please help me with this. It's urgent and I would appreciate any help. Thanks
Hello, simfonija!
You need help with the entire exam?
Should we be helping you?
Having said that, here's #1 . . .
1) Find the interval of convergence of the power series: .
Ratio Test
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For convergence,
So we have: . . . .
Then: . . . . .
Subtract 1: . .
Multiply by -2: . .
Therefore: . . . .
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We have: .
Divide top and bottom by
Then: .
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
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.
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.
#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.
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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,