Results 1 to 14 of 14

Math Help - Integrals aren't real...

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    228

    Integrals aren't real...

    Or are they?



    Guys, find the antiderivative of {1/2y dy. There are... two antiderivatives? Those are (1/2)ln(abs(2y)) + C and (1/2)ln(abs(y)) + C. Confused? So am I.

    Try solving the differential equation dy/dx = 2y(3-x) twice, using the identical antiderivatives of {1/2y dy = (1/2)ln(abs(2y)) + C and (1/2)ln(abs(y)) + C. I think you'll come up with different answers just like I did lol.


    I've another question. Is it possible to find the exact sum of the following series -

    -3/(2^2) + 6/(2^5) - 9/(2^8)....
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,740
    Thanks
    478
    (1/2)ln(abs(2y)) + C and (1/2)ln(abs(y)) + C
    these expressions differ by a constant. what does that tell you?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by Kaitosan View Post
    Or are they?



    Guys, find the antiderivative of {1/2y dy. There are... two antiderivatives? Those are (1/2)ln(abs(2y)) + C and (1/2)ln(abs(y)) + C. Confused? So am I.

    Try solving the differential equation dy/dx = 2y(3-x) twice, using the identical antiderivatives of {1/2y dy = (1/2)ln(abs(2y)) + C and (1/2)ln(abs(y)) + C. I think you'll come up with different answers just like I did lol.


    I've another question. Is it possible to find the exact sum of the following series -

    -3/(2^2) + 6/(2^5) - 9/(2^8)....
    \frac{1}{2}\ln(2y) and \frac{1}{2}\ln(y) differ by a constant.

    For \frac{1}{2}\ln(y) + C, let C=\frac{1}{2}\ln(2) and you'll get \frac{1}{2}\ln(2y)

    In fact, \frac{d}{dy}\left[\frac{1}{2}\ln(ky)\right] = \frac{1}{2y} for any k.

    The sum is 3\sum_{n=1}^{\infty}\frac{n(-1)^n}{2^{3n-1}} = 6\sum_{n=1}^{\infty}\frac{n(-1)^{3n}}{2^{3n}} = 6\sum_{n=1}^{\infty}n\left(\frac{-1}{2}\right)^{3n}

    Hmm...I'm actually blanking on how to compute the exact sum, but Maple says it's -\frac{16}{27}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,746
    Thanks
    647
    Hello, Kaitosan!

    Find the exact sum of: . -\frac{3}{2^2} + \frac{6}{2^5} - \frac{9}{2^8} + \hdots

    We have: . . . . . S \; = \; -\frac{3}{2^2} + \frac{6}{2^5} - \frac{9}{2^8} + \frac{12}{2^{11}} - \frac{15}{2^{14}} + \hdots  .[1]

    Multiply by \frac{1}{2^3}\!: . \frac{1}{8}S \;=\; \qquad\;\; -\frac{3}{2^5} + \frac{6}{2^8} - \frac{9}{2^{11}} + \frac{12}{2^{14}} - \hdots .[2]


    Add [1] and [2]: . \frac{9}{8}S \;=\;-\frac{3}{2^2} + \frac{3}{2^5} - \frac{3}{2^8} + \frac{3}{2^{11}} - \frac{3}{2^{14}} + \hdots

    \text{And we have: }\;\frac{9}{8}S \;=\;-\frac{3}{2^2}\underbrace{\bigg[1 - \frac{1}{2^3} + \frac{1}{2^6} - \frac{1}{2^9} + \frac{1}{2^{12}} - \hdots\bigg]}_{\text{geometric series}}

    The G.S. has first term a = 1 and common ratio r = -\frac{1}{8}

    . . Its sum is: . \frac{1}{1-(-\frac{1}{8})} \:=\:\frac{1}{\frac{9}{8}} \:=\:\frac{8}{9}


    Therefore: . \frac{9}{8}S \;=\;-\frac{3}{4}\left(\frac{8}{9}\right) \quad\Rightarrow\quad \boxed{S \;=\;-\frac{16}{27}}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jan 2009
    Posts
    290
    Quote Originally Posted by redsoxfan325 View Post
    In fact, \frac{d}{dy}\left[\frac{1}{2}\ln(ky)\right] = \frac{1}{2y} for any k.
    What?

    I thought \frac {d} {dx} \ln x = \frac {x'} {x}

    So wouldn't \frac {d} {dx} \ln 2x = \frac {2} {x}?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by chengbin View Post
    What?

