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# In the past 50 years, the population of honeybees in the

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Re: In the past 50 years, the population of honeybees in the [#permalink]

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14 Nov 2012, 09:07
Vips0000 wrote:
Jp27 wrote:
Vips0000 wrote:

True, option A looks for potential alternatives for pollination. But it looks for "insect pollination". But even if we find insect X, Y or Z - we still need to evaluate and find out if those insects would not be affected because of pesticides. This definitely can not help in evaluating the argument.
As far as D is concerned - it is not talking about stopping the trend - but it talks about reversal of trend. What it means is if we reduce pesticide bee population will increase -> which shows pollination would happen again -> consumer can get fruits. So it weakens argument when answered in affirmative.

The OE is definitely not correct.

I guess D is correct answer because the conclusion states "Therefore, if the honey bee population continues this drastic decline, then most fruits will no longer be available to consumers."
That is -> the conclusion is contingent on the basis only when "honey bee population continues this drastic decline". Per D, if we can reverse or not reverse the trend is irrelevant to issue of getting fruits (given the bees continue to decline)
In other words, we have evaluate all the answer choice considering what will happen if the trend continues.

Makes sense?

Cheers

cheers

Not really!

Yes conclusion is -
"if bee population continues to decline-> most fruits will no longer be available to consumers"

most basic logical meaning of this?
if population doesnt decline (or increase -reversing the trend!) -> fruits may be available to consumers.
Which is option D. So it matters!

This (bolded) is wrong logic
If I'm in Pasir Ris New Town then I'm in Singapore.
From this you conclude ->
If Im NOT in Pasir Ris New Town then I'm NOT in Singapore -

Clearly wrong. I could be at some other place in Singapore such as Ang Mo Kio
Correct Inference is
If im NOT in Singapore then I'm Definitely NOT in Pasir Ris New Town -> this is the correct inference.

Now?
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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14 Nov 2012, 09:25
Jp27 wrote:

This (bolded) is wrong logic
If I'm in Pasir Ris New Town then I'm in Singapore.
From this you conclude ->
If Im NOT in Pasir Ris New Town then I'm NOT in Singapore -

Clearly wrong. I could be at some other place in Singapore such as Ang Mo Kio
Correct Inference is
If im NOT in Singapore then I'm Definitely NOT in Pasir Ris New Town -> this is the correct inference.

Now?

Sorry dude.. but these examples are like comparing apples to oranges.

if bee population continues to decline-> most fruits will no longer be available to consumers
This is cause and its effect

While this one is not ->
If I'm in Pasir Ris New Town then I'm in Singapore.
It is just logical deduction.
It will be negated as :
If i'm not in singapore then I'm not in pasir ris new town.

I'll give u one more example -

Cause and effect:
If it rains, I'll get wet
If it doesnt rain - I'll not get wet!

Logical deduction-
Mumbai is in india.
I'm in Mumbai -> then I'm in india
In this case negation can not happen as u did, It can be done as ->
If i'm not in india -> then i'm not in mumbai.
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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14 Nov 2012, 09:36
Vips0000 wrote:
Jp27 wrote:

This (bolded) is wrong logic
If I'm in Pasir Ris New Town then I'm in Singapore.
From this you conclude ->
If Im NOT in Pasir Ris New Town then I'm NOT in Singapore -

Clearly wrong. I could be at some other place in Singapore such as Ang Mo Kio
Correct Inference is
If im NOT in Singapore then I'm Definitely NOT in Pasir Ris New Town -> this is the correct inference.

Now?

Sorry dude.. but these examples are like comparing apples to oranges.

if bee population continues to decline-> most fruits will no longer be available to consumers
This is cause and its effect

While this one is not ->
If I'm in Pasir Ris New Town then I'm in Singapore.
It is just logical deduction.
It will be negated as :
If i'm not in singapore then I'm not in pasir ris new town.

I'll give u one more example -

Cause and effect:
If it rains, I'll get wet
If it doesnt rain - I'll not get wet!

Logical deduction-
Mumbai is in india.
I'm in Mumbai -> then I'm in india
In this case negation can not happen as u did, It can be done as ->
If i'm not in india -> then i'm not in mumbai.

No dude. When you start a statement as you have done here, with IF...

Vips0000 wrote:
Cause and effect:
If it rains, I'll get wet
If it doesnt rain - I'll not get wet!

Then IT'S ALWAYS CONDITIONAL NOT CAUSALITY. you can still wet when someone pours water on you!

If I'm in Pasir Ris New Town then I'm in Singapore.
If im NOT in Singapore then I'm Definitely NOT in Pasir Ris New Town

Mumbai is in india.
If i'm not in india -> then i'm not in mumbai.

I don't see any difference between two. but you claim otherwise.

Cheers
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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14 Nov 2012, 09:41
JP..
I hope you are answering D in the right manner.
D says : Will reducing the use of pesticides in the United States reverse the decline in honeybee populations?
You will answer in either yes or no, so as to get the contrasting solutions.
If the answer is yes, then reducing the use of pesticides in US reverses the decline in honeybee population. RESULT- There will not be any further decline in honeybee population. Hence there will not be any decline in the population of major pollinators of fruits and hence fruits can be made avaliable to the consumers in US. Acts as a weakener.
If the answer is no, then reducing the use of pesticides in US will not reverse the decline in honeybee population. RESULT:- The decline will continue and hence consumers will not be getting fruits. Acts as a strengthener.
Now this makes sense
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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14 Nov 2012, 09:49
Jp27 wrote:
If I'm in Pasir Ris New Town then I'm in Singapore.
If im NOT in Singapore then I'm Definitely NOT in Pasir Ris New Town

Mumbai is in india.
If i'm not in india -> then i'm not in mumbai.

I don't see any difference between two. but you claim otherwise.

Cheers

Difference is in these statements: -

If statement is:
Mumbai is in india.

Logical deduction->
If i'm in mubai ->then i'm in india

Correct Negation
If i'm not in india -> then i'm not in mumbai.
Incorrect Negation
If i'm not in mumbai-> then i'm not in india

Having explained that - I think both of us are not going to be convinced. I rest my case
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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14 Nov 2012, 09:50
Hey marcab,
You too just proved D matters in evaluating the argument. Question asks for "least" case.
JP?
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Last edited by Vips0000 on 14 Nov 2012, 10:20, edited 1 time in total.
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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14 Nov 2012, 10:16
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Vips0000 wrote:
nelz007 wrote:
In the past 50 years, the population of honeybees in the United States has been cut in half. The decline is due primarily to the increasing use of pesticides in the United States, as well as to the introduction of two types of mites that weaken and kill the bees. Honeybees are the primary pollinators for a variety of important fruit crops, including oranges, apples, grapes, peaches, cranberries and watermelons. Therefore, if the honey bee population continues this drastic decline, then most fruits will no longer be available to consumers.

In evaluating the conclusion, which of the following questions would be LEAST useful to answer?

A Are there other insect pollinators that could pollinate these fruit crops instead of the honeybee?
B Are honeybee populations declining in other important fruit-producing regions, like Chile and New Zealand?
C Is it feasible for humans to hand-pollinate the fruits that have been pollinated by bees?
D Will reducing the use of pesticides in the United States reverse the decline in honeybee populations?
E Is it possible to genetically engineer fruit-producing plants so that they no longer require pollination?

Conclusion is: if the honey bee population continues this drastic decline, then most fruits will no longer be available to consumers.
Lets see each choice one by one.

A Are there other insect pollinators that could pollinate these fruit crops instead of the honeybee?
Even if the ans to this question is yes, chances are pesticides would eliminate those insects as well. note, we are evaluating this conclusion in the light of decline in bee population due to pesticides.
Therefore yes or no in this question does not matter in evaluating the argument.

B Are honeybee populations declining in other important fruit-producing regions, like Chile and New Zealand?
conclusion is that most fruits will no longer be available to customer. If the ans to this question is no that means there will be pollination and there will be fruits. if ans to this question is yes then we know that conclusion holds good. So this question helps.

C Is it feasible for humans to hand-pollinate the fruits that have been pollinated by bees?
Yes/no to this question provides alternate method of pollination. if yes ->pollination -> fruits. if no -> no pollination -> no fruit. So this question helps.

D Will reducing the use of pesticides in the United States reverse the decline in honeybee populations?
yes would mean if population continues to decline we can reverse the trend.-> continue pollination->fruits!
no would mean we can not reverse the trend -> no pollination->no fruits.

E Is it possible to genetically engineer fruit-producing plants so that they no longer require pollination?[/
in this choice we are just eliminating middle man pollination of our other answer choices.. yes->fruits, no->no fruits.

Hence ans A it is.

No the answer here is D.

You are concerned ONLY of pollination or something related directly to it

The least is something that talks about others stuff.

For the argument at end: pesticide is the LEAST for evaluating the argument because is not the main point

Quote:
In evaluating the conclusion, which of the following questions would be LEAST useful to answer?

A Are there other insect pollinators that could pollinate these fruit crops instead of the honeybee?
B Are honeybee populations declining in other important fruit-producing regions (the pollination is implied), like Chile and New Zealand?
C Is it feasible for humans to hand-pollinate the fruits that have been pollinated by bees?
D Will reducing the use of pesticides in the United States reverse the decline in honeybee populations?
E Is it possible to genetically engineer fruit-producing plants so that they no longer require pollination?

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Re: In the past 50 years, the population of honeybees in the [#permalink]

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14 Nov 2012, 20:23
VeritasPrepKarishma wrote:
nelz007 wrote:
In the past 50 years, the population of honeybees in the United States has been cut in half. The decline is due primarily to the increasing use of pesticides in the United States, as well as to the introduction of two types of mites that weaken and kill the bees. Honeybees are the primary pollinators for a variety of important fruit crops, including oranges, apples, grapes, peaches, cranberries and watermelons. Therefore, if the honey bee population continues this drastic decline, then most fruits will no longer be available to consumers.

In evaluating the conclusion, which of the following questions would be LEAST useful to answer?

A Are there other insect pollinators that could pollinate these fruit crops instead of the honeybee?
B Are honeybee populations declining in other important fruit-producing regions, like Chile and New Zealand?
C Is it feasible for humans to hand-pollinate the fruits that have been pollinated by bees?
D Will reducing the use of pesticides in the United States reverse the decline in honeybee populations?
E Is it possible to genetically engineer fruit-producing plants so that they no longer require pollination?

Responding to 3 pms on this question!

I couldn't recall the honeybee question offhand but was convinced that it must be tricky if people are this unsettled because of it. But mind you, in my opinion, the question is straightforward. All you need to do is focus on what the question is asking you. Do not let emotions interfere!

In all our strengthen/weaken questions we tell you to look at the conclusion. Try to strengthen/weaken that. Here the question clearly asks you to figure out what will not help in evaluating the CONCLUSION.

What is the conclusion?

"If the honey bee population continues this drastic decline, then most fruits will no longer be available to consumers."

The conclusion is that if the honey bee population continues to decline, then consumers will not get fruit. What you have to evaluate is this "If the honeybee population continues to decline, will the consumers get fruit?". Basically, we are wondering whether there are alternative ways of pollination or getting fruit.

