Results 1 to 14 of 14

Thread: Sum

  1. #1
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Sum

    Here is a fun one

    Find

    $\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n\quad\forall~|x|<1$

    Where $\displaystyle a$ is an arbitrary constant.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Thanks
    1
    $\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n = \sum_{n=0}^{\infty}(n^3+1)x^n+\sum_{n=0}^{\infty}( a-1)x^n$

    $\displaystyle \boxed{1}~\sum_{n=0}^{\infty}(a-1)x^n = \frac{a-1}{1-x}$

    Now the other one.

    $\displaystyle f(x) = \sum_{n=0}^{\infty}(n^3+1)x^n$

    $\displaystyle f(x) = \sum_{n=0}^{\infty}(n+1)(n^2-n+1)x^n$

    $\displaystyle \int f(x)~dx = F(x)$

    $\displaystyle F(x) = \sum_{n=0}^{\infty}(n^2-n+1)x^{n+1}$

    $\displaystyle F(x) = \sum_{n=0}^{\infty}(n^2-n)x^{n+1}+\sum_{n=0}^{\infty}x^{n+1}$

    $\displaystyle F(x) = \sum_{n=0}^{\infty}n(n-1)x^{n+1}+\frac{x}{1-x}$

    $\displaystyle F(x) = x^3\underbrace{\sum_{n=0}^{\infty}n(n-1)x^{n-2}}_{\text{here}}+\frac{x}{1-x}$

    Now integrate that sum two times, then apply the geometric progression formula, then differentiate two times.

    $\displaystyle F(x) = \frac{2x^3}{(1-x)^3}+\frac{x}{1-x}$

    $\displaystyle \boxed{2}~f(x)=\frac{7x^2-2x+1}{(x-1)^4}$

    Finally, $\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n = \frac{7x^2-2x+1}{(x-1)^4}+\frac{a-1}{1-x}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Thanks
    1
    The other way to solve this is insipid and not that beautiful.

    $\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n = \sum_{n=0}^{\infty}n^3 x^n+\sum_{n=0}^{\infty}a x^n$

    $\displaystyle \sum_{n=0}^{\infty}a x^n = \frac{a}{1-x}$

    $\displaystyle f = \sum_{n=0}^{\infty}n^3 x^n = x \underbrace{\sum_{n=0}^{\infty}n^3 x^{n-1}}_{g}$

    $\displaystyle G = \sum_{n=0}^{\infty}n^2 x^{n} = x \underbrace{\sum_{n=0}^{\infty}n^2 x^{n-1}}_{h}$

    $\displaystyle H = \sum_{n=0}^{\infty}n x^{n} = x\underbrace{\sum_{n=0}^{\infty}n x^{n-1}}_{t}$

    $\displaystyle T =\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$

    Now rewind it to f by differentiating and multiplying by x.


    See this page for some related info:
    Polylogarithm - Wikipedia, the free encyclopedia
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    I am going to be lazy and skip a bunch of Latexing. But here is my approach.

    I often worked these by using derivatives. A common method.

    $\displaystyle \sum_{n=0}^{\infty}n^{3}x^{n}+\sum_{n=0}^{\infty}a x^{n}$

    Of course, we know the second one is $\displaystyle \frac{a}{1-x}$.

    Let's put it aside for the time being.

    $\displaystyle \sum_{n=0}^{\infty}n^{3}x^{n}$

    Let's work with the third derivative:

    $\displaystyle =\frac{dx}{dx^{3}}\left[\frac{1}{1-x}\right]=\sum_{n=3}^{\infty}n(n-1)(n-2)x^{n-3}$

    $\displaystyle \frac{6x^{3}}{(x-1)^{4}}=x^{3}\frac{d}{dx^{3}}\left[\frac{1}{1-x}\right]=\sum_{n=0}^{\infty}n(n-1)(n-2)x^{n}=(n^{3}-3n^{2}+2n)x^{n}$

    $\displaystyle \sum_{n=0}^{\infty}n^{3}x^{n}-3\sum_{n=0}^{\infty}n^{2}x^{n}+2\sum_{n=0}^{\infty }nx^{n}$

    Now, I have a few of these written down from deriving them before. Saving me the time of doing it every time.

    So we get from the third derivative above:

    $\displaystyle \sum_{n=0}^{\infty}n^{3}x^{n}=\frac{6x^{3}}{(x-1)^{4}}-\frac{3x(x+1)}{(x-1)^{3}}-\frac{2x}{(x-1)^{2}}=\frac{x^{3}+4x^{2}+x}{(x-1)^{4}}$

    Combine with the other easier part and we joyfully get:

    $\displaystyle =\frac{x^{3}+4x^{2}+x}{(x-1)^{4}}+\frac{a}{1-x}$

    Which I am pretty sure is equivalent to wingless' solution, just in another form.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by wingless View Post
    $\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n = \sum_{n=0}^{\infty}(n^3+1)x^n+\sum_{n=0}^{\infty}( a-1)x^n$

    $\displaystyle \boxed{1}~\sum_{n=0}^{\infty}(a-1)x^n = \frac{a-1}{1-x}$

    Now the other one.

