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William Elliot
science forum Guru

Joined: 24 Mar 2005
Posts: 1906

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: differential equation help

Learn to cross post!!

On Mon, 28 Feb 2005, Carlos Romero wrote:

 Quote: Hello, I was wondering if some could help me find where I went wrong on http://www.techdigital.net/math/question.jpg I am supposed to solve the differential equation y''+4y=0 with the method 'reduction of order', and get y=c1sin(2x)+c2cos(2x); however, that is not what I am getting and cannot find where I went wrong. In my book its telling me to solve it by taking the 'higher order' differential equation that cannot be solve and reducing them to first order differential equations that can, which they call "reduction of order." Thanks Carlos R.
Triple888
science forum beginner

Joined: 24 Mar 2005
Posts: 3

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: The Beauty of Math

Anthony wrote:
 Quote: I am reading many physics books lately and the equations in them are all so interesting...the beauty of a powerful equation. So I am wondering...I want to get back into math again (I was in colllege for a couple of years about 15 years ago)...I took Calculus 1 and 2, Differential Equations, Physics, etc......your basic college core stuff. But now, 15 years later, I want to re-learn math...and to be able to understand and figure out how these physicists came up with these advanced equations. I look at it like I am setting out to learn a new foreign language just like anyone else would teach themselves French, Spanish, Italian, etc.... So does anyone have any suggestions for best going about this? Any books or study tips to use to accomplish my goal? Where even to start? My main goal is to best understand the equations in books from like Stephen Hawking, Einstein, string theory, etc.......theoretical physics stuff...dealing with cosmology and quantum mechanics. Thanks in advance.

Like in one of the other replies: best to go back to your school text
books and go from there.

If those too hard, go a bit further back. If easy maybe college books.
Also, you might want and find a school college syllabus and see what
todays topics are.

Whatever your plans are you sure need to work hard at it! Also, think
prospects? Or just to satisfy the hungry mind?
Guest

 Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: The Beauty of Math I greatly appreciated "Mathematical Thought from Ancient to Modern Times" by Morris Kline. The original text was one tome of about 1500 pages. It is now published in several volumes. It litterally starts the history of math from ancient times, but progresses to relatively modern times, and shows show problems of physical interest lead to, say, Riemannian geometry and tensor calculus. There is always a close interplay in the book between physics and math history. Alex
Michael A. Terrell
science forum beginner

Joined: 24 Mar 2005
Posts: 20

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: 40M for Bush Inauguration and 15M for Tsunami Disaster WAS Re: Bush accused of undermining the UN with aid coalition

Franz Heymann wrote:
 Quote: "James Beck" wrote in message OK, How about 3, that's THREE, hurricanes hit Florida almost back to back. How much aid did the world send? Lots of people with no roofs over their heads. You were in a position to cope. The countries most seriously affected by the present catastrophe are not in a position to cope. Franz

How the hell do you know what it was like in Florida? I lived
through all the hurricanes that passed through here. There are still
roofs on my street with blue tarps covering holes where trees fell on
the roof. Why is this? Because there is still a shortage of shingles
and plywood. The building materials are being shipped overseas to take
care of people in other countries.

Two of my neighbors lost their homes during the hurricanes and another
home burnt to the ground while being repaired. We are getting close to
the 2005 hurricane season and there are businesses waiting on
replacement signs. Some companies went out of business because of lost
business due to damage and weeks without electricity. There was very
little gasoline, no food in the stores after the first day they reopened
and very little bottled water available. What was available came from a
bottling plant up north that shut down operations and switched to

A damaged power pole on my street was just repaired a couple weeks
ago. A lot of cell sites were damaged the remaining ones were
completely overloaded. Land lines were mostly out and I went weeks at a
time without electricity. I have severe breathing problems so I sat
around gasping for breath for weeks. I had to spend time in a shelter
for the disabled, along with hundreds who couldn't go home because the
roads were closed and there was a total curfew. They couldn't even
finds out if their homes were still standing. The local radio and TV
stations were off the air so we got little news, mostly from 100 miles
or more away by AM radio.

How about the next time their is a disaster somewhere else in t he
world we just tell the UN to handle it and not give one red cent, or
send any of our military and see how well you assholes handle things on
your own? We were still delivering aid when the ungrateful bastards
told us we had to leave by a certain date, and the morons at the UN were
complaining the helicopters weren't spending a whole lot of time doing
surveys instead of delivering food, water and medicines. It would be an
interesting experiment, but it isn't the way the US treats others.

I say we should evict the UN and make them move to a third world
country for their new headquarters. They could use a little of the
money they have stolen to pay protection money to the locals so they
don't blow up tier offices every other week. They shouldn't be in one
of the safest places on earth, they should be in the worst hell hole we
can find. It might give them incentive to do what they area supposed to
do, other than take bribes and steal from programs they oversee.

BTW, did you have any brain cells that survived?
--
?

