
• [27] Konarovskiĭ, V. V. (2010). On an infinite system of diffusing particles with coalescing. Theory Probab. Appl. 55 134–144.
• [28] Konarovskii, V. V. (2011). The martingale problem for a measure-valued process with heavy diffusing particles. Theory Stoch. Process. 17 50–60.
• [29] Konarovskyi, V. (2013). Mathematical model of heavy diffusion particles system with drift. Commun. Stoch. Anal. 7 591–605.
• [30] Konarovskyi, V. V. (2014). Large deviations principle for finite system of heavy diffusion particles. Theory Stoch. Process. 19 37–45.
• [31] Lang, R. and Nguyen, X.-X. (1980). Smoluchowski’s theory of coagulation in colloids holds rigorously in the Boltzmann-Grad-limit. Probab. Theory Related Fields 54 227–280.
• [32] Le Jan, Y. and Raimond, O. (2004). Flows, coalescence and noise. Ann. Probab. 32 1247–1315.
• [33] Liptser, R. S. and Shiryaev, A. N. (2001). Statistics of Random Processes. I, expanded ed. Springer, Berlin.
• [34] Malovichko, T. V. (2009). Girsanov’s theorem for stochastic flows with interaction. Ukrainian Math. J. 61 435–456.
• [35] Norris, J. and Turner, A. (2015). Weak convergence of the localized disturbance flow to the coalescing Brownian flow. Ann. Probab. 43 935–970.
• [36] Norris, J. R. (1999). Smoluchowski’s coagulation equation: Uniqueness, nonuniqueness and a hydrodynamic limit for the stochastic coalescent. Ann. Appl. Probab. 9 78–109.
• [37] Piterbarg, V. V. (1998). Expansions and contractions of isotropic stochastic flows of homeomorphisms. Ann. Probab. 26 479–499.
• [38] Ruelle, D. (1969). Statistical Mechanics: Rigorous Results. W. A. Benjamin, Inc., New York.
• [39] Shamov, A. (2011). Short-time asymptotics of one-dimensional Harris flows. Commun. Stoch. Anal. 5 527–539.
• [40] Smoluchowski, M. V. (1916). Drei vortrage uber diffusion, brownsche bewegung und koagulation von kolloidteilchen. Physik. Zeit. 17 557–585.
• [41] Wang, H. (1997). State classification for a class of measure-valued branching diffusions in a Brownian medium. Probab. Theory Related Fields 109 39–55.
• [42] Wang, H. (1998). A class of measure-valued branching diffusions in a random medium. Stoch. Anal. Appl. 16 753–786.


text:
# If a projectile is shot at an angle of pi/12 and at a velocity of 24 m/s, when will it reach its maximum height??

Mar 6, 2016

$t \cong 0 , 63 s$

#### Explanation:

$g \cdot t = {v}_{i} \cdot \sin \alpha$
$t = \frac{{v}_{i} \cdot \sin \alpha}{g}$
$\alpha = \frac{\pi}{12} \cdot \frac{180}{\pi} = 15$
$t = 24 \cdot \sin \frac{15}{9 , 81}$
$t \cong 0 , 63 s$


text:
TSTEP

Bulk Data Entry Defines time step parameters for control and intervals at which a solution will be generated and output in transient analysis.

Format

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
TSTEP SID N1 DT1 N01 W3,1 W4,1
N2 DT2 N02 W3,2 W4,2
etc.
TMTD TC1 TC2 TC3 TC4 Alpha Beta
MREF TOL TN1 TN2

Example

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
TSTEP 2 10 .001 5
9 0.01 1

Definitions

Field Contents SI Unit Example
SID Set identification number.

No default (Integer > 0)

N# Number of time steps of value DT#.

No default (Integer ≥ 1)

DT# Time increment.

No default (Real > 0.0)

N0# Skip factor for output. Every N0i-th step will be saved for output.

Default = 1 (Integer > 0)

W3,# The frequency of interest in radians per unit time; used for the conversion of overall structural damping into equivalent viscous damping. 3

Default = blank (Real > 0.0, or blank)

W4,# The frequency of interest in radians per unit time; used for the conversion of element structural damping into equivalent viscous damping. 3

Default = blank (Real > 0.0, or blank)

TMTD Time integration method.
For Linear Direct Transient Analysis:
= blank (Default)
= 1
Newmark-Beta method.
For Nonlinear Direct Transient Analysis:
= 1 (Default)
Generalized alpha method.
= 2
Backward Euler method.

(Integer > 0) 6

TC1 Time integration parameters for transient subcases.

Default = -0.05 (-1/3 < Real < 0) 6

TC2 Time integration parameters for transient subcases.

Default= 0.25*(1-TC1-TC4)2 (Real ≥ 0.25 - 0.5*(TC4 + TC1)) 6

TC3 Time integration parameters for transient subcases. 6

Default = 0.5-TC1-TC4 (Real)

TC4 Time integration parameters for transient subcases. 6

Default = 0 (-1 < Real < 0.5)

Alpha Rayleigh damping coefficient for transient subcases. 7

Default = 0.0 (Real)

Beta Rayleigh damping coefficient for transient subcases. 7

Default = 0.0 (Real)

MREF Controls activation of automatic time-stepping for linear transient and nonlinear transient analysis.
Linear direct transient analysis (Newmark-Beta method):
= 0
Automatic time-stepping is deactivated.
= 1 (Default)
Automatic time-stepping is active via the reference displacement method.
Nonlinear direct transient analysis (Generalized alpha method):
= 0
Automatic time-stepping is deactivated.
= 1 (Default)
Automatic time-stepping is active via the reference displacement method.
Nonlinear direct transient analysis (Backward Euler method):
= 0 (Default)
Automatic time-stepping is deactivated.
= 1
Automatic time-stepping is active via the reference displacement method.

(Integer) See Comment 8

TOL Tolerance for automatic time stepping in transient subcases.

Default = 1.0 (Real > 0) 8

TN1 Control parameter for automatic time stepping in transient subcases. It specifies the maximum number of cut-backs in a single time step.

Default = 5 (Integer > 0) 8

TN2 Control parameter for automatic time stepping in transient subcases. It specifies the minimum number of time step enlargement requests required before the solver enlarges the next time step.

Default=1 (Integer > 0) 8

1. TSTEP entries must be selected with the Subcase Information command TSTEP=SID.
2. The entry permits changes in the size of the time step during the course of the solution. In the example shown, there are 10 time steps of value .001, followed by 9 time steps of value .01. Also, in the case of this example, you have requested that the output be recorded for t = 0.0, .005, .01, .02, .03, and so on.
3. W3 and W4 define frequencies used in linear transient analyses to convert structural damping to equivalent viscous damping. The W3 and W4 fields on TSTEP are not supported for Nonlinear Transient Analysis. For Nonlinear Transient, PARAM, W3 and PARAM, W4 can be used.
4. Different values for W3 and W4 may be set for each set of time increments. If any of the fields are left blank then the value is taken from the PARAM, W3 or PARAM, W4 definition.
5. Transient Response Analysis using Fourier Transformation cannot be used in a model, which also contains a Modal Frequency Response Analysis subcase. OptiStruct will error out in such cases.
6. Time integration method.
The TMTD field can be used to control the time integration scheme for both linear transient and nonlinear transient analysis.
• Linear Transient Analysis

If TMTD field is blank for linear transient analysis, the traditional time integration scheme is used. Automatic time-stepping (MREF=1) is not supported for traditional time integration.