A Are there other insect pollinators that could pollinate these fruit crops instead of the honeybee?
Alternative method of pollination

B Are honeybee populations declining in other important fruit-producing regions, like Chile and New Zealand?
Alternative method of obtaining fruit

C Is it feasible for humans to hand-pollinate the fruits that have been pollinated by bees?
Alternative method of pollination

D Will reducing the use of pesticides in the United States reverse the decline in honeybee populations?
Doesn't help in evaluating the conclusion. The conclusion clearly says that if the decline continues, fruits will not be available. We don't have to question whether there are ways to reverse the decline. The point is - if the decline happens, can we do something.

E Is it possible to genetically engineer fruit-producing plants so that they no longer require pollination?
Alternative method of obtaining fruit

Not satisfactory.

How does A give you an alternative for polliation? if bee population is declining due to pesticide, can we turn a blind eye to the fact that pesticides would eliminate insects too? And assume that we got one way of pollination?

Also the statement:
D Will reducing the use of pesticides in the United States reverse the decline in honeybee populations?
Doesn't help in evaluating the conclusion. The conclusion clearly says that if the decline continues, fruits will not be available. We don't have to question whether there are ways to reverse the decline. The point is - if the decline happens, can we do something.

"The conclusion clearly says that if the decline continues, fruits will not be available". yes it does. And that is what we have to evaluate.

We dont have to question if there are ways to reverse the decline. But 'consider'- if there is a method that can 'reverse' (note:reverse, Not STOP) the decline, would it impact the conclusion?
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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14 Nov 2012, 22:18
VeritasPrepKarishma wrote:
nelz007 wrote:
In the past 50 years, the population of honeybees in the United States has been cut in half. The decline is due primarily to the increasing use of pesticides in the United States, as well as to the introduction of two types of mites that weaken and kill the bees. Honeybees are the primary pollinators for a variety of important fruit crops, including oranges, apples, grapes, peaches, cranberries and watermelons. Therefore, if the honey bee population continues this drastic decline, then most fruits will no longer be available to consumers.

In evaluating the conclusion, which of the following questions would be LEAST useful to answer?

A Are there other insect pollinators that could pollinate these fruit crops instead of the honeybee?
B Are honeybee populations declining in other important fruit-producing regions, like Chile and New Zealand?
C Is it feasible for humans to hand-pollinate the fruits that have been pollinated by bees?
D Will reducing the use of pesticides in the United States reverse the decline in honeybee populations?
E Is it possible to genetically engineer fruit-producing plants so that they no longer require pollination?

What is the conclusion?

"If the honey bee population continues this drastic decline, then most fruits will no longer be available to consumers."

The conclusion is that if the honey bee population continues to decline, then consumers will not get fruit. What you have to evaluate is this "If the honeybee population continues to decline, will the consumers get fruit?". Basically, we are wondering whether there are alternative ways of pollination or getting fruit.

D Will reducing the use of pesticides in the United States reverse the decline in honeybee populations?
Doesn't help in evaluating the conclusion. The conclusion clearly says that if the decline continues, fruits will not be available. We don't have to question whether there are ways to reverse the decline. The point is - if the decline happens, can we do something.

Hi karishma..thanks for the reply but still I am not convinced over the explanation of D. Can you please elaborate more?
I still feel that D can be answered in as weakener and strengthener.
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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14 Nov 2012, 23:24

D will do the least to evaluate the conclusion. The catch lies in the conclusion. Lets rephrase the conclusion.

Therefore, if the honey bee population continues this drastic decline, then most fruits will no longer be available to consumers.

It is a prediction based on the condition. We have to evaluate the prediction in the light of this condition only. Thus, the objective is to evaluate whether fruits will be available if the honey bee population declines. We are already told that the Population of honey bee is important for the pollination of fruits so reversing the condition will not help in evaluating the prediction. The reversal of the population is out of scope as we need to check whether the fruits can be made available if the honey bee population keeps declining.
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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15 Nov 2012, 01:27
VeritasPrepKarishma wrote:
Vips0000 wrote:
How does A give you an alternative for polliation? if bee population is declining due to pesticide, can we turn a blind eye to the fact that pesticides would eliminate insects too? And assume that we got one way of pollination?

We do not know whether the insects get affected. It's an 'evaluate' question. You first evaluate whether someone else can pollinate. Once you find someone, then you will evaluate the status of those insects and what affects them etc. You are jumping the gun by saying that the same pesticides will affect every species of insects and hence no other possible pollinator can survive.

Vips0000 wrote:
Also the statement:
D Will reducing the use of pesticides in the United States reverse the decline in honeybee populations?
Doesn't help in evaluating the conclusion. The conclusion clearly says that if the decline continues, fruits will not be available. We don't have to question whether there are ways to reverse the decline. The point is - if the decline happens, can we do something.

"The conclusion clearly says that if the decline continues, fruits will not be available". yes it does. And that is what we have to evaluate.

We dont have to question if there are ways to reverse the decline. But 'consider'- if there is a method that can 'reverse' (note:reverse, Not STOP) the decline, would it impact the conclusion?

Here you haven't understood the conclusion. Say, I give you my opinion "If people keep fighting, the world will end." You need to evaluate my opinion. What will you evaluate? Will you evaluate whether people will keep fighting or will you evaluate what happens when people keep fighting. In my opinion, if people keep fighting, the world will end. You need to find out what happens 'if people keep fighting'. My opinion is based on a condition. If this happens, that will happen. When you evaluate my opinion, you are going to evaluate whether 'that will happen or not if this happens'. You don't have to question whether people will keep fighting or not.
Similarly, the conclusion states 'if honeybee population continues to decline...'
Don't worry about whether actually honeybee population will decline, stay same or increase. Find out what happens if it continues to decline.

Kudos Karishma....That was awesome.
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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15 Nov 2012, 03:50
hey where is my kudos

I said exactly what karishma said, but with fewer words

Thanks Karishma
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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15 Nov 2012, 05:03
This was a good question Thanks for the explanation karishma.
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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22 Nov 2012, 22:58
leigimon wrote:
For D, even if we account for the pesticide we would probably need to account for the mites as well to be definitive.

But I think it comes down to the condition of the argument, that we should accept the decline as given.

I have to agree with the first statement here. Even if pesticide is reduced, you still need to make an additional assumption that the mites don't kill the honeybees as well. D is not a question that can give a direct resolution to the availability of the fruit.
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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25 Nov 2012, 23:22
Dont agree with the OA, OA should be B

B Are honeybee populations declining in other important fruit-producing regions, like Chile and New Zealand? -- We are talking about a trend in United states and this seems to be the least important.

D Will reducing the use of pesticides in the United States reverse the decline in honeybee populations.
The answer to this questions definetely matters.
Yes : Rev the decline - > Pollination inc -> O/p increase
N0 : The same trend - > Reduced pollination -> Reduced O/p

Quote:
Somebody just gave below expln
B Are honeybee populations declining in other important fruit-producing regions, like Chile and New Zealand?
we will get fruits from other counties, who cares abt the bees!

If we were to ignore this honeybee decline altogether there is no point writing this argument at all. Just import fruits from other countries.

The conclusion cares about honey bees.
Therefore, if the honey bee population continues this drastic decline, then most fruits will no longer be available to consumers.
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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26 Nov 2012, 15:48
HiteshPunjabi wrote:
Dont agree with the OA, OA should be B

B Are honeybee populations declining in other important fruit-producing regions, like Chile and New Zealand? -- We are talking about a trend in United states and this seems to be the least important.

D Will reducing the use of pesticides in the United States reverse the decline in honeybee populations.
The answer to this questions definetely matters.
Yes : Rev the decline - > Pollination inc -> O/p increase
N0 : The same trend - > Reduced pollination -> Reduced O/p

Quote:
Somebody just gave below expln
B Are honeybee populations declining in other important fruit-producing regions, like Chile and New Zealand?
we will get fruits from other counties, who cares abt the bees!

If we were to ignore this honeybee decline altogether there is no point writing this argument at all. Just import fruits from other countries.

The conclusion cares about honey bees.
Therefore, if the honey bee population continues this drastic decline, then most fruits will no longer be available to consumers.

If you notice a pattern with the other answer choices (A, C, E), along with B, they all provide an alternative solution that directly addresses that most fruits won't be available to consumers. If you answer Yes to these 4 options, you can solve the problem and more fruits will be available to consumers. However, even if you answer Yes to B, you still have another question (what about the 2 new types of mites).

Also, even if you reduce the use of pesticide, you cannot assume that the population will drastically increase right away, or double to the totals prior to the 50 years. You're still stuck with a limited fruit supply.
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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28 Nov 2012, 22:56
I initially went with A. Seeing Karisma's explanation it is clear. Kudos!
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Re: In the past 50 years, the population of honeybees in the [#permalink]

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06 Dec 2012, 19:09
Expert's post
1
This post was
BOOKMARKED
HiteshPunjabi wrote:
Dont agree with the OA, OA should be B

B Are honeybee populations declining in other important fruit-producing regions, like Chile and New Zealand? -- We are talking about a trend in United states and this seems to be the least important.

D Will reducing the use of pesticides in the United States reverse the decline in honeybee populations.
The answer to this questions definetely matters.
Yes : Rev the decline - > Pollination inc -> O/p increase
N0 : The same trend - > Reduced pollination -> Reduced O/p

Quote:
Somebody just gave below expln
B Are honeybee populations declining in other important fruit-producing regions, like Chile and New Zealand?
we will get fruits from other counties, who cares abt the bees!

If we were to ignore this honeybee decline altogether there is no point writing this argument at all. Just import fruits from other countries.

The conclusion cares about honey bees.
Therefore, if the honey bee population continues this drastic decline, then most fruits will no longer be available to consumers.

Check out the explanation given in this post: http://www.veritasprep.com/blog/2012/11 ... onclusion/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 07 May 2012 Posts: 76 Location: United States Followers: 3 Kudos [?]: 142 [0], given: 23 Re: In the past 50 years, the population of honeybees in the [#permalink] ### Show Tags 07 Nov 2013, 06:16 Hi Karishma, I agree with what you have said above. But the reason I felt A was just as strong as D ,was , since the argument already states that honeybees are the PRIMARY POLLINATORS for these fruits. So we already know that there are other pollinators that could pollinate these fruits. So when you think on these lines, A doesn't really help much , since we already know the answer to this from the argument itself. Although I do agree that D doesn't help in evaluating the argument either . So I cant figure out a way to justifiably pick one over the other. -Jyothi _________________ Jyothi hosamani Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7177 Location: Pune, India Followers: 2161 Kudos [?]: 13987 [0], given: 222 Re: In the past 50 years, the population of honeybees in the [#permalink] ### Show Tags 07 Nov 2013, 21:38 gmacforjyoab wrote: Hi Karishma, I agree with what you have said above. But the reason I felt A was just as strong as D ,was , since the argument already states that honeybees are the PRIMARY POLLINATORS for these fruits. So we already know that there are other pollinators that could pollinate these fruits. So when you think on these lines, A doesn't really help much , since we already know the answer to this from the argument itself. Although I do agree that D doesn't help in evaluating the argument either . So I cant figure out a way to justifiably pick one over the other. -Jyothi The argument only says that the honeybees are primary pollinators for these fruits. It doesn't mean there are other insect pollinators that could be used to pollinate. If honeybees are not there, the primary pollinator will be gone and little pollination may take place through other mediums such as wind/some other insects - we don't know. What (A) is trying to evaluate is whether there are other insect pollinators which could pollinate these crops (the use of 'could' tells you that they probably don't pollinate these crops right now). We are looking for an alternative primary pollinator. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: In the past 50 years, the population of honeybees in the   [#permalink] 07 Nov 2013, 21:38

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text:
Day 12 is about moving a ship across a coordinate plane using directions and a way point that moves and rotates around the ship. There’s a bit of geometry, and I made a really dumb mistake that took me a long time to figure out.