    I thought \frac {d} {dx} \ln x = \frac {x'} {x}

    So wouldn't \frac {d} {dx} \ln 2x = \frac {2} {x}?
    \frac d{dx}\left[\ln2x\right]=\frac{d/dx[2x]}{2x}=\frac2{2x}=\frac1x
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jan 2009
    Posts
    290
    Quote Originally Posted by Reckoner View Post
    \frac d{dx}\left[\ln2x\right]=\frac{d/dx[2x]}{2x}=\frac2{2x}=\frac1x
    OMG I'm an idiot!!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Dec 2008
    Posts
    228
    Quote Originally Posted by Soroban View Post
    Hello, Kaitosan!


    We have: . . . . . S \; = \; -\frac{3}{2^2} + \frac{6}{2^5} - \frac{9}{2^8} + \frac{12}{2^{11}} - \frac{15}{2^{14}} + \hdots  .[1]

    Multiply by \frac{1}{2^3}\!: . \frac{1}{8}S \;=\; \qquad\;\; -\frac{3}{2^5} + \frac{6}{2^8} - \frac{9}{2^{11}} + \frac{12}{2^{14}} - \hdots .[2]


    Add [1] and [2]: . \frac{9}{8}S \;=\;-\frac{3}{2^2} + \frac{3}{2^5} - \frac{3}{2^8} + \frac{3}{2^{11}} - \frac{3}{2^{14}} + \hdots

    \text{And we have: }\;\frac{9}{8}S \;=\;-\frac{3}{2^2}\underbrace{\bigg[1 - \frac{1}{2^3} + \frac{1}{2^6} - \frac{1}{2^9} + \frac{1}{2^{12}} - \hdots\bigg]}_{\text{geometric series}}

    The G.S. has first term a = 1 and common ratio r = -\frac{1}{8}

    . . Its sum is: . \frac{1}{1-(-\frac{1}{8})} \:=\:\frac{1}{\frac{9}{8}} \:=\:\frac{8}{9}


    Therefore: . \frac{9}{8}S \;=\;-\frac{3}{4}\left(\frac{8}{9}\right) \quad\Rightarrow\quad \boxed{S \;=\;-\frac{16}{27}}

    Oh wow, that's awesome haha. Ok, thank you very much for showing me how to do the problem. However, I'm a bit confused. I've been doing only the basic series computation as shown in your last few steps.

    I can clearly see that your first few steps (multiplying by 1/8 and adding two series) helped to isolate a true geometric series. But what made you choose those methods? How would I recognize when and what to multiply/add series to change it up so as to use the GS formula? Suppose there's another complex GS problem like this one. How would I know exactly what to do in the first few steps in order to simplify it? I refuse to believe that you made a lucky guess of which number to multiply the series lol.

    I appreciate your help very much!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Dec 2008
    Posts
    228
    For example, how would you find the series of...

    1/3 + 2/9 + 3/27 + 4/81......

    Apparently you can't use convenient means of geometric addition. So what to multiply by?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by Kaitosan View Post
    For example, how would you find the series of...

    1/3 + 2/9 + 3/27 + 4/81......

    Apparently you can't use convenient means of geometric addition. So what to multiply by?
    \sum_{n=1}^{\infty}\frac{n}{3^n}

    You can use a similar trick:

    S = 1/3 + 2/9 + 3/27 + 4/81...~~~~~(1)

    3S = 1+2/3+3/9+4/27+5/81...~~~~~(2)

    Subtract (1) from (2):

    2S = \underbrace{1+1/3+1/9+1/27+1/81...}_{\text{geometric series}}

    For the GS: a_0 = 1; r=\frac{1}{3}

    sum of GS = \frac{1}{1-1/3} = \frac{3}{2}

    So we have 2S = \frac{3}{2}

    \boxed{S = \frac{3}{4}}
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Dec 2008
    Posts
    228
    Sigh. I completely understand the process but the problem is that I don't understand "why", "what", and "when." All I understand is that there are tricks to simplify series by adding and multiplying them by something but how do you know exactly what to multiply???
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    I think I understand what you are getting at and unfortunately there is no easy answer. When learning integrals for the first time, people constantly feel in the same boat, asking "Well how did you know to use 'x' trick or 'y' formula?". They get upset that I can look at the problem and know instantly what to do. The best answer for this I think is that you just have to lots and lots of these problems. The more you do and the more different types you encounter, the better you will get at recognizing the little differences that lead to certain methods. There is no generic answer for this. Sorry.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    I second that.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Dec 2008
    Posts
    228
    Thanks for trying though!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving that there aren't any more homomorphisms
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 23rd 2011, 06:51 PM
  2. Residue Theorem and Real integrals
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 17th 2011, 07:06 AM
  3. Contour Integrals (to Evaluate Real Integrals)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: January 17th 2011, 09:23 PM
  4. Non-real Integrals...
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 15th 2010, 02:28 PM
  5. Real Analysis: Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 28th 2008, 09:39 AM

Search Tags


/mathhelpforum @mathhelpforum