If TMTD=1, the Newmark-Beta Integration scheme is used for linear transient analysis. Automatic time-stepping (MREF=1) is supported for Newmark-Beta time integration.

For more information, refer to Linear Transient Analysis in the User Guide.

• Nonlinear Transient Analysis

If the TSTEP entry is not referenced or present in the input deck, or if the TMTD field is blank, the Generalized Alpha method (more specifically, the HHT- $\alpha$ method) will be used by default.

When TMTD =1, coefficients of Generalized Alpha method are specified using TC1, TC2, TC3, and TC4, for four non-dimensional parameters ( $\alpha ,\beta ,\gamma ,$ and ${\alpha }_{m}$ ), respectively. In general, the Generalized Alpha method should be used for most nonlinear transient analyses. In this method, numerical damping can be adjusted through the parameters $\alpha ,\beta ,\gamma ,$ and ${\alpha }_{m}$ . In particular, non-zero $\alpha$ and ${\alpha }_{m}$ introduces damping for high-frequency response components. On the other hand, Backward Euler method can be used for quasi-static analysis, such as post-buckling problem, since this method is dissipative, and therefore stable.

When TMTD =2, the backward Euler method does not require the input of the TCi fields. The Alpha and Beta fields introduce subcase-dependent Rayleigh damping, so the viscous damping matrix $C$ in a particular subcase is $C=\alpha M+\beta K$ .

For more information, refer to Nonlinear Transient Analysis in the User Guide.

7. Subcase-dependent Rayleigh damping for nonlinear direct transient subcase is issued using Alpha, and Beta. Alternatively, they can be specified using PARAM, ALPHA1 and PARAM, ALPHA2, respectively.
8. Automatic time stepping.
Automatic time-stepping is applicable for both linear and nonlinear transient subcases and is controlled using the MREF field.
MREF=0
Indicates no automatic time stepping.
MREF=1
Automatic time stepping is activated, and the reference displacement is used to normalize the acceleration error measure.
TOL
Specifies the tolerance for time step adjustment error control.
TN1
Specifies the maximum number of cutbacks in a single time step.
TN2
Specifies the minimum number of time step enlargement requests required before the solver enlarges the next time step.

If DTMIN and DTMAX are specified on the NLADAPT entry for Nonlinear Transient Analysis, they have highest priority over the time-stepping process and they are always respected.

9. This card is represented as a loadcollector in HyperMesh.


text:
Sciencemadness Discussion Board » Fundamentals » Reagents and Apparatus Acquisition » PID Controlled Cheap 'Homemade' Hotplate Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues

Author: Subject: PID Controlled Cheap 'Homemade' Hotplate
BobD1001
Hazard to Others

Posts: 182
Registered: 29-3-2013
Member Is Offline

Mood: No Mood

I recently inherited a very old ceramic resistance type hot plate. Its a very good condition functioning unit, and I would like to put it to use, but unfortunately, the controls consist of the wide temperature range of ON and OFF as seen below:

So I began thinking about mounting a relatively thick Aluminum plate onto the existing surface with a thermocouple drilled deep into the side of it to get a relatively accurate reading and controlling it using a PID and Solid State Relay for the switching. Here's a link to what I'm thinking of using:

http://www.aliexpress.com/item/Digital-PID-Temperature-contr...

I have a full machine shop, so making/drilling the aluminum plate is no concern. I just wanted to see if anyone could find a flaw in my idea before I ordered the PID, SSR, and Thermocouple Probe.

Thank you,
Bob
Magpie
lab constructor

Posts: 5939
Registered: 1-11-2003
Location: USA
Member Is Offline

Mood: Chemistry: the subtle science.

Hi Bob,

That's a ludicrously low cost for a PID + SSR.

I use a PID to control the temperature of heating baths (water, silicone oil, molten metal, etc). By using the auto-tune feature it gives very close temperature control (1-2°C).

I think your plan will work.

The single most important condition for a successful synthesis is good mixing - Nicodem
ElectroWin
Hazard to Others

Posts: 224
Registered: 5-3-2011
Member Is Offline

Mood: No Mood

dont you want to control this based on the measured temperature of the stuff you're cooking, rather than the temperature of the aluminum plate?
BobD1001
Hazard to Others

Posts: 182
Registered: 29-3-2013
Member Is Offline

Mood: No Mood

Electrowin, ill probably get an extra probe so that I can either control the plate temperature, or use the additional probe to control the temp of whatever is actually on the plate. All I would need to do is disconnect the one probe from the PID, and then connect the other. Of course I could also wire in a switch to change it from monitoring the one probe to the other.
BobD1001
Hazard to Others

Posts: 182
Registered: 29-3-2013
Member Is Offline

Mood: No Mood

Well I just ordered the PID 'kit' with the SSR, K type thermocouple, and also I grabbed a big heat sink for the SSR just in case. As soon as I get it Ill update this thread and detail the building of the homemade temperature controlled hotplate so that others may be able to do the same.

[Edited on 13-4-2013 by BobD1001]
Steve_hi
Hazard to Others

Posts: 196
Registered: 4-12-2010
Member Is Offline

Mood: No Mood

Please let us know if you get the material and if the items are described. unbelievable price.

Hope it works out
BobD1001
Hazard to Others

Posts: 182
Registered: 29-3-2013
Member Is Offline

Mood: No Mood

One Long Post.... Coming up!

Ok, so the PID controlled Hotplate has been a pretty incredible success! Its not 100% done. Still adding the variable voltage control for the stirrer, it needs some rework to make it a bit more compact, and of course another thermocouple so that instead of just measuring plate temp, one could also control the temp of solutions or baths.

Here was the PID box as recieved from China... I was worried something might have broken, but it all was in perfect condition!

Here's the contents of the box:

I got a project box from radioshack, and cut some holes in it, then fit in the elctronics:

Here was one of the preliminary tests just making sure it all worked, and could properly monitor temp and switch the old hot plate properly:

Drilling and tapping the aluminum plate for the thermocouple:

First real life action shot!

I quickly determined I didn't like the old hot plate being used as my heater, so I ordered 2 400 watt ceramic heating cartridges for dirt cheap on ebay. I also have the thermocouple hole taped off here so that when drilling the heater cartridge holes no chips get stuck in there.

Perfect fit. As always.

Now for the stand: Laid out the drill points making sure not to impede upon the heating cartridge holes, and then proceeded to drill away.

Some long bolts, washers, and nuts finish the structure nicely.

Then I decided I needed an electronic stirrer as well, so I made a nice cylindrical aluminum heatsink on my lathe, centered it upon an old CPU fan, and epoxied. Next I found the center of mass of a large old hard drive Nd magnet and used high temp epoxy to fix it to the top of the heatsink. Ill get some better pictures of it up shortly.

As you can see, it works wonderfully...

More updates on the electronic stirrer control board and pictures of the cpu fan setup as well.