## Challenge

The puzzle can be found here. The puzzle input will look like:

F10
N3
F7
R90
F11

Each line has one of seven actions: move (N)orth, (S)outh, (E)ast, (W)est or (F)orward, or turn (L)eft or (Right) in degrees, and then an integer number of spaces to move or degrees to turn.

For part one, the ship starts facing east (90 degrees), and moving does not change the direction of the ship. I need to find the manhattan distance between the ship’s starting and ending positions.

In part two, the directional move commands are actually for a wayfinding beacon that starts at 1 north and 10 east. The directional commands move the beacon, and the turn commands rotate the beacon around the ship. Only the forward command moves the ship.

## Solution

### Part 1

I’m sure there are prettier ways to do it, but I just loop over the instructions, moving the ship or changing the direction each time. A quick glance shows that all the turn instructions are in increments of 90 degrees, but the math isn’t too complicated to handle any values

#!/usr/bin/env python3

import math
import sys

with open(sys.argv[1], "r") as f:

direction = 90
x, y = 0, 0

for i in insts:
d, p = i[0], int(i[1:])
if d == "N":
y += p
elif d == "S":
y -= p
elif d == "E":
x += p
elif d == "W":
x -= p
elif d == "L":
direction = (direction - p) % 360
elif d == "R":
direction = (direction + p) % 360
elif d == "F":

print(f"Part 1: {abs(x)+abs(y)}")

This gets the correct result.

### Part 2

I think there’s a way to conceptualize part two such that I can turn part one into a solve function and get either back. But for sake of speed and clarity, I just copied all of part one and started changing bits. I’ll add a position for the waypoint, and handle rotations using geometric calculations with sin and cos. To turn right, I’ll just set the angle to 360 minus the angle and then turn left that amount.

sx, sy = 0, 0
wx, wy = 10, 1

for i in insts:
d, p = i[0], int(i[1:])
if d == "N":
wy += p
elif d == "S":
wy -= p
elif d == "E":
wx += p
elif d == "W":
wx -= p
elif d == "L":
wx_orig = wx
wy = round(
)
elif d == "R":
p = 360 - p
wx_orig = wx
wy = round(
)
elif d == "F":
sy += wy * p
sx += wx * p

print(f"Part 2: {abs(sx) + abs(sy)}")

What got me stuck for a long while was that I was forgetting to keep the original wx so when it changed in the first calculation, the calculation of wy was wrong. I added a line to print the status at the end of each loop: print(f'{i.strip():5s} {sx} {sy} {wx} {wy}') and watched how it was breaking, and eventually figured that out.

### Final Code

Script runs instantly:

\$ time python3 day12.py 12-puzzle.txt
Part 1: 759
Part 2: 45763

real    0m0.021s
user    0m0.014s
sys     0m0.007s

#!/usr/bin/env python3

import math
import sys

with open(sys.argv[1], "r") as f:

direction = 90
x, y = 0, 0

for i in insts:
d, p = i[0], int(i[1:])
if d == "N":
y += p
elif d == "S":
y -= p
elif d == "E":
x += p
elif d == "W":
x -= p
elif d == "L":
direction = (direction - p) % 360
elif d == "R":
direction = (direction + p) % 360
elif d == "F":

print(f"Part 1: {abs(x)+abs(y)}")

sx, sy = 0, 0
wx, wy = 10, 1

for i in insts:
d, p = i[0], int(i[1:])
if d == "N":
wy += p
elif d == "S":
wy -= p
elif d == "E":
wx += p
elif d == "W":
wx -= p
elif d == "L":
wx_orig = wx
wy = round(
)
elif d == "R":
p = 360 - p
wx_orig = wx

format: plain


text:
Browse Questions

The force experienced by a pole of strength $100\: A-m$ at a distance of $0.2 \: m$ from a short magnet of length $10\: cm$ and pole strength $100\: A-m$ on its axial line will be

$\begin {array} {1 1} (a)\;2.5 \times 10^{-2}N & \quad (b)\;5.0 \times 10^{-2}N \\ (c)\;2.5 \times 10^{-3}N & \quad (d)\;5.0 \times 10^{-3}N \end {array}$

$F = mB= \large\frac{\mu_0}{4 \pi}$ $\large\frac{2m'l}{x^3}$$m$
Ans : (a)


text:
# Inaccessible cardinal

In mathematics, a cardinal number k > [itex]\aleph_0[itex] (aleph-null) is called weakly inaccessible, or just inaccessible, if the following two conditions hold.

1. cf(k) = k, where cf denotes the cofinality. Such a cardinal is called a regular cardinal.
2. There is no next smaller cardinal number; i.e., for every cardinal l < k, there is another cardinal number between l and k. Such a cardinal number is called a limit cardinal.

Every transfinite cardinal number is either regular or a limit; however, only a rather large cardinal number can be both. In fact, assuming that ZFC is consistent, the existence of inaccessible cardinals provably cannot be proven in ZFC.

All inaccessible cardinals are large cardinals. Some inaccessible cardinals may be strongly inaccessible cardinals.

• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy


text:
# Reference frame

1. Oct 30, 2009

### eshia5857

1. The problem statement, all variables and given/known data

An airplane, whose air speed is 580 km/h, is supposed to fly in a straight path 38 degrees N of E. But a steady 72 km/h wind is blowing from the North. In what direction should the plane head?

Relevant Equations
Pythagorean theorem
Law of sines

Attempt at a solution
The first thing I did was to draw myself a diagram so that I could see what is happening in this problem. I tried to attach a diagram that I made myself, but that did not work out at all.
What I did was to label each one Velocity of the plane with respect to the ground was pointing in the N of E direction and was approximated to 38 degrees, Velocity of the air with respect to the ground was place so that it was head to tail of the previous vector and pointed in the southward direction and the last one was labeled velocity of the plane with respect to the air and it was placed facing in the Eastern direction located directly onto the x-axis. Once I resolve it into components I am getting Y: $$V_{pg}$$*sin($$\theta$$)-$$V_{ag}$$*sin($$\theta$$), for the x component I found $$V_{pg}$$*cos($$\theta$$).I plugged my numbers into the equations and found my y components to have a value of 312.76 km/h and my x components to have a value of 457.05 km/h. I took the arcsin(y/x) and found the direction to be 43.2 degrees this troubles me because the solution is supposed to be 43.6 degrees. I need to know how to fix this error so that I don't make mistakes like this in the future I would appreciate any assistance that can be offered. Thank you for your time.

2. Oct 30, 2009

### Sorry!

Was this a textbook question and was it the answer found in the back of the book?
And did you use exact values or rounded values... always use exact values use the calculators memory or store function.

3. Oct 31, 2009

### Andrew Mason

You have to show us how you worked it out. It strikes me that this is a rather difficult vector problem. Unless you show us exactly what you did, it is difficult to find out where you went wrong. What did you find to be the magnitude of the plane to ground velocity? How did you determine that?

AM

4. Oct 31, 2009

### eshia5857

Re:frame of reference continuation

Alright, sorry I was finally able to get an image onto this site, and built up the courage to let you see how wrong I am. I am certain that I am doing something wrong and I have looked at this problem so many times that I thought my eyes would cross. I have to know what is wrong with my logic. Please help anyone, I do not want a solution there is one in the back of my text, I want to understand. I also forgot to include the information Adrew Mason asked about. I have found on few pieces of paper that have now become garbage Vpa, I changed my diagram, used conversions then not using conversions and at some point I know I began to make up some rules of my own out of desperation. I am lost!

#### Attached Files:

• ###### photo.jpg
File size:
38.5 KB
Views:
190
Last edited: Oct 31, 2009
5. Oct 31, 2009

### eshia5857

Thank you for the last part of your comment it got me really thinking about it again and now I have something more concrete to try, fingers crossed.

6. Oct 31, 2009

### Andrew Mason

Re: frame of reference continuation

What you have to do is calculate the length of the plane-ground vector. Call it L. Call the wind velocity vector (length of the air-ground velocity vector) W and the plane-air velocity vector P.

Write out the expression for L in terms of W and P and the known angle. You will get a quadratic equation in terms of L. Find the solution of that equation and then you can determine the angle of P. The correct answer is 43.6 degrees north of east.

AM

7. Oct 31, 2009

### emyt

Re: frame of reference continuation

Hey, I was trying this as well, I was onto something but I gave up.. could you show your work please?

thanks

8. Oct 31, 2009

### Andrew Mason

Re: frame of reference continuation

You have a vector of indeterminate length going in a direction of 38 degrees north of east (plane-ground velocity - call it L). You have a vector of length 72 (km/hr) from the tail of the previous vector heading due south (W). The difference is the plane-air velocity vector (P)

W + P = L

Work out the length of P in terms of W and L:

P^2 = (W + Lsin38))^2 + (Lcos38)^2

Work that out to find L. You have to solve the quadratic equation (Hint: what is cos^2 + sin^2)?

AM

9. Oct 31, 2009

### eshia5857

There is an attachement in one of the previous parts of this post but I have figured out how to solve it now, and an putting the finishing touches on it, so that it is accurate in everyway. Also I was really confused when I started the problem, take another look at your diagram. My original diagram was causing me to solve the problem wrong. I don't know about just giving out an answer like that, they made me work for mine, now you work for yours. Nothing personal it will just make you better. If you still can't get it in a while, then I will attach my new diagram.
Sorry that message is for emyt, thank you guys very much I appreciate being able to come to a forum where people don't just blurt out answers. Thanks for all the insight, have a great week-end.

Last edited: Oct 31, 2009
10. Oct 31, 2009

### emyt

Re: frame of reference continuation

Ahhh I see, I have a diagram like this drawn, I just didn't make the connection.. thanks


text:
## Algebra 1

If a product is 0, one of the factors must be 0. Use the addition property of equality to solve for x. $x-9=0\ or\ x-8=0$ $x-9+9=0+9\ or\ x-8+8=0+8$ $x=9\ or\ x=8$


text:
# The finite difference method for the heat equation-error

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!!

I am implementing in a program the finite difference method for the heat equation.
The problem is the following:
$$u_t(x,t)=(g(x,t)u_x(x,t))_x+f(x,t), \forall (x,t) \in [0,1]x[0,1]$$
$$u(0,t)=u(1,t)=0, \forall t \in [0,1]$$
$$u(x,0)=0, \forall x \in [0,1]$$

where $f(x,t)=\pi x \cos(\pi x t) (1-x)-g(x,t)(2 \pi t \cos(\pi x t)-(x-1) \pi^2 t^2 \sin(\pi x t))+g_x(x,t) (\sin(\pi x t)+\pi (x-1) t \cos(\pi x t))$

$u(x,t)=\sin( \pi x t)(1-x)$

$g(x,t)=2+ \sin(xt)$

The method is the following: ($U_j^n$ is the approximation of $u(x_j,t_n), h=\frac{1}{J}, Dt=\frac{1}{N}, x_j=jh, t_n=nDt$)
$$U_j^0=0, j=0,...,J$$

$$n=0,...,N-1:\frac{U_j^{n+1}-U_j^n}{Dt}=\frac{1}{h}[g(x_j+\frac{h}{2},t_{n+1})\frac{U_{j+1}^{n+1}-U_j^{n+1}}{h}-g(x_j-\frac{h}{2},t_{n+1})\frac{U_j^{n+1}-U_{j-1}^{n+1}}{h}]+f(x_j,t_{n+1}), j=1,...,J-1$$
$$U_0^{n+1}=U_J^{n+1}=0$$

I found the following errors:
For $J=N=10: 0.179505$
For $J=N=20: 0.089506$

Could you tell me if these errors are correct?

