What can be done about someone who asks questions, then consistently deletes them?

I can only offer sympathy.

We are in the same boat.

Thanks for the question, same thing happened to me with this guy, except I posted the answer and saw the question deleted 2 minutes later. If I may, I will post mine as well (if I may) given it is distinct from yours (variation of parameters) and thus offers something different.

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I will work from a grassroots level, there is no need to memorize the "formulas" for these, you can just construct them each time.

Variation of parameters initially pursues seeking the homogeneous solution

y'' + y = 0 --> y = A sin(x) + B cos(x) , please let us know if more detail is needed here.

The method continues by varying the parameters A and B, such that they are a function of x. Allowing this permits inputting the solution into the differential equation and constructing two simultaneous equations for A and B (actually their derivatives) wherein you are able to solve for these coefficient functions. The results reproduce the homogeneous solution and also uncovers another part (the nonhomogeneous solution) which gives you the total solution. It sounds like such a bold way of pursuing a solution, it really is a wonder it even works. Anyway, let us vary the parameters which we write as

y = A(x) sin(x) + B(x) cos(x), which means A and B are functions of x

then (using primes to denote dA/dx and dB/dx), our derivatives, which we need to input y and y'' into the DE are calculated as:

y' = A' sin(x) + B' cos(x) + A cos(x) - B sin(x),note A and B are still functions of x but I will omit the notation for brevity

Here, the method makes another bold move, later on we are going to have only one equation (which is the differential equation itself when we insert y'' and y into it) but two unknowns A' and B'. We are "free" to choose another condition so as to have two equations (free in the sense that, we further choose a restriction [equation] and if it works, it works, if not we were too forward in our assumptions in order to get a solution). The additional condition that is often taken now is to let the terms with dA/dx and dB/dx sum to zero, i.e.

A' sin(x) + B' cos(x) = 0 ..............Eqn. (1)

Again, this is by our own choosing. I do not think we have a good reason to try this a priori, but afterward we will see it works (as noted, it still surprises me this method even works to begin with because of how forward it is). We enforce this condition now to reduce our work later on with differentiation, thus

y' = A' sin(x) + B' cos(x) + A cos(x) - B sin(x) = A cos(x) - B sin(x)

computing the second derivative, and recalling our result for y:

y'' = (A' - B) cos(x) - (A + B') sin(x)

y = A(x) sin(x) + B(x) cos(x)

inserting this into the DE:

y'' + y = tanx

(A' - B) cos(x) - (A + B') sin(x) + Asin(x) + B cos(x) = tan(x)

A' cos(x) - B cos(x) - A sin(x) - B' sin(x) + A sin(x) + B cos(x) = tan(x)

so A sin(x) cancels with a -A sin(x) term, and B cos(x) cancels with a -B cos(x) term, showing what we know all ready: y'' + y = 0, i.e. the homogeneous solution is indeed a solution. What we are left with is

A' cos(x) - B' sin(x) = tan(x)..................................… (2)

This is our second equation. In summary, to get here we made one restriction of our own (Eqn. (1)), and the DE gave us another equation, these are two equations in two unknowns A' and B'

Taking a look at them both:

A' sin(x) + B' cos(x) = 0 ..............Eqn. (1)

A' cos(x) - B' sin(x) = tan(x).........Eqn. (2)

we can organize this into a matrix equation if we like (and as is customary):

[ sin(x)...........cos(x)] [ A' ] = [ 0 ]

[ cos(x)..........-sin(x)] [ B' ] = [ tan(x) ]

You can solve this system however you want, the usual way for variation of parameters is to use Cramer's rule (look it up, it comes from linear algebra), this method is well-known and the way you do it proceeds by defining

W =

[ sin(x)...........cos(x)]

[ cos(x)..........-sin(x)]

b =

[ 0 ]

[ tan(x) ]

and we define additional notation for convenience (augmented notation):

(W | b)

=

[ sin(x)........... 0 ]

[ cos(x)..........tan(x) ]

(b | W)

=

[ 0 ..........cos(x)]

[ tan(x)..........-sin(x)]

Anyway, finding A' and B' is easy using Cramer's rule, let det(W) = determinant of matrix W (what is called a Wronskian), and so on. Cramer's rule furnishes A' and B' by formulas:

A' = det(b | W) / det(W)

B' = det(W | b) / det(W)

calculate these determinants:

det(b | W) = - cos(x)tan(x) = sin(x)

det(W | b) = sin(x)tan(x) = sin^2(x) / cos(x)

det(W) = -(sin^2(x) - cos^2(x)) = -1

Thus,

A' = det(b | W) / det(W) = sin(x) / (-1) = -sin(x)

B' = det( W | b) / det(W) = sin^2(x) cos(x) / (-1) = - sin^2(x) cos(x)

These are the derivatives of A and B, integrate to find

A = integral { A' dx} = cos(x) + C, C = constant

B = integral{ - sin^2(x) cos(x) dx}

Hello, Mathmom!