And as for the hotplate.... IT WORKS GREAT! The max temp programmed into the PID is at 400C which I may raise a couple hundred degrees since its a breeze to change.
violet sin
International Hazard

Posts: 1385
Registered: 2-9-2012
Location: Back yard staring at stars
Member Is Offline

Mood: Good

Rex-c 100? Sorry reading this from an iTouch. Any how I got what looks to be the same thing from eBay an the dang thing wouldn't drive the included SSR itself. Was a relay model w/ an SSR, no low voltage signal out for it. used a cheap LED driver through the relay terminals on PID and spent 6$for an upgraded high temp probe from another seller. Nice stainless sheath. I also found a hot water heater control box a perfect fit for it all. Ya reprogramming the upper limits was easy. Just a few tips to offer any one trying the same thing. Guess mine wasn't a true REX C-100, but a verry affordable knockoff. The pin-out provided didn't at all match the companies online manual. Thing works fine though. I really like the cartridge element idea, looks sweet . Nice craftsmanship bfesser Resident Wikipedian Posts: 2114 Registered: 29-1-2008 Member Is Offline Mood: No Mood I've been waiting for this thread to resurface so I could post this: DIY PID Controlled Soldering Hotplate (MightyOhm.com) BobD1001 Hazard to Others Posts: 182 Registered: 29-3-2013 Member Is Offline Mood: No Mood violet sin, I know that there was a lot of confusion on which ones will actually switch the SSR.... I suppose some unscrupulous some eBay sellers just include an SSR with their cheaper model to make customers think it will work. Here's a picture of the box label on my PID: The important thing to see if the "V" at the end of the model name, it means that there is a voltage output which will switch an SSR. However in your situation, with the cheaper relay controlled types, if you find the input leads for the relay, you can solder in a wire on each the + and -, which can the directly switch an SSR. bfesser, I did come across that page when designing my hotplate, some helpful information! violet sin International Hazard Posts: 1385 Registered: 2-9-2012 Location: Back yard staring at stars Member Is Offline Mood: Good Ya mine was designated "Rex-c100 fk02 m-an" and just had an SSR included . Even though I asked prior to purchasing. But the 3 diode driver i got puts out 12-18v 300ma and runs a couple light indicators + 12v cooling fans from a dead computer. Just take the needed 20ma from one LED and loop through the relay terminals. So it's not a big waste of space/money. 3.50 shipping included from china off eBay. I work a LOT away from home (3.5hr drive gone for weeks at a time, actually get to go home today, yay) so stuff usually just waiting for me when I get home otherwise think the shipping time would piss me off, 3+ weeks. Again nice work. Think the seller was " easebon" on eBay. Deff gotta read the serial #'s before buying even if you ask them. -Violet Sin- I also read about unsoldering the relay to use its 12v signal to switch, and am proficient in soldering, but chose to take a more creative attack. The power on indicator lights n pair of <1w fans were handy. [Edited on 12-9-2013 by violet sin] [Edited on 12-9-2013 by violet sin] BobD1001 Hazard to Others Posts: 182 Registered: 29-3-2013 Member Is Offline Mood: No Mood Well I just uploaded a video on my youtube channel showing the workings of the hotplate and newly implemented PWM stirrer controller. Check it out, and let me know what you think. Still some refining and finishing is needed, but it really turned out better than expected! Sorry for the slight shakiness, couldn't find my tripod.  [Edited on 23-10-2013 by BobD1001] Dariusrussell Hazard to Self Posts: 70 Registered: 27-6-2013 Location: Southie Member Is Offline Mood: Organometallics If you haven't already done this, one thing would be to use the program button on the SSR. It should be somewhere on there or in the chinglish instructions. Once you do that set it like normal and leave it for a few hours (4ish) and it should do a much better job with keeping in range. bismuthate International Hazard Posts: 803 Registered: 28-9-2013 Location: the island of stability Member Is Offline Mood: self reacting That's amazing. How much did it cost to make and how long did it take? P.S. this reminds me of when I made a centrifuge out of Legos. I'm not a liar, I'm just an enthusiastic celebrant of opposite day. I post pictures of chemistry on instagram as bismuthate. http://iconosquare.com/bismuthate or this viewer if you don't have an instagram (it sucks though) http://web.stagram.com/n/bismuthate BobD1001 Hazard to Others Posts: 182 Registered: 29-3-2013 Member Is Offline Mood: No Mood It ran me about$65 total, as I had quite a few of the parts already laying around, and it wouldn't have been much more if buying everything new if one knew where to look.
Organikum
resurrected

Posts: 2250
Registered: 12-10-2002
Location: Europe
Member Is Offline

Mood: lonely

REX-C100, I have some of these, for 12\$ inclusive shiping a bargain and they do a pretty good job on standard setting already.

I got the kind with relais though, just removed this and drive a SSR with the 12V for the relais - this works flawless.

Some thermocouples are just not good I found out, if there is a problem its always on contacts or thermocouple.
There is a hiden setting to change the input from thermocouple type K to PTC 100, this changes also the reading to 10th of degrees Celsius, say not 000°C but 000,0°C and it can be set as such. Just if someday a more detailed regulation is wanted. The PTC has also a smaller risk of misreadings for any reasons, more robust and professional and they got pretty cheap too nowdays, I prefer them.

If you want a fast heating up and a better regulation its possible to use a second relais connected to the alarm output. Alarm set to low so the relais opens and serves full power to the heating element up to a certain temperature, then the PID takes over and regulates only the last 20% - thats in many situations a good thing as the REX has no "ramp" feature.
A other variation which makes even more sense is to have two heating coils, one to heat up by relais and one to hold and fine-regulate by SSR - the second much less powerful then the first of course. In this configuration a heatsink on the SSR gets virtually obsolete, no need to stress the part too much and even if it should burn out - having 150W full on is a big difference to having 1500W blasting, what can get unpleasant rather fast.

Great work!
Split pole motors (from microwave fans for example) are better for stirrers IMHO , more powerful and they couple so very well to the stirbar. IKA uses them too and they are free (s.o.)

I wish I had and could work a lathe too, oh yes! I am jealous lol

/ORG

[Edited on 15-11-2013 by Organikum]

Irgendwas is ja immer
violet sin
International Hazard

Posts: 1385
Registered: 2-9-2012
Location: Back yard staring at stars
Member Is Offline

Mood: Good

Nice, I like your alarm for coarse tuning and SSR for fine tuning. Great idea
Cattmy
Harmless

Posts: 2
Registered: 25-12-2013
Member Is Offline

Mood: No Mood

Hi there!

I just recieved the last part for my Hotplate, and I just finished the machining. Now, I have the M*AN controler and was wondering how I can wire it up. I saw you wrote something on this before, but I dont understand. Can you please explain?

Cattmy
Harmless

Posts: 2
Registered: 25-12-2013
Member Is Offline

Mood: No Mood

I found a method on Youtube.

Its a really easy way of making a M*AN control an SSR.

Thanks, and Merry Christmas!

[Edited on 26-12-2013 by Cattmy]

 Sciencemadness Discussion Board » Fundamentals » Reagents and Apparatus Acquisition » PID Controlled Cheap 'Homemade' Hotplate Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues


text:
## Conditional Distributions

Conditional probability is a key part of Baye’s theorem. In plain language, it is the probability of one thing being true given that another thing is true. It differs from joint probability, which is the probability that both things are true without knowing that one of them must be true. Before diving deeper into the math part, let’s see some practical application of conditional distributions.

Assume that a law enforcement department is looking into the connection between road accidents and intoxication among automobile drivers. On one hand, the department could come up with the probability that a driver is intoxicated and involved in an accident. That would be a joint probability. On the other, the department could determine the probability that a driver is involved in an accident given that they are intoxicated (It’s already known that the driver is intoxicated). This would be a conditional probability. The probabilities of these two events cannot be the same. Adding information (in the conditional case) alters probability.

In the medical field, a team of doctors could gather data to help them analyze and predict cases of kidney failure. On one hand, the probability that a patient’s left and right kidneys are both infected is a joint probability, whereas a conditional probability is the probability that the left kidney is infected if we know that the right one is infected. We can use a Euler diagram to demonstrate the difference. In the diagram, each large square has area 1, and the smaller squares represent probabilities.