Last edited:

#### mathmari

##### Well-known member
MHB Site Helper
I made some modifications at my program, and now I find these errors:
for $J=N=10: 0.174628$
for $J=N=20: 0.086412$

But when we double $J$ and $N$, shouldn't the error get halved?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey!!!

where $f(x,t)=\pi x \cos(\pi x t) (1-x)-g(x,t)(2 \pi t \cos(\pi x t)-(x-1) \pi^2 t^2 \sin(\pi x t))+g_x(x,t) (\sin(\pi x t)+\pi (x-1) t \cos(\pi x t))$
Can it be that it should be:
\begin{aligned}
f(x,t)=\pi x \cos(\pi x t) (1-x)&+\ g(x,t)(2 \pi t \cos(\pi x t)-(x-1) \pi^2 t^2 \sin(\pi x t)) \\
&+\ g_x(x,t) (\sin(\pi x t)+\pi (x-1) t \cos(\pi x t))
\end{aligned}
With a $+$ instead of a $-$?

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!!

Can it be that it should be:
\begin{aligned}
f(x,t)=\pi x \cos(\pi x t) (1-x)&+\ g(x,t)(2 \pi t \cos(\pi x t)-(x-1) \pi^2 t^2 \sin(\pi x t)) \\
&+\ g_x(x,t) (\sin(\pi x t)+\pi (x-1) t \cos(\pi x t))
\end{aligned}
With a $+$ instead of a $-$?
Yes you are right!!!
I replaced the functions at the differential equation and it should be a $+$ instead of a $-$...

#### mathmari

##### Well-known member
MHB Site Helper
After changing the sign of f, I ran again the program and get the following:
for $J=N=4: 0.261844$
for $J=N=8: 0.137994$

Are these errors correct?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
After changing the sign of f, I ran again the program and get the following:
for $J=N=4: 0.261844$
for $J=N=8: 0.137994$

Are these errors correct?
Ha! I think I finally got my expressions right.

With $J=N=4$, I get a maximum deviation in $U_j^N$ of $0.035486$.
With $J=N=8$, that error is $0.00765$.
With $J=N=10$, it is $0.004647$.
With $J=N=20$, it is $0.000768$.

Or were you talking about an other error??

#### mathmari

##### Well-known member
MHB Site Helper
Or were you talking about an other error??
I'm talking about the error that is calculated by $$\underset{1 \leq n \leq N}{max} \underset{1 \leq j \leq J-1}{max}|U_j^n-u(x_j,t_n)|$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I'm talking about the error that is calculated by $$\underset{1 \leq n \leq N}{max} \underset{1 \leq j \leq J-1}{max}|U_j^n-u(x_j,t_n)|$$
Oh! Of course.

Anyway, with $J=N=4$, I still get $0.035486$.

#### mathmari

##### Well-known member
MHB Site Helper
Oh! Of course.

Anyway, with $J=N=4$, I still get $0.035486$.
Because my errors differ from yours, I post the approximations that I found for $J=N=5$.. Could you tell if they are also different from yours?

n=0:
0.092009
0.131979
0.119732
0.056693

n=1:
0.185583
0.257637
0.224462
0.106287

n=2:
0.285664
0.383913
0.318851
0.150719

n=3:
0.394673
0.508693
0.395790
0.186965

n=4:
0.509846
0.621373
0.441645
0.208845

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Because my errors differ from yours, I post the approximations that I found for $J=N=5$.. Could you tell if they are also different from yours?
I've got:
$$u = \begin{bmatrix} 0 & 0.100267 & 0.198952 & 0.2945 & 0.385403 & 0.470228 \\ 0 & 0.149214 & 0.289052 & 0.410728 & 0.506597 & 0.570634 \\ 0 & 0.14725 & 0.273819 & 0.361931 & 0.399211 & 0.380423 \\ 0 & 0.096351 & 0.168866 & 0.199605 & 0.180965 & 0.117557 \\ \end{bmatrix}$$
$$U = \begin{bmatrix} 0 & 0.099991 & 0.199442 & 0.298364 & 0.395847 & 0.490025 \\ 0 & 0.148563 & 0.289159 & 0.415125 & 0.518709 & 0.591527 \\ 0 & 0.146334 & 0.273258 & 0.364719 & 0.406867 & 0.390194 \\ 0 & 0.095615 & 0.168179 & 0.2005 & 0.183009 & 0.116612 \\ \end{bmatrix}$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Here's what I did:
$$U^{n+1} = (A^{n+1})^{-1}(U^n + \Delta t \cdot f^{n+1})$$

What did you do?

#### mathmari

##### Well-known member
MHB Site Helper
Here's what I did:
$$U^{n+1} = (A^{n+1})^{-1}(U^n + \Delta t \cdot f^{n+1})$$

What did you do?
So you solved a system of the form $Ax=b$, right? I did the same by solving this equation with the Gaussian elimination for tridiagonal systems.
I calculated $b$ as followed:
Code:
for(j=1; j<J-1; j++){
b[j-1]=f(j*h,(n+1)*Dt)+U[SUB]j[/SUB][SUP]n[/SUP]/Dt;
}
But why do we have then different results? Have I maybe done something wrong with the matrix A?

The diagonal of the matrix is:
Code:
for(j=0; j<J-1; j++){
Diag[j]=(g((j+1)*h+h/2, (n+1)*Dt)+g((j+1)*h-h/2, (n+1)*Dt))/(pow(h,2))+1.0/Dt;
}
The elements that are under the main diagonal are:
Code:
LDiag[0]=0;
for(j=1; j<J; j++){
LDiag[j]=-g((j+1)*h-h/2, (n+1)*Dt)/(pow(h,2));
}
And the elements that are above the main diagonal are:
Code:
for(j=0; j<J-2; j++){
UDiag[j]=-(g((j+1)*h+h/2, (n+1)*Dt)/(pow(h,2)));
}

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So you solved a system of the form $Ax=b$, right? I did the same by solving this equation with the Gaussian elimination for tridiagonal systems.
I calculated $b$ as followed:
Code:
for(j=1; j<J-1; j++){
b[j-1]=f(j*h,(n+1)*Dt)+U[SUB]j[/SUB][SUP]n[/SUP]/Dt;
}
But why do we have then different results? Have I maybe done something wrong with the matrix A?
Hold on just there.
You are effectively iterating from j=1 to J-2.
Shouldn't that be from j=1 to J-1?

#### mathmari

##### Well-known member
MHB Site Helper
Hold on just there.
You are effectively iterating from j=1 to J-2.
Shouldn't that be from j=1 to J-1?
With for(j=1; j<J-1; j++) I get for n=0:
0.092009
0.131979
0.119732
0.056693

With for(j=1; j<J; j++) I get for n=0:
0.099991
0.148563
0.146334
0.095615

I calculated it by hand and for n=0 I get(I hope I have done right the calculations):
0.0914
0.1315
0.1270
0.0818

(For $n=0$, $b=f(x_j,t_1)$ since $U^0=0$)

That means that I have made something wrong at the calculation of $b$, or not?

Last edited:

#### mathmari

##### Well-known member
MHB Site Helper
I think that the way that I calculate the function $f$ is wrong, because I get different results as I got when I calculated it by hand..

With the following code I get
3.048142
3.361183
3.327608
2.973833

But it should be
3.00855798
3.30227614071
3.269475746
2.93579504
or not?

Code:
int main(){
int j,J,N;
printf("Give N and J:\n");
scanf("%d %d", &N, &J);

for(j=1; j<J; j++) {
printf("\n %lf\n", f(xx(j,J),tn(1,N)));
}
return 0;
}

double xx(int j,int J){
double x,h=1.0/(double)J;
x=j*h;
return x;
}

double tn(int n, int N){
double t, Dt=1.0/N;
t=n*Dt;
return t;
}

double g(double x, double t){
double gx;
gx=2+sin(x*t);
return gx;
}

double dg(double x, double t){
double dgx;
dgx=t*cos(x*t);
return dgx;
}

double f(double x,double t){
double fx;
double pi=4.0*atan(1.0);
fx=pi*x*cos(pi*x*t)*(1-x)+g(x,t)*(2*pi*t*cos(pi*x*t)-(x-1)*pi*pi*t*t*sin(pi*x*t))+dg(x,t)*(sin(pi*x*t)+pi*(x-1)*t*cos(pi*x*t));

return fx;
}

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
With for(j=1; j<J-1; j++) I get for n=0:
0.092009
0.131979
0.119732
0.056693

With for(j=1; j<J; j++) I get for n=0:
0.099991
0.148563
0.146334
0.095615

I calculated it by hand and for n=0 I get(I hope I have done right the calculations):
0.0914
0.1315
0.1270
0.0818

(For $n=0$, $b=f(x_j,t_1)$ since $U^0=0$)

That means that I have made something wrong at the calculation of $b$, or not?
With your correction of $J$ instead of $J-1$, you get the same result as me, so I think that is correct.
With your calculation by hand, I suspect you made a mistake somewhere.

I think that the way that I calculate the function $f$ is wrong, because I get different results as I got when I calculated it by hand..

With the following code I get
3.048142
3.361183
3.327608
2.973833

But it should be
3.00855798
3.30227614071
3.269475746
2.93579504
or not?
Your code looks fine and its result matches mine.
I get:
$$f(x_j, t_1) = \begin{bmatrix} 3.048142 \\ 3.361183 \\ 3.327608 \\ 2.973833 \end{bmatrix}$$

What is your reason to think it should be otherwise?

#### mathmari

##### Well-known member
MHB Site Helper
With your correction of $J$ instead of $J-1$, you get the same result as me, so I think that is correct.
With your calculation by hand, I suspect you made a mistake somewhere.

Your code looks fine and its result matches mine.
I get:
$$f(x_j, t_1) = \begin{bmatrix} 3.048142 \\ 3.361183 \\ 3.327608 \\ 2.973833 \end{bmatrix}$$

What is your reason to think it should be otherwise?
I had only the same results as yours only for $U_j^1$ .
Now I changed some values of the integers at the for loop and I get the same results as yours for all n.

Thank you very much for your help!!!


text:
# M10

Question: Find the size of the angle marked z

Answer: The angles in the diagram go all the way around a point, so they must add up to $360\degree$. One of the angles doesn’t have a measurement, but it is marked with a square which means it’s a right angle ( $90\degree$ )

We can write the following equation:

$63+57+62+68+90+z=360$

Then we just simplify it by adding together the number on the left hand side:

$340+z=360$

Then subtract 340 from both sides of the equation to give us the answer:

$z=20\degree$


text:
January 21, 2022

# Solved examples on rational numbers

Here we are giving some practice  questions based  on  rational numbers.

