mm-252
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Subject: Re: Late Math REUs?Look at the NSF site, get
the REU list, call them. Problem solved.cc )> said:> ,> Can
anyone recommend any math rela REUs that are still considering
> student applications? In case it changes matteI'm a minority
junior > Math/EE double major.> Alex-- Bill StevensonEditor
in ChiefACM Crossroads
Magazinehttp://www.acm.org/crossroadshttp://
www.billstevenson.orgApplied Cognitive Science
LaboratoryPennsylvania State
===
REUs?> Look at the NSF site, get the REU list, call them.
Problem solved.Am doing so. Are there any that aren't funded
===
that was NIH funded, but that wasn't in math. I have doubts
that there are non-NSF math ones, or at least particularly
many of them. You should just as well try to be a professor's
summer slave (wage payroll) or something at a local school.cc
)> said:>> Look at the NSF site, get the REU list, call them.
Problem solved.> Am doing so. Are there any that aren't funded
by the NSF?> Alex-- Bill StevensonEditor in ChiefACM
Crossroads
Magazinehttp://www.acm.org/crossroadshttp://
www.billstevenson.orgApplied Cognitive Science
LaboratoryPennsylvania State
===
neededHello. I'm having a possible issue with a diffeq
problem, but myanswer may be right. I'm looking for a little
guidance and I have topreface my question with a summary of my
current work to lend itcontext.A 2 kg mass is dropped from rest
from 30m above water. When the mas s the water the momentum
is unchanged. Wind resistance isproportional to the velocity
and has a proportionality constant kwhich is 10 N*sec/m.For
this part I setup the equation like so:F = ma = m*(dv/dt) =
-mg - kv, so dv/dt = -(g - pv ) where p = k/m = 5and g =
9.8After moving things around... and letting v(0) = 0, v(t) =
-1.96 - 1.96*e^(-5t)integrating with respect to t and solving
for the initial value of x =0,x(t) = .392 - .392*e^(-5t) -
1.96tsolving numerically for x(t) = -30, t is approx
15.506--------------------------------------------------------
--------Now I have to examine the conditions after it hits
water. Here'swhere my problem comes in I think. I'm trying to
come up with onefunction that will be used for this entire
problem. It'll have twoparts (air/water) with the air part
used when t <= 15.506 and thewater part used when t >= 15.506,
so I assume t = 15.506 is theinitial condition and at 15.506
the velocity is -1.96. Is this a badidea?First note that no
momentum is lost. The proportionality constantunder water is
100 N*sec/m. Buoyancy is half of the weight (.5*mg).dv/dt = -g
- pv + .5g = -.5g - pv, where p is k/m or 50 this time.Moving
stuff around and setting v(15.506) = -1.96 yieldsv(t) =
-9.516*10^336*e^(-50t) - .098Perhaps you see why I think I am
making a mistake now. 10^336? Thatconstant is boggling. I
think I setup my initial conditionsincorrectly. I do NOT want
an exact answer to the problems, just alittle guidance if
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diffeq help needed> Hello. I'm having a possible issue with a
diffeq problem, but my> answer may be right. I'm looking for a
little guidance and I have to> preface my question with a
summary of my current work to lend it> context.> A 2 kg mass
is dropped from rest from 30m above water. When the mass> hits
the water the momentum is unchanged. Wind resistance is>
proportional to the velocity and has a proportionality
constant k> which is 10 N*sec/m.> For this part I setup the
equation like so:> F = ma = m*(dv/dt) = -mg - kv, so dv/dt =
-(g - pv ) where p = k/m = 5> and g = 9.8You've made a sign
error: dv/dt = -g - pv = -(g + pv)In any case, leaving this as
dv/dt + pv = -g makes maniplutaions easier.(It would also be of
great help if you refrained from substitutingnumerical values
of p,g, etc. until *after* you have comple yourmanipulations;
then you can easily change p and g (as, for instance,you
require below) without having to repeat the entire analysis.)>
After moving things around... and letting v(0) = 0,> v(t) =
-1.96 - 1.96*e^(-5t)Which means, since v(0) = 0, that +1.96 =
-1.96, indicating that youhave made an error.> integrating
with respect to t and solving for the initial value of x =>
0,> x(t) = .392 - .392*e^(-5t) - 1.96tWhich is at least
consistent with the initial condition.> solving numerically
for x(t) = -30, t is approx 15.506>
--------------------------------------------------------------
--> Now I have to examine the conditions after it hits water.