Let X be the probability that a patient’s left kidney is infected, and let Y be the probability that the right kidney is infected. On the left side of the diagram, the green area represents the probability that both of the patient’s kidneys are infected. This is the joint probability (X,Y). If Y is  true (e.g., given that the right kidney is definitely infected), then the space of everything not Y is dropped and everything in Y is rescaled to the size of the original space. The rescaled green area on the right hand side is now the conditional probability of X given Y, expressed as P(X|Y). Put different, this is the probability that the left kidney is infected if we know that the right kidney is infected. It’s also important to note that the conditional probability of X given Y is not necessarily equal to the conditional probability of Y given X

simple terms, we define conditional distribution as the distribution of one random variable given the value of another random variable.

### Discrete Conditional Functions

The conditional probability mass function of $$X$$, given that $$Y = y$$, is defined by

$$g(x|y) = \frac{f(x,y)}{f_Y(y)}, \qquad \text{provided that } f_Y(y) > 0.$$

Similarly, the conditional probability mass function of $$Y$$, given that $$X = x$$, is defined by

$$h(y|x) = \frac{f(x,y)}{f_X(x)}, \qquad \text{provided that } f_X(x) > 0.$$

If we compare this definition with the univariate case we had

$$P(A|B) = \frac{P(A \cap B)}{P(B)}, \qquad \text{provided that } P(B) > 0.$$

where $$P(B)$$ happened first and had an impact on how A occurred (because if their intersection is empty the occurence of B wouldn’t have any effect on the probability of A). In the bivariate case, the intersection is given by joint pmf $$f(x,y)$$ and the event that would have an effect on how $$Y$$ occurs is $$X$$, and values of $$X$$ can be got from its marginal pmf, $$f_X(x)$$.

#### Example 1: Discrete conditional probability function

A bivariate distribution has the following probability function:

Determine:

(I) the marginal distribution of X, and the marginal distribution of Y

The marginal distribution of X can be found by summing the columns in the
table

$$P(X = 0) = 0.4, P(X = 1) = 0.3 , P(X = 2) = 0.3$$

The marginal distribution of Y can be found by summing the rows in the
table

$$P(Y = 1) = 0.2, P(Y = 2) = 0.4 , P(Y = 2) = 0.4$$

(II)  the conditional distribution of $$Y|X=2$$

Using the definition of conditional probability:

$$P(Y=1|X=2)=\cfrac { P(Y=1,X=2) }{ P(X=2) } =\cfrac { 0 }{ 0.3 } =0$$

$$P(Y=2|X=2)=\cfrac { P(Y=2,X=2) }{ P(X=2) } =\cfrac { 0.2 }{ 0.3 } =0.67$$

$$P(Y=3|X=2)=\cfrac { P(Y=3,X=2) }{ P(X=2) } =\cfrac { 0.1 }{ 0.3 } =0.33$$

#### Example 2: Discrete conditional probability function Let $$X$$ and $$Y$$ have the joint pmf

$$f(x,y) = \frac{5x+3y}{81}, \quad x=1,2, \quad y=1,2,3.$$

We proceed to find their marginal functions

$$f_X(x)= \frac{15x + 18}{81}, \quad x=1,2,$$

and

$$f_Y(y) =\frac{6y+15}{81} \quad y=1,2,3.$$

Thus, the conditional pmf of $$Y$$, given that $$X=x$$, is equal to

$$h(y|x) = \frac{(5x+3y)/81}{(15x+18)/81} = \frac{5x+3y}{15x+18}, \quad y=1,2,3, \text{when}\quad x=1\text{ or }2$$

Then finding the probabilities of an event in the space:

$$P(Y=1|X=2)= h(1|2)=\frac{13}{48}$$

The candidate can practice this concept by finding the “conditional” opposite, $$g(x|y)$$.

If we find all the probabilities for this conditional probability functions we would see that they behave as the joint probability mass functions seen in last chapter. Let’s keep $$X=2$$ fixed and check this:

$$P(Y=2|X=2)= h(2|2)=\frac{16}{48} \text{ and } P(Y=3|X=2)= h(3|2)=\frac{19}{48}$$

Then if we sum these values with X fixed

$$P(Y|X=2) = h(1|2) + h(2|2) + h(3|2) = \frac{13}{48} + \frac{16}{48} + \frac{19}{48} = 1$$

This condition is fulfilled for each value of X summed by each value of Y, and the same occurs for the $$g(x|y)$$ conditional function.

Thus, $$h(y|x)$$ and $$g(x|y)$$ both satisfy the conditions of a probability mass function, and we can do the same operations we did on a joint pmf, such as:

Computing conditional probabilities such as

$$P(a < Y < b| X = x) = \sum_{\{y:a < y < b\}} h(y|x)$$

and conditional expectations such as

$$E[u(Y)|X = x] = \sum_{y}u(y)h(y|x)$$

in a manner similar to those associated with unconditional probabilities and expectations.

Two special conditional moments are the conditional mean of $$Y$$, given that $$X = x$$ defined by

$$\mu_{Y|x}=E(Y|x)=\sum_{y}yh(y|x),$$

and the conditional variance of $$Y$$, given that $$X = x$$, defined by

$$\sigma_{Y|x}^2 = E\{[Y – E(Y|x)]^2|x\} = \sum_{y}[y-E(Y|x)]^2h(y|x),$$

We can compute this as:

$$\sigma_{Y|x}^2 = E(Y^2|x) – [E(Y|x)]^2.$$

We would use the same logic to find the conditional mean $$\mu_{X|y}$$ and the conditional variance $$\sigma_{X|y}^2$$

Using the values we found in example 2, let’s compute the conditional mean and variance:

\begin{align} \mu_{Y|2} & = E(Y|X=2) = \sum_{y=1}^{3}yh(y|2)\\ & = \sum_{y=1}^{3}y\left(\frac{10+3y}{48}\right) = 1 \left(\frac{13}{48}\right) + 2 \left(\frac{16}{48}\right) + 3 \left(\frac{19}{48}\right)\\ & = \frac{17}{8} \end{align}

and

\begin{align} \sigma_{Y|2}^2 & = E\left[\left(Y – \frac{17}{8}\right)^2\bigg|X=2\right] = \sum_{y=1}^{3}\left(y-\frac{17}{8}\right)^2\left(\frac{10+3y}{48}\right)\\ & = \left(1-\frac{17}{8}\right)^2\frac{13}{48}+\left(2-\frac{17}{8}\right)^2\frac{16}{48}+\left(3-\frac{17}{8}\right)^2\frac{19}{48}\\ & = \left(\frac{81}{64}\right)\frac{13}{48}+\left(\frac{1}{64}\right)\frac{16}{48}+\left(\frac{49}{64}\right)\frac{19}{48} \approx 0.651. \end{align}

## Continuous Conditional Functions

If X and Y have a joint probability density function $$\text{f}(\text{x},\text{y})$$, then the conditional probability density function of X given that $$\text{Y}=\text{y}$$ is defined, for all values of y such that $$\text{f}_\text{y}(\text{y})>0$$, by

$$\text{g}\left(\text{x}|\text{y}\right)=\frac{\text{f}\left(\text{x},\text{y}\right)}{\text{f}_{\text{Y}}(\text{y})}$$

Similarly, the conditional probability density function of Y given X, $$\text{h}(\text{y}|\text{x})$$, similarly. Such that;/p>

$$\text{h}\left(\text{y}|\text{x}\right)=\frac{\text{f}\left(\text{x},\text{y}\right)}{\text{f}_\text{X}\left(\text{x}\right)}$$