The same has happened to me several times, at first I didn't pay any attention, but once I answered one by one 4 very interesting questions, asked by the same Asker, the last one took me half an hour or so to type the answer and the next day all of them were missing from my profile page. Viewing the browser history of the previous day I saw 4 times 'This question has been deleted' with 'Sometimes a question is deleted in accordance with Y!A Community Guidelines' added below.

I agree, such behavior is irritating and highly unethical, but Upward Bound Precalc Tutor is right: it is NOT a violation of the Y!A Community Guidelines, so it is perfectly LEGAL and that is what matters. Y!A is not perfect and we must take that into account, playing this game - we must obey the rules, trying to help others and some of them will not be grateful for the help we provide.

So, what can be done? As Tutor suggests, we can try to recognize the user and not answer any more of his questions. Once recognized, you can block him (I have already 5 persons with improper behavior in my Blocked List). Another possibility is to send him an email via the Internal Yahoo Service if this option on his profile page is active. But many have deactivated this option, then it remains to answer his next question in a manner like this:

http://answers.yahoo.com/question/index;_ylt=AuwKP...

Such an answer risks to be reported though - reported answers are removed almost automatically, Yahoo! staff can not read all of them. Yet I have pasted the above link in some questions of persons, who have deleted my previous answers and twice got apologies from them along with promises to abandon the wrong practice in the future.

This is the price people like us are ready to pay for our willingness to help sharing our knowledge. I am sure that the improper behavior of some participants in Y!A Mathematical Forum will not be a reason for you to quit answering. Many of Your answers will get the valuable words 'Thank You' for sure, so keep on answering!

You are right. When you put this much work into an answer, you would hope the person would

leave the question so you could display it and help them. The only thing I know of is to try

to recognize the user and not answer any more of their questions. That is tough as these people

can create new accounts, change usernames and repeat the behavior.

However, there is no true incentive to do this. Yahoo! charges askers five points each time they

ask a question and YOU don't get your points returned if you delete the question.

I also don't think deleted question impact your best answer percentage either. If you get to

answer a question and post the answer before the person deletes it, you get to keep your two points.

However, I understand that two points is not that big of a deal once you have worked your rear end

off to provide an accurate detailed response to a user's question, especially once you have made it

to level 7.

You are right, it is irritating and aggravating to get a question deleted before you can post your

answer. It is not, to my knowledge, a violation of the Yahoo! Answers community guidelines

to delete questions so we have to take this sometimes, much like a snowball in the face

during junior high school days.

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You will be surprised but when I first attempted to post this, I got the famous

"Yahoo Answers is currently unavailable" Please try again later.

6 times before It finally allowed me to post my response.

Yahoo itself sometimes does things.

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I REALLY FEEL YOUR PAIN NOW. I just spent an hour and a half writing this answer

to get "This question has been deleted"

Some HOT DOG just posted a question

REAL ANALYSIS -

Given sequences an and bn of positive numbers

such that an is bounded and limit an/bn =∞

prove that lim bn = 0

My Answer

My style of proof is to break complex proofs into sequences of

simple statements so that they can be completed with the

near simplicity of what you had to do in your high school

course in Euclidean Geometry. No shorthand or high falutin

abbreviations and jargon which makes these upper level

math courses multiple times more difficult than they ought to be.

We are given that

....{a(n)} is a positive real number. Given 1

....(b(n)} is a positive real number. Given 2

....{a(n)} is bounded. Given 4

and

..........a(n)

lim...▬▬▬ =∞.. Given 5

n→∞...b(n)

....0<a(n) by Given 1 and the definition of sequence of positive real numbers→ Statement 6

....0<b(n) by Given 2 and the definition of a sequence of positive real numbers→Statement 7

...a(n) < M(i) for ALL positive integers i

by Given 4 and the definition of a bounded sequence of real numbers→Statement 8

and of course M(i) is a sequence finite real numbers forming the set of upper bounds for a(n)

Let M = The infinum{least upper bound} of M(i)→Statement 9

so a(n) ≤ M from Statements 8 and 9→ Statement 10

From Given 5 and Statement 10 we can conclude

..........M

lim...▬▬▬ =∞.. Statement 11

n→∞...b(n)

Now, let us suppose for contradiction that

lim...b(n) = K where K is a nonzero real number.

n→∞...

from Given 1 and Given 2 K>0 Statement 12

lim..........M(n).... Let n be in place of i in the comments after statement 7

n→∞

..........▬▬▬ =∞.. By Statement 11 and one of the properties of sequences of limits

lim........b(n)

n→∞ .........................................................Statement 13

Substituting from Statements 9 and 12 into 13, we get

..........M

........▬▬▬ =∞.. Conclusion 14 ????

...........K....

However we supposed that K was positive real number and M is a positive real number

and we realize by closure of division within the positive real numbers that M/K has to be

a positive real number so Conclusion 14 must be false.

From that we must conclude that K is not positive. However it is a limit point of b(n), a sequence

of positive real numbers, SO it is the infinum {greatest lower bound} of the positive real numbers

and that is ZERO.

----Then this question was deleted!! This person's Mom needs to get the belt!

Aww man that sucks. That looks like so much working out and typing!