1. Are the following statements are true or false? Give reason  for your answers.

(i)  every whole number is a natural numbers.

(ii) every integer is a rational number .

(iii)There are infinitely many rational numbers between any two given rational numbers.

(iv) 0 is a rational  number.

(V)Every rational number is a whole number.

ANS. (i) False , beacuse ‘0’ is a whole  number not a natural number.

(ii)True,  because every integer $m$ can be expressed in the form $\frac{m}{1}$, so it is a rational number.

(iii) True

(iv) True , since we can write  $0=\frac{0}{1}$

(v)false

2. Find six rational numbers between 3 and 4.

Sol.  To find a rational  number between  $r$ and  $s$ we can add $r$ and $s$ and divide the sum by 2 that is  $\frac{r+s}{2}$  lies between  $r$ and  $s$ .

Now let  $r$ =3  and  $s$ =4 , then $\frac{r+s}{2}$  = $\frac{7}{2}$ is a number lying between  3 and 4 .

Next let  $r$ =3  and  $s$ =$\frac{7}{2}$, then $\frac{3+\frac{7}{2}}{2}=&space;\frac{13}{4}$  is also a number lying between  3 and 4.

Rational no. between $\frac{7}{2}&space;\;&space;and&space;\;&space;4$ is  $\frac{\frac{7}{2}+4}{2}=&space;\frac{15}{4}$  .

Proceeding in the same manner  the six  rational  numbers  between 3 and 4 are $\frac{7}{2},$  $\frac{13}{4}$ , $\frac{15}{4},\frac{27}{8},&space;\frac{28}{8},$ $\frac{29}{8}$.

3. Express the decimal    $32.12\bar{35}$    in the form $\frac{p}{q}$  where $p$  and $q$ are integers and $q\neq&space;0$.

Sol. let    $x=32.12353535......$.  Since two digits are repeating, we multiply by $x$ by 100 to get

$100x=&space;3212.353535..........$

So,      $100x=3180.23+32.123535...........&space;\:&space;=3180.23+x$

Therefore           $100x-x&space;=3180.23$

i.e. $99x=&space;\frac{318023}{100}$  , which gives  $x=\frac{318023}{9900}$

Hence , $32.12\bar{35}$ =$\frac{318023}{9900}$.

#### Bina singh

View all posts by Bina singh →
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text:
# WhenEvent in NDSolve

How come this doesn't work as I intended?

s[{x0_, y0_}, a_, v_] := NDSolveValue[{
x'[t] == vx[t], WhenEvent[x[t] > 0, vx[t] -> -vx[t]],
y'[t] == vy[t] - t, WhenEvent[y[t] < 0, vy[t] -> -vy[t]],
x[0] == x0, vx[0] == v Cos@a,
y[0] == y0, vy[0] == v Sin@a},
{x[t], y[t]}, {t, 0, 10},
DiscreteVariables -> {vx, vy}]

Simulating falling and bouncing of a 2D mass point.

With[{r0 = {-5, 7}},
ParametricPlot[Evaluate@s[r0, -.9, 2],
{t, 0, 10},
PlotRange -> {{-10, 10}, {-10, 10}},
Epilog -> {Point@r0}]]

I can't understand why it doesn't bounce on the x-axis first, and what does happen there anyway? Event x[t]>0 works fine.

-
Have you tried removing the WhenEvent for x[t]>0? Is it the other When Event still ignored? –  Spawn1701D Apr 10 '13 at 21:13
Try this WhenEvent[{x[t] > 0, y[t] <0}, If[x[t]>0,vx[t] -> -vx[t],vy[t] -> -vy[t] ]] I think you can have only one WhenEvent included. –  Spawn1701D Apr 10 '13 at 21:16
@Spawn1701D Gee, now the same "refraction" occurs at y-axis as well. –  BoLe Apr 10 '13 at 21:20
I read the manual and I thing that if you have more than one WhenEvents you have to put them in a list, e.g { WhenEvent[x[t] > 0, vx[t] -> -vx[t]], WhenEvent[y[t] < 0, vy[t] -> -vy[t]]} –  Spawn1701D Apr 10 '13 at 21:22

Here you have a bouncing ball simulation using WithEvents[]:

s[{x0_, y0_}, a_, v_] := NDSolveValue[
{
y''[t] == -1,
x'[t] == vx[t],
y[0] == y0,
x[0] == x0,
y'[0] == v Sin@a,
vx[0] == v Cos@a,
vy[0] == v Sin@a,
WhenEvent[y[t] < 0, y'[t] -> - y'[t]],
WhenEvent[x[t] > 0, vx[t] -> -vx[t]]},
{x, y}, {t, 0, 8},
DiscreteVariables -> {vx, vy}] ;

With[{r0 = {-1, 1}},
ParametricPlot[Through[s[r0, 1, 1][u]], {u, 0, 5}, PlotRange -> All,
Epilog -> {Point@r0}]]

-
That is weird. If anything, vy picks up, and what does vx[t] -> - 3 vx[t] mean? I'm running a physical system here! :) –  BoLe Apr 10 '13 at 22:11
Thanks. I made a conceptual mistake, I can now understand what was going on above. –  BoLe Apr 11 '13 at 19:09

Following help I started with lowering the order of equations (and so doubling the number of variables). Here I wrote the equations in a more familiar 2nd order; everything works as expected. I put the function below in a tiny loop which bisected the right initial velocity v0 that returns the ball in its starting position.

Could this be done within NDSolve? I looked Wolfram tutorial for Boundary Value Problems but haven't been able to figure it out.

sstop[{x0_, y0_}, a_, v0_] := NDSolveValue[{
x''[t] == 0, WhenEvent[x[t] > 0, x'[t] -> -x'[t]],
y''[t] == -1, WhenEvent[y[t] < 0, y'[t] -> -y'[t]],
WhenEvent[x[t] < x0, Sow[{t, y[t]}]; "StopIntegration"],
x[0] == x0, x'[0] == v0 Cos[a],
y[0] == y0, y'[0] == v0 Sin[a]},
{x[t], y[t]}, {t, 0, 20}] // Reap

-
If you have a new question, you should post it as a new question instead of an "answer" –  rm -rf Apr 12 '13 at 18:13
@rm-rf Even though it is sort of continuation of this question? I think I'll think about it for a bit before, don't want to pose a stupid question. –  BoLe Apr 12 '13 at 20:59
Well, in that case you can update your question. Note that where the line is drawn between a "continuation" and a "totally new question" differs from person to person, but updating the post is the way to go –  rm -rf Apr 12 '13 at 21:26


text:
# The longitudinal stress in a thin cylinder pressure vessel of diameter d and internal pressure p is:

This question was previously asked in
ISRO VSSC Technical Assistant Mechanical held on 25/02/2018
View all ISRO Technical Assistant Papers >
1. $$\frac{{pd}}{t}$$
2. $$\frac{{pd}}{{2t}}$$
3. $$\frac{{pd}}{{4t}}$$
4. $$\frac{{pd}}{{8t}}$$

Option 3 : $$\frac{{pd}}{{4t}}$$
Free
ISRO VSSC Technical Assistant Mechanical held on 09/06/2019
2168
80 Questions 320 Marks 120 Mins

## Detailed Solution

By thin walled cylinder, we mean that the thickness ‘t' is very much smaller than the radius Ri and we may quantify this by stating that the ratio t / Ri of the thickness of radius should be less than 0.1.

In a thin shell circumferential stress is $${\sigma _c} = \frac{{Pd}}{{2t}}\;$$ and

Longitudinal stress will be half of the circumferential stress i.e. $${\sigma _l} = \frac{{Pd}}{{4t}}$$.

 Hoop stress $${\sigma _h} = \frac{{Pd}}{{2t}} = {\sigma _1}$$ Hoop strain $${\epsilon_h} = \frac{{Pd}}{{4tE}}\left( {2 - \mu } \right)$$ Longitudinal stress $${\sigma _L} = \frac{{Pd}}{{4t}} = {\sigma _2} = \frac{{{\sigma _1}}}{2}$$ Longitudinal strain $${\epsilon_L} = \frac{{Pd}}{{4tE}}\left( {1 - 2\mu } \right) = \frac{{{\sigma _2}}}{E}\left( {1 - 2\mu } \right)$$


text:
# nLab Voevodsky motive

A. Suslin and V. Voevodsky have realized a triangulated category ${D}^{b}\left({\mathrm{ℳℳ}}_{k}\right)$ which is supposed to be the bounded derived category of the hypothetical abelian category of mixed motive?s over $k$, predicted by Grothendieck–Beilinson–Deligne. There are variants developed by Hanamura and M. Levine. There is a different “derived” approach to mixed motives, namely the ${A}^{1}$-homotopy theory of F. Morel and V. Voevodsky. Voevodsky used the derived category of mixed motives to solve Milnor's conjecture in algebraic K-theory.


text:
Document type:
Zeitschriftenaufsatz
Author(s):
Niklowitz, P. G.; Dunsiger, S. R.; Pfleiderer, C.; Link, P.; Schneidewind, A.; Faulhaber, E.; Vojta, M.; Huang, Y.-K.; Mydosh, J. A.
Title:
Role of Commensurate and Incommensurate Low-Energy Excitations in the Paramagnetic to Hidden-Order Transition of ${\mathrm{URu}}_{2}{\mathrm{Si}}_{2}$
Abstract:
We report low-energy inelastic neutron scattering data of the paramagnetic (PM) to hidden-order (HO) phase transition at T0=17.5K in URu2Si2. While confirming previous results for the HO and PM phases, our data reveal a pronounced wave-vector dependence of low-energy excitations across the phase transition. To analyze the energy scans we employ a damped harmonic oscillator model containing a fit parameter 1/$\Gamma$ which is expected to diverge at a second-order phase transition. Counter to expe...    »

Journal title:
Physical Review B
Year:
2015
Journal volume:
92
Month:
sep
Journal issue:
11
Pages contribution:
115116
Fulltext / DOI:


text:
# Square root of a positive $C^\infty$ function.

Suppose $f$ is a $C^\infty$ function from the reals to the reals that is never negative. Does it have a $C^\infty$ square root? Clearly the only problem points are those at which $f$ vanishes.

-
A natural extension of the question is whether every non-negative smooth function is the sum of finitely many squares of smooth functions. The standard anwser from the literature seems to be "there is an unpublished counterexample by S.Cohen and D.Epstein". Does anybody know what this counterexample look like ? –  Guillaume Aubrun Aug 27 '12 at 15:11

The answer is "no". This is covered in great detail here:

http://www.math.polytechnique.fr/~bony/BBCP_jfa.pdf

-
There is a famous open problem, whose solution is attributed to Paul Cohen, but no published paper seems to be available: there exists $f\in C^\infty(\mathbb R,\mathbb R_+)$ such that $f$ is not a finite sum of squares of smooth functions. That could be a relevant question for MO. –  Bazin Aug 31 '12 at 20:18

The function $$f(x) = \begin{cases} \sin^2 \left(\frac{1}{x} \right) e^{-1/x} + e^{-2/x} & \text{if x > 0,}\\ 0 & \text{if x \leq 0,} \end{cases}$$ is $C^\infty$ but has no $C^2$ square root. I found this example in the paper Choosing roots of polynomials smoothly by Alekseevsky, Kriegl, Losik, and Michor (available freely here). This example appears to have come from Frank Warner's (unpublished) 1963 dissertation.