### Example: Continuous Conditional Functions

The random variables X and Y have a joint density function:

$$\text{f}\left(\text{x},\text{y}\right)=\frac{1}{20}\left(2\text{x}+5\text{y}\right) \ \ \ 0 < \text{x} < 2,\ 0<\text{y}$$

Find the conditional density function of X given $$\text{Y}=\text{y}$$

#### Solution

We know that:

$$\text{g}\left(\text{x}|\text{y}\right)=\frac{\text{f}\left(\text{x},\text{y}\right)}{\text{f}_\text{Y}(\text{y})}$$

Now,

\begin{align*} \text{f}_\text{Y}\left(\text{y}\right)&=\int_{-\infty}^{\infty}{\text{f}\left(\text{x},\text{y}\right)\text{dx},\ \ \ \ \ \ \ \text{y}\epsilon \text{S}_\text{y}}, \\ &=\int_{\text{x}=0}^{2}{\frac{1}{20}(2\text{x}+5\text{y})\text{dx}} \\ &=\frac{1}{20}\left[\frac{2}{2}\text{x}^2+5\text{xy}\right]_{\text{x}=0}^2 \\ &=\frac{1}{20}(\left[2^2+5\text{y}\times2\right]-{[2}^0+5\text{y}\times0]) \\ &=(4+10\text{y})/20 \\ \therefore \text{f}_\text{Y}\left(\text{y}\right) & =\frac{4+10\text{y}}{20}=\frac{2+5\text{y}}{10} \end{align*}

So,

\begin{align*} \text{g}\left(\text{x}|\text{y}\right)&=\frac{\frac{1}{20}\left(2\text{x}+5\text{y}\right)}{\frac{1}{20}\left(4+10\text{y}\right)}=\frac{2\text{x}+5\text{y}}{2(2+5\text{y})} \\ &=\frac{2\text{x}+5\text{y}}{2(2\text{x}+5\text{y})},\ \ 0<\text{x} < 2 \end{align*}

The conditional pdf of Y given that $$\text{X}=\text{x}$$ is;

$$\text{h}\left(\text{y}|\text{x}\right)=\frac{\text{f}_{\text{XY}}\left(\text{x},\text{y}\right)}{\text{f}_\text{X}\left(\text{x}\right)},\ \ \ \ \ \text{ provided that } \text{f}_\text{X}\left(\text{x}\right)>0;$$

Learning Outcome

Topic 3.b: Multivariate Random Variables – Determine conditional and marginal probability functions, probability density functions, and cumulative distribution functions.


text:
# Homework Help: Still don't get Impulse

1. Dec 10, 2008

### harvellt

1. The problem statement, all variables and given/known data
A ball is moving toward a batter at 32m/s and weighs 0.15kg. After the batter hits it goes 105.0m at an initial angle of 52 degrees. What is the magnitude of the impulse the ball receives in the collision with the bat, and what is the direction of the impulse?

[b2. Relevant equations[/b]
I=MV+MV'

Xmax=(2V2 sin $$\phi$$ cos $$\phi$$)/g

3. The attempt at a solution
I am having trouble finding the initial velocity of the ball after the collision. I tried using the second equation and found the velocity to be 32m/s. Is it wrong to use that formula to find the original velocity?

2. Dec 10, 2008

### mukundpa

What it means ? velocity of displacement?

3. Dec 10, 2008

### harvellt

The ball leaves the bat at a angle of 52 degrees and travels a distance of 105 meters befor it lands, neglecting the height of the collision.

4. Dec 10, 2008

### mukundpa

OK, now can you find the velocity of projection. you have the angle of projection and the range.

5. Dec 10, 2008

### rl.bhat

Use projectile equation. In that angle of projection is 52 degree and range is 105 m. And find velocity of the ball with which it leaves the bat.

6. Dec 10, 2008

### rl.bhat

Yes. Velocity of the ball hitting the bat and leaving the bat is nearly equal. (32 m/s and 32.6 m/s). Now what is angle with which the ball hits the bat?

Last edited: Dec 10, 2008
7. Dec 10, 2008

### mukundpa

yes using equation of X maximum you can find the

8. Dec 10, 2008

### harvellt

Thank you all so much I got it my major problem after I found the V0 was a sign mistake. I forgot the fact that the x velocity needed to have opposite signs, and now I have it.


text:
# Formulate mathematical equation

I have a function:

$$f(n) = 1 - \left(\frac{1}{n}\right)$$

I want to add another variable $k$ to this equation such that: if $k=n$ , then $f(n,k)=1$
if $n=1$, then $f(n,k)=0$ How can we do that ?

-

Consider the function

$g(n,k) = 1 - \frac{1}{n} + \frac{1}{k}$

Then if $k = n$, you have $g(n,k) = 1$. (Note that $n$ and $k$ can not be equal to $0$.)

As Ross Millikan suggested, for your new question, you can consider the function:

$$h(n,k) = \begin{cases} g(n,k) & \quad n \neq 1 \\ 0 & \quad n = 1 \end{cases}$$

$h$ is a perfectly good function and satisfy the property that $n \neq 1$ and $k = n$, then $h(n,k) = 1$; and if $n = 1$, then for all $k$, $h(n,k) = 0$.

-
I am sorry I did not notice your answer, I've changed the question. I need also if n=1 , then f(n,k) = 0 –  M.A Aug 24 '12 at 3:45
@Mohammed: What is wrong with if n=1, f(n,k)=0, else what William proposed? –  Ross Millikan Aug 24 '12 at 4:06
if n=1 and k=50 for example , it do not return 0 –  M.A Aug 24 '12 at 4:08
@Mohammed: my if statement will solve that. Are $n$ and $k$ naturals or reals? +1 for William –  Ross Millikan Aug 24 '12 at 4:24

For the first part, a key point is that if $ab=0$, then $a=0$ or $b=0$. So this suggests that whatever you add on should be multiplied by what you already have, since the equation as you give it solves $f(1,k)=0$. Multiplying all that by something won't change the result.

Similarly, for the second part, what simple relationship gives $1$ when two numbers are equal? There's a basic operation that does exactly what you want for that part. Stick them together and you should have a suitable solution.

-

Assuming that the following statement will never be true - n=k=1 - you can also do this without an "if" statement. Try $(1-\frac{1}{n})^{n-k}$. If n=1, then the fraction $\frac{1}{n}$ will be $\frac{1}{1}=1$, and $1-1=0$. $0^{anything} = 0$. The other issue is if $n=k$. In this case, $n-k = 0$ and $anything^0 = 1$.

I think this equation is also good because at $f(1,1)$, your constraints have given both of the values $0$ and $1$, so it is undefined at $f(1,1)$. At $f(1,1)$, my equation gives you $0^0$, which is undefined.

-
your function works well , but It did not return a value between [0,1] –  M.A Aug 25 '12 at 14:24
@Mohammed None of these answers do, if $n,k \in \mathbb{R}$.For the other answer, take n=.00001. –  mathguy Aug 25 '12 at 14:31
You should specify that constraint in your question. –  mathguy Aug 25 '12 at 14:32


text:
8:04 AM
@JohnRennie Good Morning sir 😊 How are you?

@user8718165 hi :-)

@JohnRennie hello sir, can I ask you a little qn? I asked about it a few days ago..but a little different one sir.

8:22 AM
@user8718165 I'm busy for about 15 minutes more. Then I'll be happy to answer.

@JohnRennie sure sir :-) No worries...

