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Math Help - calculus and ode's

  1. #1
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    calculus and ode's

    please help me with this. It's urgent and I would appreciate any help. Thanks
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  2. #2
    dan
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    you need help with the whole test?
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  3. #3
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    Yes, I don't have any notes and have real problems to solve it. Thanks in advance
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  4. #4
    dan
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    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

    dan
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  5. #5
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    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: . \sum^{\infty}_{n=0}\frac{(1 - \frac{x}{2})^n}{\sqrt{n+1}}

    Ratio Test

    r \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{(1 - \frac{x}{2})^{n+1}}{\sqrt{n+2}} \cdot \frac{\sqrt{n+1}}{(1 - \frac{x}{2})^n} \;=\;\sqrt{\frac{n+1}{n+2}}\cdot\left(1 - \frac{x}{2}\right)

    R\;=\;\lim_{n\to\infty}|r|\;=\;\lim_{n\to\infty}\l  eft|\sqrt{\frac{n+1}{n+2}}\cdot\left(1 - \frac{x}{2}\right)\right| \;= \;\left|1 - \frac{x}{2}\right| **


    For convergence, R < 1

    So we have: . . . . \left|1 - \frac{x}{2}\right| \; < \; 1

    Then: . . . . . -1 \; < \; 1 - \frac{x}{2} \; < \; 1

    Subtract 1: . . -2 \; < \; -\frac{x}{2} \; < \; 0

    Multiply by -2: . . 4 \; > \; x \; > \; 0


    Therefore: . . . . \boxed{0  \: < \: x \: < \: 4}

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

    **
    We have: . \lim_{n\to\infty}\sqrt{\frac{n+1}{n+2}}

    Divide top and bottom by n:\;\;\sqrt{\frac{\frac{n}{n} + \frac{1}{n}}{\frac{n}{n} + \frac{2}{n}}} \;=\;\sqrt{\frac{1 + \frac{1}{n}}{1 + \frac{2}{n}}}

    Then: . \lim_{n\to\infty} \sqrt{\frac{1 + \frac{1}{n}}{1 + \frac{2}{n}}} \;= \;\sqrt{\frac{1 + 0}{1 + 0}} \;=\;\sqrt{1} \;= \;1

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  6. #6
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    Hi and thanks. I know I am asking a lot but with no notes I am lost. And again please help with the rest
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by simfonija View Post
    Hi and thanks. I know I am asking a lot but with no notes I am lost. And again please help with the rest
    Please forgive me, but I am curious. What happened to your notes?

    -Dan
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  8. #8
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    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
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by simfonija View Post
    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
    My apologies. I was not intending to attack you, I was simply asking a question.

    -Dan
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  10. #10
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    I have something to add to Soroban's post.
    He did not find the entire interval of convernge.
    The endpoints are,
    x=0,x=4
    The two respective series we obtain are,
    \sum_{n=0}^{\infty} \frac{1}{\sqrt{n+1}}
    This can the zeta function \zeta(1/2), alternatively you can use the p-series test for exponent 1/2.
    And also, the function,
    f(x)=\frac{1}{\sqrt(x+1)} is a positive continous decreasing function and the improper integral:
    \int_0^{\infty} \frac{dx}{\sqrt{x+1}} diverges. Thus, x\not =0

    Checking the point at x=4 we have,
    \sum_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{n+1}}
    We have that \lim_{n\to \infty}a_k=0
    The series is strictly decreasing and alternating. Thus, by Leibniz's test it is convergent.
    0<x\leq 4
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  11. #11
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    Does anybody have idea how to solve 2 and 3 please
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  12. #12
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    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?
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  13. #13
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    Hi
    Task 2 I have no idea how to solve. Task number 3 I tried and I did some calculation but 3 (a) I donít know why 6^1/2=2(1+1/2)^1/2 is better than the other. For 4 I also donít 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.
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  14. #14
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    #4)
    Solve, for x>0.
    x^2y''-xy'+y=0.
    Since, this is equivalent to,
    y''-\frac{1}{x}y'+\frac{1}{x^2}y=0 where p(x),q(x) are continous on some open interval there exists a basis of dimension two. Hence two linearly independant solutions.
    We are given that y_1=x is a solution.
    To find, the second linearly independent solution we use reduction of order (which leads to):
    y_2=y_1\int \frac{W(y_1,y_2)}{y_1^2} dx
    The Wronskian (we can remove the constant multiplier) by Abel's theorem is,
    \mbox{exp} \left( -\int p(x) dx \right)=e^{\ln  |x|}=x
    Thus,
    y_2=x\int \frac{x}{x^2}dx=x\ln x
    We see that this is a solution to this differencial equation which is linearly independent to y_1=x
    Thus, the set of all solutions can be expressed as (general solution),
    C_1 x+C_2 x\ln x
    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,
    W(y_1,y_2)=\left| \begin{array}{cc} x&x\ln x\\1&1+\ln x \end{array} \right|=x

    Now we use Variation of Parameters.
    The formula is,
    -y_1\int \frac{y_2 g(x)}{W(y_1,y_2)}dx+y_2\int \frac{y_1 g(x)}{W(y_1,y_2)}dx
    Thus,
    -x\int \frac{x\ln x\cdot x^{5/2}}{x} dx+x\ln x\int \frac{x\cdot x^{5/2}}{x} dx
    Thus,
    -x\int x^{5/2}\ln x dx+x\ln x\int x^{5/2} dx
    Integrate use by parts on first integral,
    -x\left( \frac{2}{7}\ln x x^{7/2} - \frac{4}{49} x^{7/2} \right) +x\ln x \left( \frac{2}{7}x^{7/2} \right)
    Thus,
    \frac{4}{49}x^{9/2}-\frac{2}{7}x^{9/2}\ln x+\frac{2}{7}x^{9/2}\ln x
    Thus,
    \frac{4}{49}x^{9/2}
    Thus, all solutions can be expressed as,
    C_1 x+C_2 x\ln x+\frac{4}{49}x^{9/2}
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