Here's> where my problem comes in I think. I'm trying to come
up with one> function that will be used for this entire
problem. It'll have two> parts (air/water) with the air part
used when t <= 15.506 and the> water part used when t >=
15.506, so I assume t = 15.506 is the> initial condition and
at 15.506 the velocity is -1.96. Is this a bad> idea?Depends
how accurate you want to be. If 5*t0 is large, where t0 is
thecritical value of t, then you can make this approximation;
I don't thinkthat 5*15.5 < 100 is all that large, however, so
I'd work out the actualvalue according to the first part and
use that.> First note that no momentum is lost.That's an
assumption; there is going to be some momentum lost in
thecollision with the water (resulting in a splash), but if
you wantto ignore it then fair enough.> The proportionality
constant> under water is 100 N*sec/m. Buoyancy is half of the
weight (.5*mg).> dv/dt = -g - pv + .5g = -.5g - pv, where p is
k/m or 50 this time.> Moving stuff around and setting v(15.506)
= -1.96 yields> v(t) = -9.516*10^336*e^(-50t) - .098> Perhaps
you see why I think I am making a mistake now. 10^336? That>
constant is boggling. I think I setup my initial conditions>
incorrectly. I do NOT want an exact answer to the problems,
much.Bearing in mind that t >= 15.506, the exp(50*15.506)
(which I assume iswhere you've got 10^336 from, I can't be
of exp(-50t), so the final answer willnot be O(10^336). This
is an example of my earlier point: you put numbersin before
absolutely necessary, and have thus confused yourself.For
computational purposes, you should in any case
calculateexp(-50*(t - 15.506)) rather than
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in throughout theprocess, but I decided to put the numbers in
for brevity. I alsomistyped those incorrect equations; it was
===
formula for this simple sequence.Maybe someone here will find
one and get the credit.The Encyclopedia of Integer Sequences
website:http://www.research.att.com/~njas/sequences/
I(n) IS the same sequence you found at EIS:Number of cells of
square lattice of edge 1/n inside quadrant of unitcircle
centered at 0.although I would not have sta it like that
===
applaud your impulse to be polite, it's counterproductive> to
start a new thread with it.While I applaud your impulse to
condemn counterproductivity, I would havereserved my
vituperation for the individual who replied to the
OP'squestion in an entirely new thread and with no indication
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Stan, end your flaming rampage Honey, when you answer as many
questions as I do -- even a tenth of that number -- then your
opinion will have value.-- Stan Brown, Oak Road Systems,
Cortland County, New York, USA http://OakRoadSystems.comAn
expense does not have to be required to be considered
===
Honey, when you answer as many questions as I do -- even a
tenth of> that number -- then your opinion will have value.Of
course, what else did I expect than yet another flame from
you. You canpost a million posts a day, but if 999,999 of them
are flames, then I don'tsee your point.OK I now see your point.
I forgot that *you* are the standard by whichUsenet posting is
===
questionsolving it.Let U be an open set in R^n.Considerf in
C^{infty}(U, R)such thatexists lim_{x -> x_0} f(x) /
||x-x_0||^{k-1} = 0 .Then f belongs to I^k_{x_0}(U, R)
.-------------------------------------------------------------
-----------------I^k_{x_0}(U, R) denotes the product of the
ideal I_{x_0}(U, R)k-times with itself and I_{x_0}(U, R)
denotes the ideal in C^{infty} (U,R)of function vaniscing at
p.In other words, f belongs to I^k_{x_0}(U, R) if and only if
fis of the form:f = h_0 g_01 * g_02 * .. * g_0{k-1}*g_0k +
+...+ + h_m g_m1 * g_02 * .. * g_m{k-1}*g_0kwhere m is a
natural number , say m=1 or m >1 ,h_i belongs to C^{infty}(U,
R)andg_ij belongs to I_{x_0}(U,R) <==> g_ij(x_0)= 0 in R.Note
that for m=0 , the proposition above is false,consider for
example k=2 and f(x,y)=x^2+y^2.Then, exists lim_{(x,y) ->
(00)} (x^2+y^2) / ||(x,y)||^{2-1} = lim_{(x,y) ->(00)}
(x^2+y^2)^{1/2}=0,on the other hand, it's impossible to write
f asf = h_0 g_01 * g_02 with g_01 and g_02 in C^{infty}(U, R)
and vanishingat p.My ask is:Is the proposition above true with
===
Integral> No:> sec^2(v) = 1 + tan^2(v)Oh, gosh. I swear I knew
that - really. I worked on that thing for ahalf hour and never
caught that. It's amazing how my brain can go sodense for so
===
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simplesince everyone pay 10$ and total they paid 30$when the
man return to them 3$ ,1$ for each onethe paid
3*(10-1)$=27$but the 27$ is nothing but the real price=25$+the
2$ in which the mantake. then it's wrong to say that the total
is 27$+2$=29$if u wanna make the correct calculation u should
do as follow27$+the sum in which they got back =3$ ant totally
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the green house is a white house insidethe white house is a red
===
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===
a Green house inside the green house is a white house
inside>the white house is a red house inside the red house is
alot of babies.The morons seem thick on the ground these
days.What is it about alt, undergrad, and math that attracts
people with non-mathematical riddles?-- Stan Brown, Oak Road
Systems, Cortland County, New York, USA
http://OakRoadSystems.comAn expense does not have to be
required to be considered necessary. -- IRS Form 1040 line 23
===
riddlewww.brainbashers.com - Spoilerwww.brainbashers.com -
Spoilerwww.brainbashers.com - Spoilerwww.brainbashers.com -
Spoilerwww.brainbashers.com - Spoilerwww.brainbashers.com -
===
help me solve this riddle by support1.mathforum.org
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but do not walk, I tell you things but do not talk.what am
===
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answer is a CLOCK.(It runs and it tells time.)