Michael A. Terrell
Central Florida
Frank Bemelman
science forum beginner

Joined: 24 Mar 2005
Posts: 5

 Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: 40M for Bush Inauguration and 15M for Tsunami Disaster WAS Re: Bush accused of undermining the UN with aid coalition "Michael A. Terrell" schreef in bericht news:423256FF.3AF8557A@earthlink.net... [snip] And I don't have to read yet another fucking idiot. PLONK! -- Thanks, Frank. (remove 'q' and 'invalid' when replying by email)
Mark Jones
science forum beginner

Joined: 24 Mar 2005
Posts: 11

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: 40M for Bush Inauguration and 15M for Tsunami Disaster WAS Re: Bush accused of undermining the UN with aid coalition

Frank Bemelman wrote:
 Quote: "Michael A. Terrell" schreef in bericht news:423256FF.3AF8557A@earthlink.net... [snip] And I don't have to read yet another blah,blah,blah,blah...

In many ways, Michael makes a point. It is easy for everyone else to criticize
the US, and with all the #&@^%\$-up things going on here, it's no wonder. But
when it comes down to it, Americans make a lot of sacrifices and ALWAYS give
generously. It is very discouraging to hear others criticize you when you are
the one giving. It makes you want to NOT be so nice. And can you blame us?

Like I said before, 40M for Bush inaguration and 15M for Tsunami relief - does
that 15M include all the helicopter costs? We figured that each one costs about
\$1200/hr to operate (fuel, maintainance, personnel, insurance, etc.), and there
were a hundred of them operating 24h/day for weeks... add that figure up! The
cost of operating the desalination equipment on the navy vessels relocated there
for the same timeframe, was that also included? What about all the other
military men's pay? The Sailors? The Marines? Who paid for that?

And did anyone ever say THANK YOU?

Yeah, Americans might have money. That doesn't mean we HAVE to piss it all
away, especially to ungrateful countries. As Michael pointed out, we've got our
own problems at home which have been backing up. We'll give away what we deem is
fair (and graciously so) - have some respect!

I'm sorry Michael has had such a hard time in Florida. I'd help him out
physically or financially, except I'm in Ohio, without a pot to piss in myself
like most of my neighbors and the rest of the working class. Been unemployed for
over a year now, barely scratching by thanks to Bush and his outsourcing tax
incentives. And the world thinks Americans are all rich... no, I'm afraid we're
all stupid instead. Stupid for A) putting other countries ahead of our own
welfare, and B) electing and then, miraciously, re-electing Bush.

-- "Perseverance, and perseverance alone, is omnipotent." MCJ 1998
Stan Brown
science forum Guru Wannabe

Joined: 06 May 2005
Posts: 279

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: IIT screening test question

 Quote: On 12 Feb 05 03:20:19 -0500 (EST), pH wrote: Q: A bear is chasing a man, who reaches the edge of a cliff and jumps down. The man reaches the ground in 5s. The height of the cliff is 123 metres. What is the colour of the (fur of the) bear? white because the value of g=9.8 which is at north pole and south pole which is polar region

Nonsense. The time of 5 s contains one significant digit. The
variation in the acceleratikon of gravity over the earth's surface
is FAR too small to show at that level of precision.

--

Stan Brown, Oak Road Systems, Tompkins County, New York, USA
Ashutsoh
science forum beginner

Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: Sum of series

On 27 Feb 05 03:01:08 -0500 (EST), john wrote:
 Quote: For sigma (k=1,oo) k/(2^k) You should have figured this out for yourself.

Anyway.

 Quote: k/(2^k)= (1/2)[(k-1)/(2^(k-1)) + 1/(2^(n-1))]. Thus, sum(k=1,n) k/(2^k) = 1/2[ (0/1 + 1/1)+(1/2 + 1/2) + (2/4 + 1/4) + (3/8 + 1/ + (4/16 +1/16) + (5/32 + 1/32)+...+ {(n-1)/(2^(n-1)) + 1/(2^(n-1))}] = 1/2[1+1+ 3/4 + 4/8 + 5/16 + 6/32 + ...+ n/2^n] = 1/2[1+1+1+ 16/32+...+ n/2^n] = 1/2[3 + 1/2 + ...+ n/2^n]

Let the nth partial sum be x(n),
then x(n) = (1/2)( x(n-1) + 2 - 1/(2^(n-1)) ), or
2*x(n) - x(n-1) = 2 - 1/(2^(n-1)) ), or
taking limits as n->oo and noting that limits of all terms exist,
let sum of original series be x,
then 2*x - x = 2 - 0, or
x = 2.