8:37 AM
@user8718165 OK. I'm just going to make a coffee, then I'm free.

@JohnRennie hello sir. Sure :-)

@user8718165 hi

@JohnRennie hello sir...Did you have your coffee? :)

8:52 AM
Yes :-)

@JohnRennie sir suppose a +ve ion and an e- are initially at rest in a vacuum separated by distance d with the COM at the origin and then they accelerate towards the origin...upon combining what will happen sir?

Unless they emit a photon to carry away the excess energy they can't combine. They will just fly past each other and travel back out to the same distance $d$.

@JohnRennie what sir? I mean how can they again fly back? don't they attract?
@JohnRennie hello sir

As the ion and electron move towards each other their total energy has to start constant.
As they approach each other the PE falls (becomes more negative) so the KE increases. When they are very close they both have a large KE i.e. they are moving rapidly.
So they fly past each other and out the other side.

@JohnRennie okay sir...got it...that's it
@JohnRennie Thank you very much sir :)
@JohnRennie sir...isn't there the slightest chance that they hit and break apart? just asking.

9:09 AM
Remember that these are quantum particles so they are not little spheres. They are more like fuzzy clouds and those clouds can pass through each other.

@JohnRennie 😊😊
@JohnRennie thank you very much sir...

They can interact. For example they can interact and emit a photon. That would carry away the excess energy and allow the neutral atom to form.

@JohnRennie got it sir...thank you very much

1 hour later…
10:34 AM
Hi sir :) -- Are mass and moment of inertia the only types of inertia or are there even more types?

You should avoid using the term inertia as it isn't well defined in physics.

@JohnRennie Ok sir :)
I was unaware of this fact.

In general life people tend to use the term inertia to mean that quality that makes a body hard to get moving or hard to stop if it is already moving.
In physics that property is just mass.
Or for rotational motion moment of inertia - yes, the word inertia is used there, but that's a historical accident :-)

@JohnRennie Thank you sir. As far as I know only linear and rotational motions are possible and so only linear and rotational analogues of mass are present. Are there even more types of motion in classical mechanics sir?

I can't think of anything else ...

10:41 AM
@JohnRennie Ok sir. Thank you. :)

11:08 AM
hello sir @JohnRennie

@user8718165 hi :-)

@JohnRennie sir I'm thinking of making a little rc heli a month or so later...do you think its reasonable?

@user8718165 from a kit, or making it from scratch?

@JohnRennie scratch sir...

@user8718165 wow, tht's a big project ...

11:18 AM
@JohnRennie I mean sir...I'll buy the parts and then assemble them.
@JohnRennie sir let's chat in the other room :)

11:36 AM
@JohnRennie, Hi sir. If possible could you please clarify one doubt from Ellipse?

@Intellex Yes?

@JohnRennie, Thank you sir :)
0

The following statement is given in my book under the topic Tangents to an Ellipse: The eccentric angles of the points of contact of two parallel tangents differ by $\pi$ In case of a circle, it is easy for me to visualise that two parallel tangents meet the circle at two points which are a...

Symmetry

@JohnRennie Could you please explain this point sir?

The ellipse is symmetric about its centre i.e. if you invert it though the centre it is unchanged.
That means the gradients at opposite points has to be the same.

11:40 AM
@JohnRennie I understand that the line joining the points will pass through the centre. But the eccentric angle is defined not wrt to the ellipse but wrt to the auxiliary circle and so I am unable to understand this sir.
For your reference, Auxiliary circle is the circle with diameter as the major axis of the ellipse sir.

@Intellex The point opposite to $(x,y)$ on the ellipse is $(-x,-y)$. Yes?

Yes sir.

And the angle is defined by:
$$\tan \theta = \frac{ay}{bx}$$

Understood till this sir.

Assuming we parameterise using $x = a\cos\theta$ and $y = b\sin\theta$
So the question is how does $\tan\theta$ change under the transformation $x\to -x, y \to -y$ ?

11:46 AM
@JohnRennie Thank you sir. This transformation doesn't make any change for $\tan\theta$ as the angles will lie in opposite quadrants. For example if $\theta$ is in first quadrant then the other will lie in the third. And so no change. Understood now sir :)

:-)

Is it not possible to upvote chat messages :) ?

No :-)
But that's OK. I have lots of rep already.

@JohnRennie Fine. Thank you again, sir :)

6 hours later…
5:52 PM
hello @JohnRennie sir...


text:
# Figure 6

 Ratio of kaon, proton and \pphi integrated yields to pion integrated yield as a function of the charged-particle multiplicity density for \xexe collisions at \snnXeXe and \pbpb collisions at \snnT~ and 5.02~TeV~ The statistical and systematic uncertainties are shown as error bars and boxes around the data points Predictions from the canonical statistical model (CSM) are shown as bands considering different correlation volumes~ (based on ) and chemical freeze-out temperatures~ The correlation volume indicates the volume over which the strangeness conservation is imposed


text:
# How to get the variance of the residual variance in a simple linear regression model

How do you get the variance of the residual variance in a simple linear regression model? In a book I see that ${\rm Var}(S_R^2)=\frac{2\sigma^4}{n-2}$, where $S_R^2=\sum\frac{e_i^2}{n-2}$ and $e_i=y_i-\hat{y}_i$. How can I get this result?

I already know that $\frac{\sum e_i^2}{\sigma^2}\sim\chi^2_{n-2}$, but I need to know how I get the variance of $S_R^2$.

• Could you add the self-study tag? I think it applies here. – Xi'an Feb 4 '15 at 20:24
• Hint: the variance of a $\chi^2_{n-2}$ distribution is $2(n-2)$. That's the only additional information you need. – whuber Feb 4 '15 at 20:24
• Have you visited the Wikipedia page on the $\chi^2$ distribution? It contains all the relevant items of information. – Xi'an Feb 4 '15 at 20:25
• @will198, if you've figured it out, why not answer your own question? – gung Feb 4 '15 at 20:38
• Maybe you could post this as an answer instead of a comment. – Richard Hardy Feb 4 '15 at 20:51

• $\frac{\sum e_i^2}{\sigma ^2}\sim \chi^2_{n-2}$
• and knowing that $var(\chi^2_{n-2}=2(n−2)$
• I can get that $var[\frac{\sum e_i^2}{\sigma ^2}]=2(n−2)$ dividing for $(n−2)^2$ and multipliying for $\sigma^4$ I get $var[\frac{\sum e_i^2}{(n-2)}]=\frac{2\sigma^4}{n-2}=var[S^2_R]$. I applied that $a^2 \cdot Var(x)=Var[a \cdot x]$


text:
# Re-parameterization of an asymmetric s-shaped function

Is anyone aware of a re-parameterization of any asymmetric s-shaped function (like, but not necessarily the 5 parameter logistic curve), where one of the parameters is the first inflection point of the first derivative (i.e. the maximum of the second derivative).

I mean point 1 in the upper figure:

(The picture shows the point mentioned above for the of case of a symmetric logistic function.)

So far, I had a look at various references (among others Ratkowsky, D. A. (1990). Handbook of Nonlinear Regression Models. New York, Dekker). However, I have only found parameterizations, where one of the parameters is the inflection point of the function (point 2 in the upper figure).

Unfortunately, calculating this point after the estimation is not a possible solution in my case as this parameter should be integrated in another equation that is estimated simultaneously.