-
Yes, and functions of this type are discussed in section 2 of the reference I gave in my answer. –  Dick Palais Aug 25 '12 at 4:17
Both answers were helpful. It appears that the answer is yes if you add the additional assumption that $f^3$ has a square root. Of course, a square root of $f$ is a square root of $f^3$ divided by $f$, but while you can easily show such a function is continuous, it is not so easy to show it $C^\infty$. But thanks to both of you. I wish I could check both. –  Michael Barr Aug 25 '12 at 11:15
See also the follow-up mat.univie.ac.at/~michor/roots2.pdf –  Andres Caicedo May 20 '13 at 3:47


text:
# A neat probability rule-of-thumb

Disclaimer: I am NOT a probabilist. Not only have I never taught probability, the last time I took a course in probability was in my sophomore year of college. So if this is well known (or totally wrong), forgive me.

A non-mathematician friend of mine shared this link with me. It compares the lifetime risk of dying by various means—cancer, heart disease, shark attack, etc. There are many problems with the analysis presented on this web page (for example, you are not equally likely to die from the flu in each of your 77.6 years (the average lifespan), conditional probability would be a more useful measure of risk for some of these, etc.), but I will ignore all of that. I want to focus on the last line. It says:

Lifetime risk is calculated by dividing 2003 population (290,850,005) by the number of deaths, divided by 77.6, the life expectancy of a person born in 2003.

For example, for drowning the risk is 1 in $290850005/(3306\cdot 77.6)=1133.7$

Stated another way, they are claiming that if $D$ people die each year from a given cause, the total population is $P$, and the life expectancy is $L$, then the probability of dying from the given cause is $DL/P$. I saw this and I thought, “Surely this is wrong. Why would that formula give the probability?”

So I tried to calculate it myself. Here is my back-of-the-envelope calculation. The chance of dying from this cause in one year is $D/P$. The chance of not dying from this cause in one year is $1-D/P$, the chance of not dying from this cause for $L$ years is $(1-D/P)^L$, and so the chance of dying from the cause in $L$ years is $1-(1-D/P)^L$. (Of course, this leaves open the possibility of dying several times in those $L$ years, but we’ll ignore that.)

Let’s use this formula with the drowning example. I get $1-(1-3306/290850005)^{77.6}=0.000881671\ldots$, or 1 in $1134.2$.

What?!?! I was shocked to see an answer almost identical to the one using the “wrong” technique. There must be more to this than I first thought. Let’s look a little closer.

First, notice that $1-(1-D/P)^L=1-((1+(-D)/P)^P)^{L/P}$. Sitting inside this expression is a sub-expression that looks a lot like the limit definition of $e^x$. In particular, because $P$ is a large number, this expression is very nearly $1-(e^{-D})^{L/P}=1-e^{-DL/P}$. Aha! There’s the $DL/P$ term! But we still don’t quite have what we want.

What we’ve shown is that if the probability someone dies of a given cause in one year is $x$, then the probability that they will die from it in $L$ years is approximately $1-e^{-Lx}$. Now suppose the probability $x$ is small (like the probability of dying by drowning). We will compute the linear approximation to this function at $x=0$. We see that $d(1-e^{-Lx})/dx=Le^{-Lx}$. At $x=0$, that derivative is $L$. So the linear approximation at $x=0$ is simply $Lx$. In particular, if we evaluate it at our specific annual probability value $D/P$, we obtain $DL/P$. And there it is! [Update: thank you to the commenters for pointing out that the introduction of the exponential function, while fine, is unnecessary. Quicker: just use the linear approximation for $1-(1-x)^L$ at $x=0$.]

Again, I’ve never seen this before. Perhaps it is well known. For example, maybe it is a good rule-of-thumb that all good actuaries know.

I’d be happy to hear people’s thoughts about this formula and my reasoning. Maybe there’s another, different way to see this.

[I’d like to thank my colleague Jeff Forrester for talking through this with me.]

1. dwees says:

Here’s an interesting follow up. If the actuaries have to assign a dollar value to a policy based on one of these probabilities, and they have 20,000,000 people signed up to a policy for 50 years each, how much money do they gain or lose depending on which approximation they use to calculate this probability?

2. Yan Zhang says:

Everything you’ve done is correct – here’s an equivalent though maybe mentally quicker way:

(1-x)^t = 1 – (t choose 1) x + (t choose 2) x^2 + … etc. (binomial theorem)

the interesting here is your x is D/P, which is typically a small number, in your particular case on the order of 10^{-5}. Thus, by the time you get to x^2 you’re already facing a 10^{-10} multiplier, whereas your t=L is only going to give you about 100 at most.

Therefore 1-(1-x)^t is roughtly 1 – (1-tx) = tx. Here your t=L and x = D/P, so you get your desired thing.

@dwee: since 10^{-10}*100 = 10^{-8}, and you have about 2*10^7 people, I wouldn’t see your total \$ being off by more than a couple of orders of magnitude away from the dollar (on the safe side I may go with 100 dollars. I hope I’m not totally wrong with this since I really need to sleep).

3. Dude says:

If x is small, (1-x)^n = 1 – x*n + O(x^2).

So, (1-D/P)^L can be approximated by 1 – L*D/P. Your expression follows.

4. Thanks, Yan and Dude. You’re right. The introduction of the exponential function is unnecessary.

5. Very well known result, and quite useful. I’m surprised your sophomore course didn’t include it.

1. Well, maybe we did learn it. But that was 20 years ago…


text:
# Any Graph is a Model (! or ?)

I know this could be considered a pointless question. However despite I am quite convinced that any possible model (i.e. UML, SysML, natural language, math, etc.) can be defined by means of a graph I do not know if there is a way to formally prove there exist an isomorphism between the set of Models and the set of Graphs.

I guess one should first define what a model is and I guess a possible answer is that a model can be considered as a sentence of the language defined by its meta-model. Thus models are languages. The question is therefore if there is an isomorphism between the set of languages and the set of graphs or, less formally, if it is provable that any possible language can be represented by means of a graph. I guess that the Chomsky works on languages could help but I can not find a reference on the Internet of such demonstration.

• Please add a comment when you downvote. This will help the original poster improve the question and make it more useful for the community. This is also our site policy. Oct 26 '11 at 16:02
• @SureshVenkat: I do not think that it has ever become a site policy. Oct 30 '11 at 13:04
• @TsuyoshiIto: It came up in a community-wide townhall chat a while back as a major irritant, and the mods agreed to encourage people to leave comments so as to improve the question/answer being downvoted Oct 30 '11 at 16:59
• @SureshVenkat: I think that it is silly to call something a “policy” just because the moderators agree that it is a major irritant, but do whatever you like. Oct 30 '11 at 19:20

I would like to add to Dave Clarke's answer. The binary relations are sufficient to express higher-arity relations and functions (by adding new non-logical symbols if necessary). Hence, if you allow multiple directed edges, which can take on different colors, then this is already sufficient to encode any $L$-structure. Furthermore, this encoding can be described in a first-order way.


text:
Making a one-pass compiler by generating fexprs that generate code

I'm starting to write a simple compiler that transcompiles a simple scripting language for general game playing (at least for chess-like and a few other board games) into C which will be compiled in memory with the Tiny C library.

I noticed that there's a mismatch between the order in which parser generators trigger actions and the order in which tree nodes need to be visited in order that identifiers can be type checked and used to generate code.

Parser generators trigger actions from the bottom up. Ideally when you generate code, you visit nodes in the tree in whatever order gives you the type information before you see the identifiers used in an expression. Since fexprs let you control what order you visit parts of the inner expression, they're perfect for that.

So my parser is being written so that the parse generates an s-expression that contains fexprs that when run semantically checks the program and transcompiles in a single pass.

This also suggests a new version of Greenspun's 10th rule:
A sufficiently complex C++ program will contain an informally specified, buggy and incomplete implementation of John Shutt's Kernel language.

A refutation of Gödel's first incompleteness theorem

Contradictions in Gödel's First Incompleteness Theorem.

Notes for the second edition. I have edited this post a bit, sorry if it is against the etiquette. The original post can be found here. I have removed a couple of -periopheral- mistakes. However I want to also warn that the argument is a lot clearer when adapted to a modern rendition of Gödel's theorem, based on Turing machines, as suggested in i this comment. That adaptation is carried out in the comments that hang from the previous referenced comment, this an onwards.

Intro

This refers to Kurt Gödel's "On formally undecidable propositions of principia mathematica and related systems". The notation used here will be the same as that used by Gödel in that paper.

In that work, Gödel starts with a description of the formal system P, which, according to himself, "is essentially the system obtained by superimposing on the Peano axioms the logic of PM".

Then he goes on to define a map Phi, which is an arithmetization of system P. Phi is a one-to-one correspondence that assigns a natural number, not only to every basic sign in P, but also to every finite series of such signs.

One

There are alternative arithmetizations of system P. I will later delve on how many.

This is obvious from simply considering a different order in the basic signs when they are assigned numbers. For example, if we assign the number 13 to "(" and the number 11 to ")", we obtain a different Phi.

If we want Gödel's proof to be well founded, it should obviously be independent of which arithmetization is chosen to carry it out. The procedure should be correct for any valid Phi chosen. otherwise it would **not** apply to system P, but to system P **and** some particular Phi.

To take care of this, in Gödel's proof we may use a symbol for Phi that represents abstractly any possible valid choice of Phi, and that we can later substitute for a particular Phi when we want to actually get to numbers. This is so that we can show that substituting for any random Phi will produce the same result.

The common way to do this is to add an index i to Phi, coming from some set I with the same cardinality as the set of all possible valid Phi's, so we can establish a bijection among them - an index. Thus Phi becomes here Phi^i.

Two

Later on, Gödel proceeds to spell out Phi, his Phi, which we might call Phi^0, with his correspondence of signs and numbers and his rules to combine them.

And then Gödel proceeds to define a number of metamathematical concepts about system P, that are arithmetizable with Phi^0, with 45 consecutive definitions, culminating with the definition of provable formula.

Definition of provable formula means, in this context, definition of a subset of the natural numbers, so that each number in this set corresponds biunivocally with a provable formula in P.

Let's now stop at his definition number (10):

E(x) === R(11) * x * R(13)

Here Gödel defines "bracketing" of an expression x, and this is the first time Gödel makes use of Phi^0, since:

Phi^0( '(' ) = 11

Phi^0( ')' ) = 13

If we want to remain general, we may rather do:

E^i(x) === R(Phi^i( '(' )) * x * R(Phi^i( ')' ))

Two little modifications are made in this definition. First, we substitute 11 and 13 for Phi^i acting on "(" and ")". 11 and 13 would be the case if we instantiate the definition with Phi^0.

And second, E inherits an index i; obviously, different Phi^i will define different E^i. And so do most definitions afterwards.

Since, for the moment, in the RHS of definitions from (10) onwards, we are not encoding in Phi^i the index i, which has sprouted on top of all defined symbols, we cease to have an actual number there (in the RHS); we now have an expresion that, given a particular Phi^i, will produce a number.

So far, none of this means that any of Gödel's 45 deffinitions are in any way inconsistent; we are just building a generalization of his argument.

