please help me with this. It's urgent and I would appreciate any help. Thanks

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- October 20th 2006, 03:56 AMsimfonijacalculus and ode's
please help me with this. It's urgent and I would appreciate any help. Thanks

- October 20th 2006, 04:37 AMdan
you need help with the whole test?

- October 20th 2006, 04:44 AMsimfonija
Yes, I don't have any notes and have real problems to solve it. Thanks in advance

- October 20th 2006, 04:46 AMdan
I need to go to work now but if no one esle helps I can try later.

though I might need help on some of it:eek:

dan - October 20th 2006, 04:52 AMSoroban
Hello, simfonija!

You need help with the*entire*exam?

Should we be helping you?

Having said that, here's #1 . . .

Quote:

1) Find the interval of convergence of the power series: .

Ratio Test

******

For convergence,

So we have: . . . .

Then: . . . . .

Subtract 1: . .

Multiply by -2: . .

Therefore: . . . .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

******

We have: .

Divide top and bottom by

Then: .

- October 20th 2006, 05:27 AMsimfonija
Hi and thanks. I know I am asking a lot but with no notes I am lost. And again please help with the rest

- October 20th 2006, 05:46 AMtopsquark
- October 20th 2006, 06:35 AMsimfonija
Ok, are you playing a teacher and I am bad student??? I was away for more than month (serious family reason) and hoping to manage to pass exam somehow. This is assignment which gives me 3% of total mark. On Monday I ll get copy of the notes. Just help if you can (no provocation please), if can't I also understand... Thanks

- October 20th 2006, 07:00 AMtopsquark
- October 20th 2006, 07:32 AMThePerfectHacker
I have something to add to Soroban's post.

He did not find the entire interval of convernge.

The endpoints are,

The two respective series we obtain are,

This can the zeta function , alternatively you can use the p-series test for exponent 1/2.

And also, the function,

is a positive continous decreasing function and the improper integral:

diverges. Thus,

Checking the point at we have,

We have that

The series is strictly decreasing and alternating. Thus, by Leibniz's test it is convergent.

- October 20th 2006, 02:01 PMsimfonija
Does anybody have idea how to solve 2 and 3 please :confused:

- October 20th 2006, 05:21 PMAfterShock
You're not going to learn how to do them if you have someone do ALL of them. Obviously this is a homework assignment that they expected you should be able to do. Did you even attempt any of these?

- October 20th 2006, 06:14 PMsimfonija
Hi

Task 2 I have no idea how to solve. Task number 3 I tried and I did some calculation but 3 (a) I don’t know why 6^1/2=2(1+1/2)^1/2 is better than the other. For 4 I also don’t have example of such diff equations. I am not so bad in maths but as I said this assignment is for Monday and on Monday I will copy the notes from the colleague. - October 21st 2006, 06:20 PMThePerfectHacker
#4)

Solve, for .

.

Since, this is equivalent to,

where are continous on some open interval there exists a basis of dimension two. Hence two linearly independant solutions.

We are given that is a solution.

To find, the second linearly independent solution we use reduction of order (which leads to):

The Wronskian (we can remove the constant multiplier) by Abel's theorem is,

Thus,

We see that this is a solution to this differencial equation which is linearly independent to

Thus, the set of all solutions can be expressed as (general solution),

Congratulation we solved the homogenous equation.

----

Now the non-homogenous equation.

For the purposes of the solution we need to find the Wronskian between these two solutions which is,

Now we use Variation of Parameters.

The formula is,

Thus,

Thus,

Integrate use by parts on first integral,

Thus,

Thus,

Thus, all solutions can be expressed as,