• Two comments: (i) In general it is not possible to reparameterise any asymmetric s-shaped curve in terms of this three points due to identifiability (e.g. for a one-parameter curve, these three points are a function of it); (ii) There may be many cases in which it is possible but these points cannot be found in closed-form, making the reparameterisation impractical. – user10525 Aug 14 '12 at 19:32

Pace Procrastinator, this sort of thing can be done.

Consider the five-parameter logistic model. It has many parameterizations, but it's simple and natural to reduce them to something like

$$y = \nu +\tau \left(1+e^{\frac{x-\mu }{\sigma }}\right)^{-\rho }$$

with $\rho \gt 0$. We can interpret $\mu$ and $\nu$ as $x$ and $y$ locations and $\sigma$ and $\tau$ as $x$ and $y$ scales; $\rho$ is the shape or asymmetry parameter.

Let $x_{+}$ be the location of one extremum of the second derivative and $x_{-}$ be the location of the other extremum. Then, solving for the zeros of the third derivative, I find them at

$$x_{\pm} = \mu + \sigma \log \left(\frac{(3 \rho +1) \pm\sqrt{(\rho +1) (5 \rho +1) }}{2 \rho ^2}\right).$$

Whence, setting

$$\kappa = \exp \frac{x_{+}-x_{-}}{\sigma}$$

we find

$$\rho = \frac{3 \kappa \pm\sqrt{\kappa ^3+2 \kappa ^2+\kappa }}{\kappa ^2-7 \kappa +1},$$

taking the positive sign when $\kappa$ exceeds the larger root of $1-7x+x^2=0$ (about $6.8541$) and the negative sign when $\kappa$ is less than the smaller root (about $0.145898$)--other values of $\kappa$ will not give a sigmoidal curve--and

$$\mu = \frac{x_{+} + x_{-}}{2} + \sigma \log(\rho).$$

This allows a parameterization in terms of $(\sigma, \nu, \tau, x_{-}, x_{+})$ (with some significant restrictions on $\sigma$ needed to make the results valid).

Here is a plot of $y$ (in blue) and its third derivative (in red) based on these formulas with the parameters set to $(-1/4, 1/2, 1, 1, 0)$:

Indeed, the ascending zero of the third derivative occurs at $1$ and the descending zero at $0$, as specified.

This parameterization is not necessarily so messy. If your data are within a narrow range, for instance, the asymptotic values might not matter and you could just use a suitable polynomial. Without knowing more about the particular problem, it's hard to provide a specific recommendation.

• Thanks for your help. As I am not sure about the parameter interpretation in your formula. Do you have any reference for me? I am curently using the following parameterization of the Richards curve, because of its interpretability of the parameters: en.wikipedia.org/wiki/Generalised_logistic_function If I would put the term you derived for c in the very first equation, this seems pretty messy to estimate, right? Do you see any potential for simplification? – majom Aug 15 '12 at 1:26
• BTW: In the very end, I aim to be able to use this function to estimate a nonlinear mixed effects model. Thereby, alpha and the coefficient for the first inflection point of the first derivative should be allowed to vary on the group-level. – majom Aug 15 '12 at 1:28
• @whuber Not in general (which is my point btw), it is easy to prove because there are no diffeomorphisms for larger dimensions to smaller dimensions (or viceversa). In the particular case you explain, I agree. (+1) – user10525 Aug 15 '12 at 9:16
• @Proc That's a good point of view, but I think it does not apply here. The question asks for a change of coordinates for a five-dimensional manifold in which two of the five coordinate functions are prespecified, that's all. Locally and generically there will be a large family of answers. – whuber Aug 15 '12 at 12:49
• Sorry that it took me a while to look into this. So far your solution worked great for a toy example, which I created for test purposes. Let's see if it also converges in a real world setting. Again, thanks for your help. – majom Sep 3 '12 at 7:18


text:
GATE ME
Strength of Materials
Pure Bending
Previous Years Questions

## Marks 1

Consider a beam with circular cross-section of diameter $$d.$$ The ratio of the second moment of area about the neutral axis to the section modulus of...
The cross-sections of two solid bars made of the same material are shown in the figure. The square cross-section has flexural (bending) rigidity $${{\... The beams, one having square cross section and another circular cross-section,are subjected to the same amount of bending moment. If the cross section... ## Marks 2 A simply-supported beam of length$$3L$$is subjected to the loading shown in the figure. It is given that$$P = 1N,L = 1m$$and Youn... For the component loaded with a force$$F$$as shown in the figure, the axial stress at the corner point$$P$$is ... A simply supported beam of span length$$6$$m and$$75$$mm diameter carries a udl of$$1.5$$kN/m What is the maximum value of bending moment? A simply supported beam of span length$$6$$m and$$75$$mm diameter carries a udl of$$1.5$$kN/m What is the maximum value of bending stress?... A mass less beam has a loading pattern as shown in figure. The beam is rectangular cross-section with a width of 30 mm and height of 100 mm. The ma... A cantilever beam has the square cross section of 10 mm$$ \times $$10 mm. It carries a transverse load of 10 N. Considering only the bottom fibres o... A mass less beam has a loading pattern as shown in figure. The beam is rectangular cross-section with a width of 30 mm and height of 100 mm. The ma... A tapered cantilever beam of constant thickness is loaded as shown in the sketch below. The bending stress will be: ... A 6 metre long supported wooden beam of rectangular section 10 cm$$ \times $$20 cm deep is strengthened by mild steel plates 0.5 cm$$ \times  10 ...
EXAM MAP
Joint Entrance Examination


text:
# Article

Full entry | PDF   (0.2 MB)
Keywords:
generalized $MV$-algebra; representability; congruence relation; unital lattice ordered group
Summary:
A generalized $MV$-algebra $\mathcal A$ is called representable if it is a subdirect product of linearly ordered generalized $MV$-algebras. Let $S$ be the system of all congruence relations $\rho$ on $\mathcal A$ such that the quotient algebra $\mathcal A/\rho$ is representable. In the present paper we prove that the system $S$ has a least element.
References:
[1] B. Banaschewski: On lattice-ordered groups. Fund. Math. 55 (1964), 113–122. MR 0168672 | Zbl 0129.01803
[2] G. Birkhoff: Lattice Theory. Third Edition, Providence, 1967. MR 0227053 | Zbl 0153.02501
[3] P. Conrad: Lattice Ordered Groups. Tulane University, 1970. Zbl 0258.06011
[4] A. Dvurečenskij,: Pseudo MV-algebras are intervals of $\ell$-groups. J. Austral. Math. Soc. 72 (2002), 427–445. DOI 10.1017/S1446788700036806 | MR 1902211
[5] A. Dvurečenskij, S. Pulmannová: New Trends in Quantum Structures. Kluwer Academic Publishers, Dordrecht, 2000. MR 1861369
[6] G. Georgescu, A. Iorgulescu: Pseudo $MV$-algebras: a noncommutative extension of $MV$-algebras. In: The Proceedings of the Fourth International Symposium on Economic Informatics, INFOREC, Bucharest, 6–9 May, Romania, 1999, pp. 961–968. MR 1730100
[7] G. Georgescu, A. Iorgulescu: Pseudo $MV$-algebras. Multiple-Valued Logic 6 (2001), 95–135. MR 1817439
[8] J. Jakubík: Normal prime filters of a lattice ordered group. Czech. Math. J. 24 (1974), 91–96. MR 0347702
[9] J. Jakubík: Subdirect product decompositions of MV-algebras. Czech. Math. J. 49 (1999), 163–173. DOI 10.1023/A:1022472528113 | MR 1676813
[10] J. Rachůnek: A non-commutative generalization of $MV$-algebras. Czech. Math. J. 52 (2002), 255–273. DOI 10.1023/A:1021766309509

Partner of


text:
# Characteristics - What's Going On?