 Quote: I'm thinking that the sum of the series will be 2. But I'm stuck here, I don't know what else to do. john On 26 Feb 05 21:06:37 -0500 (EST), Ashutosh wrote: On 25 Feb 05 21:51:10 -0500 (EST), john wrote: Find the sum of: sigma (k=1,oo) 1/[k(k+1)(k+2)] I know that this 1/[k(k+1)(k+2)] converges to zero, but have a hard time coming up with what the sum of the series actually converges to sigma (k=1,oo) k/(2^k) this one I'm suppose to find the sum of the series, but I'm sure that k/(2^k) converges to 1/2, but still can't find the sum of the series. Is there some trick that I should know about? Write 1/[k(k+1)(k+2)] = (1/2)[V(k)-V(k+1)] (Telescoping series), where V(k) = 1/k(k+1). For the second series, write k/(2^k) = (1/2)[(k-1)/(2^(k-1)) + 1/(2^(k-1))]. The trick lies in simplifying the expression for the partial sums, before taking the limit.
henry

Joined: 24 Mar 2005
Posts: 56

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: Trigonomentry equatrions

On 10 Sep 01 09:57:21 -0400 (EDT), Saleem Ahmed wrote:
 Quote: Would appreciate if could help me solve the following. I tried it a few times but was stuck. Solve 1. cotx * cosx = 1+sinx 2. 7 secx = 3(cosx-tanx) Thanks, Saleem

2
7sec x=3(cos x-tan x)
7=3cos(cos^2 x-sin x)
1=3/7 *cos x(1-sin^2-sin x)

1= sin 25 sin (90-x)(1-sin^2 x-sin x)

sin (90-x)=sin 2.8 or sin x(-sin x-1)=0
__________________________________________
Annette Carter
science forum beginner

Joined: 24 Mar 2005
Posts: 1

 Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: Statistic and probability problem I know that there are several ways to solve a problem, but I am trying to get a simple understanding of assessomg probability options.
henry

Joined: 24 Mar 2005
Posts: 56

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: Financial + math problem

On 24 Feb 05 19:12:58 -0500 (EST), Roxanne wrote:
 Quote: Hi I have problem that I can't solve . Its concerning a mortgage and the interest rate that would allow the person to repay in 10 years (monthly payment) to clear the mortgage. This person wants to repay with the following payment: 1st month 701\$ , 2nd month 702\$.... 60th month 760\$. Then for the following 60 payment 700\$ per month. What is the interest rate that would allow the person to pay in such payments

Splits on margin,is a good tool.Or do you need to go to the core?.
john q. public
science forum beginner

Joined: 24 Mar 2005
Posts: 1

 Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: Algebra Problem solvers www.gomath.com is worthless; any other good suggestions
john11116
science forum beginner

Joined: 24 Mar 2005
Posts: 2

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: Sum of series

For sigma (k=1,oo) k/(2^k)

k/(2^k)= (1/2)[(k-1)/(2^(k-1)) + 1/(2^(n-1))].
Thus, sum(k=1,n) k/(2^k) = 1/2[ (0/1 + 1/1)+(1/2 + 1/2) + (2/4 + 1/4)
+ (3/8 + 1/ + (4/16 +1/16) + (5/32 + 1/32)+...+ {(n-1)/(2^(n-1)) +
1/(2^(n-1))}] = 1/2[1+1+ 3/4 + 4/8 + 5/16 + 6/32 + ...+ n/2^n]
= 1/2[1+1+1+ 16/32+...+ n/2^n]
= 1/2[3 + 1/2 + ...+ n/2^n]
I'm thinking that the sum of the series will be 2. But I'm stuck here,
I don't know what else to do.

john

On 26 Feb 05 21:06:37 -0500 (EST), Ashutosh wrote:
 Quote: On 25 Feb 05 21:51:10 -0500 (EST), john wrote: Find the sum of: sigma (k=1,oo) 1/[k(k+1)(k+2)] I know that this 1/[k(k+1)(k+2)] converges to zero, but have a hard time coming up with what the sum of the series actually converges to sigma (k=1,oo) k/(2^k) this one I'm suppose to find the sum of the series, but I'm sure that k/(2^k) converges to 1/2, but still can't find the sum of the series. Is there some trick that I should know about? Write 1/[k(k+1)(k+2)] = (1/2)[V(k)-V(k+1)] (Telescoping series), where V(k) = 1/k(k+1). For the second series, write k/(2^k) = (1/2)[(k-1)/(2^(k-1)) + 1/(2^(k-1))]. The trick lies in simplifying the expression for the partial sums, before taking the limit.
Dave11125

Joined: 04 Feb 2005
Posts: 80

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: properties of i

Dale,

No, it cannot, because the reciprocal of i is actually -i. This is
because:

1/i = (-i * 1)/(-i * i) = -i / -(i * i) = -i / -(-1) = i

Basically you take 1/i and multiply denominator and numerator by -i.
To find the reciprocal of any complex number, a + bi, write

1/(a + bi)

and multiply numerator and denominator by (a - bi) and then it's just
basic algebra with the extra rule that i * i = -1.

From,

Dave

On 26 Feb 05 02:20:08 -0500 (EST), Dale Heffernan wrote:
 Quote: What is the inverse of i? Can it be moved from denominator to numerator just as if it were 1 or -1? Thanks, Dale
Veronica Howse
science forum beginner

Joined: 24 Mar 2005
Posts: 2

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