Expert: Paul Klarreich - 9/9/2007

Question
QUESTION: hello sir,
I am mahima from India. I came to know about this site by search engine. I am so happy to post my question(s).
I am unable to type questions becoz of mathematical symbols. If you dont mind, please give me your email ID, sothat I can post my questions. I will wait for your reply.
Thanking you,
mahima

ANSWER: Questioner:   mahima
Category:  Calculus
 
Subject:  calculus
Question:  hello sir,
I am mahima from India. I came to know about this site by search engine. I am so happy to post my question(s).
I am unable to type questions because of mathematical symbols. If you dont mind, please give me your email ID, s so that I can post my questions. I will wait for your reply.
Thanking you,
mahima
..........................
Hi, Mahima,

I am not permitted to supply the email address, but you can manage quite well even with this clumsy interface:

Use the ^ symbol for powers, such as x^2 for  "x squared"

Write fractions like this:

x + 3
------
2x - 8

In trigonometry, write out 'theta' or use the @ symbol for it.

If you have an integral, type something like:

{
| (x^2 + 2x) dx
}

Look at some of the previous questions -- all my answers and many by other 'experts' are listed.  

Use the "Browse Past Answers" link at the top, then look under Science for Mathematics, then look under Calculus or Advanced Mathematics.

You will see plenty of math symbols.  Cut and paste, if you like, but you won't have any trouble.  I look forward to answering your questions.

Oh, yes, don't forget to parenthesize when needed.

---------- FOLLOW-UP ----------

QUESTION: Thank you ver much for shoowing intrest for answering my questions.
1. Find the general solution of the given system of equation:- -  = 1  1  2  -
         x    1  2  1  x
         2  1  1  
Note: Rhs numbers are given in the matrix format.

2) Find the Fourier series for the given function
 f(x) = x, -pi < or = x < o
        0 < or = x < pi
f(x+2pi)= f(x).
3) Use Laplace Transform to solve given initial value problem, y^2 - 2y^1 + 4y = 0, y(0) =2, y^1 (0) = 0.
Note: y^2 means y duuble dash and y^1 means y dash.
4)(a)f(x) = 1(0<x < infinity)  has no Fourier cosine or sine transform. Give reasons.
(b) Does the cosine transform of x^-1 sinx  exist? Of x^-1 cosx  ? Give reasons.
Please answer my questions and encourage me in studies.
thanking you,
withlove,
mahima


Answer
Hi, Mahima,

That's a lot of questions all at once, so I'll see what I can do right now and you may have to wait or resubmit a couple of these.

1. Find the general solution of the given system of equation:- -  = 1  1  2  -
         x    1  2  1  x
         2  1  1  
Note: Rhs numbers are given in the matrix format.

Sorry, but this didn't come through correctly.  I don't understand the question.  If you want to write a matrix, try writing:

[ a b c ]
[ d e f ]
[ g h i ]

for the matrix.  What you wrote did not make sense.  
................................
2) Find the Fourier series for the given function
f(x) = x,     -pi <= x < 0
      what???  0 <= x < pi

I think you left something out here.  It is certainly proper to define a function differently in your two intervals, and you are trying to define  f(x) in each of the intervals, but you left out the definition for the second half of the interval.  

f(x+2pi)= f(x).
.................................


3) Use Laplace Transform to solve given initial value problem, y'' - 2y' + 4y = 0,
y(0) = 2, y'(0) = 0.

Use the standard table of LT's which tells you that, if  L(f(x)) = F(s), then:

L(y') =  sF(s) - f(0)
L(y'') = s^2 F(s) - s f(0) - f'(0)

Now use your equation:

y'' - 2y' + 4y = 0  to obtain:

L(y'' - 2y' + 4y) = L(0)

s^2 F(s) - s f(0) - f'(0) -2(sF(s) - f(0)) + 4F(s) = 0

Put in your initial values:

s^2 F(s) - s(2)  -   (0) - 2(sF(s) - 2) + 4F(s) = 0

s^2 F(s) - 2s  - 2sF(s) + 4 + 4F(s) = 0

s^2 F(s) - 2sF(s) + 4F(s) = 2s - 4

F(s)(s^2 - 2s + 4) = 2s - 4
         2s - 4
F(s) =  -------------
       s^2 - 2s + 4

Now invert the transform to find f(x).  (or  f(t), whatever)  You will do the following:

1. Complete the square in the bottom:  s^2 - 2s + 4 = s^2 - 2s + 1 + 3 = (s - 1)^2 + 3.

2. Separate the fraction into two parts:

         2s - 4          2s          4
F(s) =  ------------- =  -------------- - ---------------
       s^2 - 2s + 4     (s - 1)^2 + 3     (s - 1)^2 + 3


Now check the table of transforms at this site:

http://www.vibrationdata.com/Laplace.htm

and find the one that applies.  You can do that.