Three

There is something to be said of the propositions Gödel labels as (3) and (4), immediately after his 1-45 definitions. With them, he establishes that, in his own words, "every recursive relation [among natural numbers] is definable in the [just arithmetized] system P", i.e., with Phi^0.

So in the LHS of these two propositions we have a relation among natural numbers, and in the RHS we have a "number", constructed from Phi^0 and his 45 definitions. Between them, we have an arrow from LHS to RHS. It is not clear to me from the text what Gödel was meaning that arrow to be. But it clearly contains an implicit Phi^0.

If we make it explicit and generalized, we must add indexes to all the mathematical and metamathematical symbols he uses: All Bew, Sb, r, Z, u1... must be generalized with an index i.

Then, if we instantiate with some particular Phi^i, it must somehow be added in both sides: in the RHS to reduce the given expression to an actual number, and in the LHS to indicate that the arrow now goes from the relation in the LHS **and** the particular Phi^i chosen, to that actual number.

Obviously, if we want to produce valid statements about system P, we must use indexes, otherwise the resulting numbers are just talking about P and some chosen Phi^i, together.

Only after we have reached some statement about system P that we want to corroborate, should we instantiate some random Phi^i and see whether it still holds, irrespective of any particularity of that map.

These considerations still do not introduce contradiction in Gödel's reasoning.

Four

So we need to keep the indexes in Gödel's proof. And having indexes provides much trouble in (8.1).

In (8.1), Gödel introduces a trick that plays a central role in his proof. He uses the arithmetization of a formula y to substitute free variables in that same formula, thereby creating a self reference within the resulting expression.

However, given all previous considerations, we must now have an index in y, we need y^i, and so, it ceases to be a number. But Z^i is some function that takes a natural number and produces its representation in Phi^i. It needs a number.

Therefore, to be able to do the trick of expressing y^i with itself within itself, we need to convert y^i to a number, and so, we must also encode the index i with our 45 definitions.

The question is that if we choose some Phi^i, and remove the indexes in the RHS to obtain a number, we should also add Phi^i to the LHS, for it is now the arithmetic relattion **plus** some arizmetization Phi^i which determine the number in the RHS, and this is not wanted.

Five

But to encode the index, we ultimately need to encode the actual Phi^i. In (3) and (4), If in the RHS we are to have a number, in the LHS we need the actual Phi^i to determine that number. If we use a reference to the arithmetization as index, we'll also need the "reference map" providing the concrete arithmetizations that correspond to each index. Otherwise we won't be able to reach the number in the RHS.

Thus, if we want definitions 1-45 to serve for Gödel's proof, we need an arithmetization of Phi^i itself -with itself.

This may seem simple enough, since, after all, the Phi^i are just maps, But it leads to all sorts of problems.

Five one

Now, suppose that we can actually arithmetize any Phi^i with itself, and that we pick some random Phi^i, let's call it Phi^0: we can define Phi^0 with Phi^0, and we can use that definition to further define 10-45.

But since Phi^0 is just a random arithmetization of system P, if it suffices to arithmetize Phi^0, then it must also suffice to arithmetize any other Phi^i equally well. However, with Phi^0, we can only use the arithmetization of Phi^0 as index to build defns 10-45.

This means that, as arithmetizations of system P, the different Phi^i are not identical among them, because each one treats differently the arithmetization of itself from the arithmetization of other Phi^i.

Exactly identical arithmetical statements, such as definition (10) instatiated with some particular Phi^i, acquire different meaning and truth value when expressed in one or another Phi^i.

Among those statements, Gödel's theorem.

Five two [This argument does not hold if we have to consider recursively enumerable arithmetiazations.]

A further argument that shows inconsistency in Gödel's theorem comes from considering that if we are going to somehow encode the index with Phi^i, we should first consider what entropy must that index have, since it will correspond to the size of the numbers that we will need to encode them. And that entropy corresponds to the logarithm of the cardinality of I, i.e., of the number of valid Phi^i.

To get an idea about the magnitude of this entropy, it may suffice to think that variables have 2 degrees of freedom, both with countably many choices. Gödel very conveniently establishes a natural correspondence between the indexes of the variables and the indexes of primes and of their consecutive exponentiations, but in fact any correspondence between both (indexes of vars, and indexes of primes and exponents) should do. For example, we can clearly have a Phi^i that maps the first variable of the first order to the 1000th prime number exponentiated to the 29th power.

This gives us all permutations of pairs of independent natural numbers, and so, uncountably many choices for Phi^i; so I must have at least the same cardinality as the real line. Therefore y^i doesn't correspond to a natural number, since it needs more entropy than a natural number can contain, and cannot be fed into Z^i, terminating the proof.


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# Studies of Molecular Oxygen Adsorbed on Cu Surfaces

Prabhakaran, K and Sen, P and Rao, CNR (1986) Studies of Molecular Oxygen Adsorbed on Cu Surfaces. In: Surface Science, 177 (2). L971-L977.

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## Abstract

Molecular oxygen adsorbed on (110) and polycrystalline Cu surfaces has been investigated by UPS, XPS, AES, HREELS and LEED. Molecularly adsorbed $O_2$ on the (110) surface shows the characteristic three-peak HeII spectrum due to $\pi_g$, $\pi_u$ and $\sigma_g$ orbitals, accompanied by an O-O stretching frequency at 660 $cm^{-1}$. On the polycrystalline Cu surface, adsorbed $O_2$ shows the three peak HeII spectra with a considerably smaller separation between the $\pi_u$ and $\sigma_g$ band and two O-O stretching bands at 610 and 880 $cm^{-1}$. $O_2$ adsorbed on the Cu(110) surface gives rise to a (1 × 1) LEED pattern and characteristic K $\pi^*\pi^*$ transition in the Auger spectrum.

Item Type: Journal Article Copyright of this article belongs to Elsevier Science B.V. Division of Chemical Sciences > Solid State & Structural Chemistry Unit 14 Mar 2008 19 Sep 2010 04:43 http://eprints.iisc.ernet.in/id/eprint/13392


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# Article

Full entry | PDF   (1.0 MB)
Keywords:
nearly disjoint sequence; strong convergence; convergence $\ell$-group
Summary:
For an abelian lattice ordered group $G$ let $\conv G$ be the system of all compatible convergences on $G$; this system is a meet semilattice but in general it fails to be a lattice. Let $\alpha_{nd}$ be the convergence on $G$ which is generated by the set of all nearly disjoint sequences in $G$, and let $\alpha$ be any element of $\conv G$. In the present paper we prove that the join $\alpha_{nd}\vee\alpha$ does exist in $\conv G$.
References:
[1] L. Fuchs: Partially Ordered Algebraic Systems. Pergamon Press, Oxford, 1963. MR 0171864 | Zbl 0137.02001
[2] J. Jakubík: Sequential convergences in l-groups without Urysohn's axiom. Czechoslovak Math. J. 42 (1992), 101-116. MR 1152174 | Zbl 0770.06008
[3] J. Jakubík: Disjoint sequences in Boolean algebras. Math. Bohem 123 (1998), 411-418. MR 1667113
[4] E. P. Shimbireva: On the theory of partially ordered groups. Matem. Sbornik 20 (1947), 145-178. (In Russian.) MR 0020558 | Zbl 0029.10301

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Math Help - function for Taylor series and find the 100th derivative of this function

1. function for Taylor series and find the 100th derivative of this function

Hey, again ... My neighbour send me again one exercise for his exam for Calculus 1. He needs to find the Taylor series and the 100th derivative of this function at x = -1:
$f(x) = (x^2 + 3x + 1)*e^{2x-1}$. Actually, I really forgot a little bit what you must do here, so please help me to solve...

2. Re: function for Taylor series and find the 100th derivative of this function

The Taylor's series for $e^x$, about x= 0, is $\sum_{n=0}^\infty \frac{x^n}{n!}$. Also $e^{2x-1}= e^{-1}e^{2x}$ so the Taylor's series for $e^{2x-1}$ is $\sum_{n=0}^\infty e^{-1}2^n \frac{x^n}{n!}$. To get the Taylor's series for $(x^2+ 3x+ 1)e^{2x-1}$ multiply that series by $x^2+ 3x+ 1}$:
$x^2(\sum_{n=0}^\infty e^{-1}2^n\frac{x^n}{n!})+ 3x(\sum_{n=0}^\infty e^{-1}2^n\frac{x^n}{n!})+ (\sum_{n=0}^\infty e^{-1}2^n\frac{x^n}{n!})$ $= (\sum_{n=0}^\infty e^{-1}2^n\frac{x^{n+2}}{n!})$ $+ (\sum_{n=0}^\infty 3e^{-1}2^n\frac{x^{n+1}}{n!})$ $+ (\sum_{n=0}^\infty e^{-1}2^n\frac{x^n}{n!})$.

In order to write that as a single sum, you will want to "adjust" the indices. In the first sum let j= n+ 2 so that n= j- 2 and the sum becomes $\sum_{j= 2}^\infty e^{-1}2^{j- 2}(j)(j-1)\frac{x^j}{j!}$. For the second, let j= n+ 1 so that n= j- 1 and the sum becomes $\sum_{j=1}^\infty 3e^{-1}2^{j- 1}j\frac{x^j}{j!}$. In the final sum let j= n so it is the same sum, just indexed by j rather than n.

3. Re: function for Taylor series and find the 100th derivative of this function

Ok, thanks for that. But actually, you gave me that for $x$ = 0. I need for $x = -1$. Would it be the same way?
How do I find the 100th derivative $f^{100}(-1)$ from this Taylor series? Or I need to find first the n-th derivative and then insert the value n = 100 or can I get the 100-th derivative directly from Taylor series?
I will discuss with my neighbour tomorrow.
For $x = -1$ I get: $e^{2x-1} = e^{2(x + 1) - 3}$. Can this even help here?

4. Re: function for Taylor series and find the 100th derivative of this function

And my neighbour just send me a couple of exercises for his exam (examples for previous exams):

1. We have the rational function $f(x) = \frac{x^2 + x + 1}{-x^3 + 2x^2 - 3x + 6}$. Also we have to find the Taylor series for this rational function at point $x = 0$ and find the 88-th derivative $f^{(88)}(0)$.

2. We have to find the indefinite integral for: $\int{\sqrt{\frac{ch^2(x) - 1}{ch(x)}}}dx$, where is the $ch(x) = \frac{e^x + e^{-x}}{2}$.

3. We have to find the indefinite integral for: $\int{\frac{6x^3 + 9x^2 + 15x + 5}{\sqrt{x^2 + x + 2}}}dx$.

My neighbour just wrote me: "If you know, how to find the solution, please let me know..." If anybody here on the forum knows, please show me way ...

5. Re: function for Taylor series and find the 100th derivative of this function

$e^{2x- 1}= e^{2(x- 1)+ 1}= e(e^{2(x- 1)})$. So just change my "x" to "x- 1" and multiply by e, which will just cancel the " $e^{-1}$" in my formula.