1. Aug 16, 2013

### bolbteppa

What's really going on as regards characterstics?

For second order PDE's, in this video in which characteristics were defined as lines in the domain along which the highest order partial derivatives were discontinuous, then using this idea, one can naturally use matrices & determinants to derive the conditions under which the second order terms are equal to zero - deriving the $B^2 - 4AC$ classification of partial differential equations.

For those who are interested, the method is that given

$\mathcal{L}[\phi] = A \phi_{xx} + B \phi_{xy} + C \phi_{yy} + H(x,y,\phi, \phi_x, \phi_y) = 0$

we basically want $\vec{v} = (\frac{\partial ^2 \phi}{\partial x^2},\frac{\partial ^2 \phi}{\partial x \partial y},\frac{\partial ^2 \phi}{\partial y^2}) = (0,0,0)$ & since this is a vector, we construct an equation of the form $\mathcal{T}(\vec{v}) = \vec{w}$ & examine the case where a solution $\vec{v}$ doesn't exist, i.e. a case where an infinite number of solutions exist.

Using

$$d(\frac{\partial \phi}{\partial x}) = \frac{\partial}{\partial x} \frac{\partial \phi}{\partial x}dx + \frac{\partial }{\partial y} \frac{\partial \phi}{\partial x} dy = \frac{\partial ^2 \phi}{\partial x^2 }dx + \frac{\partial ^2 \phi}{\partial y \partial x} dy$$

$$d(\frac{\partial \phi}{\partial y}) = \frac{\partial}{\partial x} \frac{\partial \phi}{\partial y}dx + \frac{\partial }{\partial y} \frac{\partial \phi}{\partial y} dy = \frac{\partial ^2 \phi}{\partial x\partial y }dx + \frac{\partial ^2 \phi}{ \partial ^2 y} dy$$

we arrive at

$$\mathcal{T}(\vec{v}) = \vec{w} \rightarrow \left[ \begin{matrix} A & B & C \\ dx & dy & 0 \\ 0 & dx & dy \end{matrix} \right] \left[ \begin{matrix} \phi_{xx} \\ \phi_{xy} \\ \phi_{yy} \end{matrix} \right] = \left[ \begin{matrix} -H \\ d( \frac{\partial \phi}{\partial x}) \\ d( \frac{\partial \phi}{\partial y}) \end{matrix} \right]$$

so that $\det{\mathcal{T}} = Ady^2 - Bdxdy + Cdx^2 = 0 \rightarrow A(\frac{dy}{dx})^2 - B(\frac{dy}{dx}) + C = 0$ implies $\frac{dy}{dx} = \frac{B \pm \sqrt{B^2 - 4AC}}{2A}$
These are the differential equations of the characteristics, splitting into three cases: two real equations, one real equation or two complex equations... Along the characteristics we then have to solve $H(x,y,\phi, \phi_x, \phi_y) = 0$, & my guess is that the solution for this may not be the same solution for the general second order PDE, thus there will be discontinuities in the solution?

But then in this note characteristics were those curves along which the coefficient of the first term vanishes, so that a power series solution may be constructed (is that tiny discussion given an actual proof of the Cauchy-Kovalevsky theorem, just without ε-δ?'s???). Here we apparently have two conflicting descriptions of what a characteristic is, definitions that don't seem to merge as equivalent.

Thirdly, in this note characteristics are shown to arise via a change of variables aimed at eliminating higher order derivatives, how does a change of variables relate to all of this? I imagine it's something to do with choosing coordinates so that the characteristics in the domain look nicer, but I have no concrete idea as to how it applies geometrically...

Finally, how does all of this, whatever the right way of looking at it actually is, relate to characteristics of first order PDE's exactly? On pages 5 - 6 a nice geometric interpretation of first order characteristics is given, but it looks like it has nothing to do with what has been described above.

What's going on?

2. Aug 19, 2013

### lurflurf

The basic idea is we can factor a partial differential equation into ordinary differential equations. A usual example is
$$\dfrac{\partial ^2}{\partial t^2}-c^2\dfrac{\partial^2}{\partial x^2}=\left( \dfrac{\partial }{\partial t}-c\dfrac{\partial}{\partial x}\right) \left( \dfrac{\partial }{\partial t}+c\dfrac{\partial}{\partial x}\right)=\dfrac{\partial }{\partial u}\dfrac{\partial }{\partial v} \\ \text{where} \\ \dfrac{\partial }{\partial u}=\dfrac{\partial }{\partial t}-c\dfrac{\partial}{\partial x} \\ \dfrac{\partial }{\partial v}=\dfrac{\partial }{\partial t}+c\dfrac{\partial}{\partial x} \\$$
Thus we have reduced the solution of a partial differential equation to the solution of two ordinary differential equations.

Last edited: Aug 19, 2013
3. Aug 19, 2013

### bolbteppa

Thanks for the input, unfortunately I don't yet see how that explains the situation as regards characteristics. In one case characteristics are defined by the fact that the highest partials are all simultaneously zero, in the next they are defined by the fact that (only one of? or all of) the coefficients are zero. If both conditions hold then it looks like you'd get two sets of completely different characteristics, one associated to the coefficients, the other to the partials... How a change of variables affects this is beyond me.

Also, I don't really understand why it is that the characteristics being parallel to the solution curve is a bad thing? In the picture I linked to above for first order PDE's, surely the fact you know your initial curve lies along the characteristic curve gives you the information you get otherwise thinking of normal vectors - no? Further, how does this picture tie into 2nd order PDE's? It just doesn't make sense to me because even the basic definitions seem to contradict or go against one another...

Finally, I don't have any intuitive feel for why curves partitioning the domain lead to such drastically different solutions, i.e. why the characteristics cause such drastically different solutions


text:
## Problem of the week

### 2022-24 Hey, who turned out the lights?

There are light bulbs $$\ell_1,\dots, \ell_n$$ controlled by the switches $$s_1, \dots, s_n$$. The $$i$$th switch flips the status of the $$i$$th light and possibly others as well. If $$s_i$$ flips the status of $$\ell_j$$, then $$s_j$$ flips the status of $$\ell_i$$. All lights are initially off. Prove that it is possible to turn all the lights on.


text:
# How to show that all roots of the derivative of a polynomial are real

Case 1: all roots are distinct

In this case, assume $f$ has total of $n$ real roots. I can take any consecutive pair of roots $a,b$, $a < b$ and say that since $f(a) = f(b) = 0$, using Rolle's theorem, $\exists c \in \mathbb{R}$ such that $f'(c) = 0, a < c < b$. Then I end up with $n-1$ real roots in $f'$. How do I show that $f'$ has total of $n-1$ roots?

Case 2: repeated roots

Then for some order $k$ at $x_0$, $f(x) = (x-x_0)^kq(x)$ where $q(x_0)\neq 0$. Then $f'(x) = k(x-x_0)^{k-1}q(x) + (x-x_0)^k q'(x)$. How many root are there in $f'$ and how do I show that they are all real?

• What is the assumption on$f$?. It is not true that roots of the derivative of any polynomial are real. Ex. $f(x)=3x+x^{3}$. – Kavi Rama Murthy Apr 30 '18 at 9:35
• that $f$ is a polynomial with real roots – MoneyBall Apr 30 '18 at 9:36

There is a theorem of Guass and Lucas which says that for any polynomial $p$ the roots of $p'$ are all convex combinations of the roots of $p$.