6. Re: function for Taylor series and find the 100th derivative of this function

Let $g(x) = (ax^2+bx+c)e^{2x-1}$. Then $g'(x) = (2ax+b)e^{2x-1}+2(ax^2+bx+c)e^{2x-1} = \left( 2ax^2+(2a+b)x+b+2c \right)e^{2x-1}$. In other words, $g'(x)$ has the same terms with different coefficients as $g(x)$. So, let:

$\dfrac{d^n}{dx^n}\left(f(x)\right) = \left(a_nx^2+b_nx+c_n\right)e^{2x-1}$

The problem gives us that $a_0=1, b_0=3, c_0=1$. Then:

\begin{align*} \left( a_{n+1}x^2+b_{n+1}x+c_{n+1} \right)e^{2x-1} & = \dfrac{d^{n+1}}{dx^{n+1}}\left( f(x) \right) \\ & = \dfrac{d}{dx}\left[ \dfrac{d^n}{dx^n}\left( f(x) \right) \right] \\ & = \dfrac{d}{dx}\left[\left( a_nx^2 + b_nx + c_n \right)e^{2x-1} \right] \\ & = a_n\dfrac{d}{dx}\left(x^2e^{2x-1}\right) + b_n\dfrac{d}{dx}\left(xe^{2x-1}\right) + c_n\dfrac{d}{dx}\left(e^{2x-1}\right) \\ & = a_n\left( 2x^2e^{2x-1}+2xe^{2x-1} \right) + b_n\left( 2xe^{2x-1} + e^{2x-1} \right) + c_n\left( 2e^{2x-1} \right) \\ & = \left(2a_nx^2 + (2a_n+2b_n)x + (b_n+2c_n)\right)e^{2x-1}\end{align*}

Equating coefficients gives the following relations:

$a_{n+1} = 2a_n$
$b_{n+1} = 2a_n+2b_n$
$c_{n+1} = b_n+2c_n$

Since $a_0=1$, we have $a_n = 2^n$.
Plugging in, we get:

$b_{n+1} = 2\cdot 2^n+2b_n$

Since $b_0 = 3$, we have $b_n = 2^n(n+3)$

Plugging that in, we get:

$c_{n+1} = 2^n(n+3) + 2c_n$

Since $c_0=1$, we have $c_n = 2^{n-2}(n+1)(n+4)$

This gives:

\begin{align*}\dfrac{d^{100}}{dx^{100}}\left( f(x) \right) & = \left(a_{100}x^2 + b_{100}x + c_{100}\right)e^{2x-1} \\ & = \left( 2^{100}x^2 + 103\cdot 2^{100}x + 101\cdot 104\cdot 2^{98} \right)e^{2x-1} \\ & = 2^{100}\left( x^2 + 103x + 2626 \right)e^{2x-1}\end{align*}

So, $f^{(100)}(-1) = 2^{100}\left( (-1)^2 + 103(-1) + 2626 \right)e^{2(-1)-1} = 2524\cdot 2^{100}e^{-3}$

7. Re: function for Taylor series and find the 100th derivative of this function

Ok, thanks. I understand all the ways to the solution. What about the other 3 exercises? Does anybody know?

8. Re: function for Taylor series and find the 100th derivative of this function

Originally Posted by lebdim
And my neighbour just send me a couple of exercises for his exam (examples for previous exams):

1. We have the rational function $f(x) = \frac{x^2 + x + 1}{-x^3 + 2x^2 - 3x + 6}$. Also we have to find the Taylor series for this rational function at point $x = 0$ and find the 88-th derivative $f^{(88)}(0)$.

2. We have to find the indefinite integral for: $\int{\sqrt{\frac{ch^2(x) - 1}{ch(x)}}}dx$, where is the $ch(x) = \frac{e^x + e^{-x}}{2}$.

3. We have to find the indefinite integral for: $\int{\frac{6x^3 + 9x^2 + 15x + 5}{\sqrt{x^2 + x + 2}}}dx$.

My neighbour just wrote me: "If you know, how to find the solution, please let me know..." If anybody here on the forum knows, please show me way ...
1. $f(x) = \dfrac{x^2+x+1}{-x^3+2x^2-3x+6} = \dfrac{x^2+x+1}{(2-x)(x+i\sqrt{3})(x-i\sqrt{3})} = \dfrac{A}{2-x} + \dfrac{B}{x+i\sqrt{3}} + \dfrac{C}{x-i\sqrt{3}}$

Solving this partial fractions, you find:

$f(x) = \dfrac{1}{2-x} - \dfrac{i\sqrt{3}}{6(x+i\sqrt{3})} + \dfrac{i\sqrt{3}}{6(x-i\sqrt{3})}$

$\dfrac{d^n}{dx^n}\left( \dfrac{1}{2-x} \right) = \dfrac{n!}{(2-x)^n}$

$\dfrac{d^n}{dx^n}\left( \dfrac{ i\sqrt{3} }{6(x+i\sqrt{3})} \right) = \dfrac{i\sqrt{3}}{6}\cdot \dfrac{(-1)^n n!}{(x+i\sqrt{3})^n}$

$\dfrac{d^n}{dx^n}\left( \dfrac{ i\sqrt{3} }{6(x-i\sqrt{3})} \right) = \dfrac{i\sqrt{3}}{6}\cdot \dfrac{(-1)^n n!}{(x-i\sqrt{3})^n}$

So, plugging in, this gives:

$f^{(n)}(x) = \dfrac{n!}{(2-x)^n} + \dfrac{i\sqrt{3}}{6}\left( \dfrac{(-1)^{n+1}n!}{(x+i\sqrt{3})^n} + \dfrac{(-1)^n n!}{(x-i\sqrt{3})^n} \right)$

\begin{align*}f^{(n)}(0) & = \dfrac{n!}{2^n} + \dfrac{ i\sqrt{3} }{6}\left( \dfrac{ (-1)^{n+1} n! }{ (i\sqrt{3})^n } + \dfrac{ (-1)^n n! }{ (-i\sqrt{3})^n } \right) \\ & = \dfrac{n!}{2^n} + \dfrac{ i\sqrt{3} }{6}\left( (-1)^n n! \dfrac{ (i\sqrt{3})^n-(-i\sqrt{3})^n }{ (i\sqrt{3})^n(-i\sqrt{3})^n } \right) \\ & = \dfrac{n!}{2^n} + \dfrac{ i\sqrt{3} }{6}\left( (-1)^n n! \dfrac{ (i\sqrt{3})^n-(-i\sqrt{3})^n }{3^n} \right) \\ & = \begin{cases}\dfrac{n!}{2^n} & n\equiv 0\pmod{2} \\ \dfrac{n!}{2^n} + \dfrac{n!}{ (\sqrt{3})^{n+1} } & n\equiv 1\pmod{4} \\ \dfrac{n!}{2^n} - \dfrac{n!}{ (\sqrt{3})^{n+1} } & n\equiv 3\pmod{4} \end{cases}\end{align*}

Hence,
\begin{align*}f(x) & = \sum_{n\ge 0}\dfrac{f^{(n)}(0)}{n!}x^n \\ & = \sum_{n\ge 0} \dfrac{(2n)!}{2^{2n} (2n)!}x^{2n} + \sum_{n\ge 0}\dfrac{\dfrac{(2n+1)!}{2^{2n+1}} + (-1)^n\dfrac{(2n+1)!}{ 3^{n+1} } }{ (2n+1)! } x^{2n+1} \\ & = \sum_{n\ge 0} \dfrac{x^{2n}}{4^n} + \sum_{n\ge 0}\dfrac{2\cdot 3^{n+1} + (-1)^n4^{n+1}}{12^{n+1}}x^{2n+1}\end{align*}

Using complex numbers, we have:

$f(x) = \sum_{n\ge 0}\dfrac{2\cdot 3^{(n+1)/2}-2^n(i^{n+1}+(-i)^{n+1})}{12^{(n+1)/2}} x^n$

Oh, and $f^{(88)}(0) = \dfrac{88!}{2^{88}}$ (see above)

2. You need to simplify the expression. $\text{ch}^2(x) - 1 = \text{sh}^2(x)$ where $\text{sh}(x) = \dfrac{e^x-e^{-x}}{2}$. So, the expression becomes:

$\int \sqrt{ \dfrac{ \text{ch}^2(x) - 1 }{ \text{ch}(x) } }dx = \int \dfrac{ \text{sh}(x) }{\sqrt{ \text{ch}(x) } }dx$

Let $u = \text{ch}(x)$. Then $du = \text{sh}(x)dx$. This gives:

$\int \dfrac{ \text{sh}(x) }{ \sqrt{ \text{ch}(x) } }dx = \int u^{-1/2}du = 2u^{1/2}+C = 2\sqrt{\text{ch}(x)}+C$

3. First, we want to use the division algorithm to find $q(x)$ such that $6x^3+9x^2+15x+5 = q(x)(x^2+x+2)+r(x), \text{deg}(r(x))<2$. $6x^3+9x^2+15x+5 - 6x(x^2+x+2) = 3x^2+3x+5$, so let's try $q(x) = 6x+3$. Then $6x^3+9x^2+15x+5 = (6x+3)(x^2+x+2) - 1$, so $r(x)=-1$ is definitely a polynomial with degree less that 2. Let's rewrite the integral:

\begin{align*}\int \dfrac{6x^3+9x^2+15x+5}{\sqrt{x^2+x+2} }dx & = \int \dfrac{(6x+3)(x^2+x+2)-1}{\sqrt{x^2+x+2}}dx \\ & = \int 3(2x+1)\sqrt{x^2+x+2}dx - \int \dfrac{dx}{\sqrt{\left(x+\dfrac{1}{2}\right)^2 + \dfrac{7}{4} } }\end{align*}

For the first integral, use the substitution $u = x^2+x+2, du = (2x+1)dx$. For the second, use the substitution $x+\dfrac{1}{2} = \dfrac{\sqrt{7}}{2}\tan \theta, dx = \dfrac{\sqrt{7}}{2}\sec^2\theta d\theta$.

\begin{align*}\int 3(2x+1)\sqrt{x^2+x+2}dx - \int \dfrac{dx}{\sqrt{\left(x+\dfrac{1}{2}\right)^2 + \dfrac{7}{4} } } & = 3\int u^{1/2}du - \dfrac{\sqrt{7}}{2}\int \dfrac{\sec^2\theta}{\sqrt{\dfrac{7}{4}\left( \tan^2 \theta +1\right) } }d\theta \\ & = 2u^{3/2} - \int \sec \theta d\theta \\ & = 2(x^2+x+2)^{3/2} - \ln|\sec \theta + \tan \theta| + C\end{align*}

Now, $\tan \theta = \dfrac{\text{opp} }{\text{adj} } = \dfrac{2x+1}{\sqrt{7}}$, so $\sec \theta = \dfrac{\text{hyp}}{\text{adj} } = \dfrac{2\sqrt{x^2+x+2}}{\sqrt{7}}$. Plugging in, we get:

\begin{align*}& 2(x^2+x+2)^{3/2} - \ln\left| \sec \theta + \tan \theta \right| + C \\ = & 2(x^2+x+2)^{3/2} - \ln\left| \dfrac{2x+1 + 2\sqrt{x^2+x+2} }{\sqrt{7} } \right| + C\end{align*}

Simplifying (using logarithm rules), we have:

\begin{align*}& 2(x^2+x+2)^{3/2} - \ln\left| \dfrac{2x+1 + 2\sqrt{x^2+x+2} }{\sqrt{7} } \right| + C \\ = & 2(x^2+x+2)^{3/2} - \ln\left| 2x+1 + 2\sqrt{x^2+x+2} \right| + C'\end{align*}

where $C' = C+\ln(\sqrt{7})$.