    $\displaystyle f(x) = \sum_{n=0}^{\infty}(n^3+1)x^n$

    $\displaystyle f(x) = \sum_{n=0}^{\infty}(n+1)(n^2-n+1)x^n$

    $\displaystyle \int f(x)~dx = F(x)$

    $\displaystyle F(x) = \sum_{n=0}^{\infty}(n^2-n+1)x^{n+1}$

    $\displaystyle F(x) = \sum_{n=0}^{\infty}(n^2-n)x^{n+1}+\sum_{n=0}^{\infty}x^{n+1}$

    $\displaystyle F(x) = \sum_{n=0}^{\infty}n(n-1)x^{n+1}+\frac{x}{1-x}$

    $\displaystyle F(x) = x^3\underbrace{\sum_{n=0}^{\infty}n(n-1)x^{n-2}}_{\text{here}}+\frac{x}{1-x}$

    Now integrate that sum two times, then apply the geometric progression formula, then differentiate two times.

    $\displaystyle F(x) = \frac{2x^3}{(1-x)^3}+\frac{x}{1-x}$

    $\displaystyle \boxed{2}~f(x)=\frac{7x^2-2x+1}{(x-1)^4}$

    Finally, $\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n = \frac{7x^2-2x+1}{(x-1)^4}+\frac{a-1}{1-x}$
    Good job!

    Here was my solution (pretty similar to yours)

    $\displaystyle x^3\cdot\frac{d}{dx}\bigg[\frac{1}{1-x}\bigg]=\sum_{n=0}^{\infty}n(n-1)(n-2)x^n=\sum_{n=0}^{\infty}n^3x^n-3\sum_{n=0}^{\infty}n^2x^n+2\sum_{n=0}^{\infty}nx^ n$


    Ok, so firstly we see that

    $\displaystyle \boxed{1}\quad{x\cdot\frac{d}{dx}=\sum_{n=0}^{\inf ty}nx^n}$

    Next we see that

    $\displaystyle x^2\cdot\frac{d^2}{dx^2}\bigg[\frac{1}{1-x}\bigg]=\sum_{n=0}^{\infty}n^2x^n-\sum_{n=0}^{\infty}nx^n$

    So now using $\displaystyle \boxed{1}$ we can see that

    $\displaystyle \boxed{2}\quad{x^2\cdot\frac{d}{dx}\bigg[\frac{1}{1-x}\bigg]+x\cdot\frac{d}{dx}\bigg[\frac{1}{1-x}\bigg]=\sum_{n=0}^{\infty}n^2x^n}$

    So now combinging $\displaystyle \boxed{1}$ and $\displaystyle \boxed{2}$

    We get


    $\displaystyle x^3\cdot\frac{d^3}{dx^3}\bigg[\frac{1}{1-x}\bigg]=\sum_{n=0}^{\infty}n^3x^n-3\bigg[x^2\cdot\frac{d^2}{dx^2}\bigg[\frac{1}{1-x}\bigg]+x\cdot\frac{d}{dx}\bigg[\frac{1}{1-x}\bigg]\bigg]+2x\frac{d}{dx}\bigg[\frac{1}{1-x}\bigg]$

    Simplifying gives

    $\displaystyle \boxed{3}\quad\frac{x(x^2+4x+1)}{(x-1)^4}=\sum_{n=0}^{\infty}n^3x^n$


    And finally seeing that

    $\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n=\sum_{n=0}^{\infty}n ^3x^n+a\sum_{n=0}^{\infty}x^n$

    And utilizing $\displaystyle \boxed{3}$

    as well as the commonly known

    $\displaystyle a\sum_{n=0}^{\infty}x^n=\frac{a}{1-x}$

    We get

    $\displaystyle \boxed{\sum_{n=0}^{\infty}(n^3+a)x^n=\frac{x(x^2+4 x+1)}{(x-1)^4}+\frac{a}{1-x}}$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    Hey Mathstud. Looks me and you were thinking a like. Good to see we checked out.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by galactus View Post
    Hey Mathstud. Looks me and you were thinking a like. Good to see we checked out.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by wingless View Post
    The other way to solve this is insipid and not that beautiful.

    $\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n = \sum_{n=0}^{\infty}n^3 x^n+\sum_{n=0}^{\infty}a x^n$

    $\displaystyle \sum_{n=0}^{\infty}a x^n = \frac{a}{1-x}$

    $\displaystyle f = \sum_{n=0}^{\infty}n^3 x^n = x \underbrace{\sum_{n=0}^{\infty}n^3 x^{n-1}}_{g}$

    $\displaystyle G = \sum_{n=0}^{\infty}n^2 x^{n} = x \underbrace{\sum_{n=0}^{\infty}n^2 x^{n-1}}_{h}$

    $\displaystyle H = \sum_{n=0}^{\infty}n x^{n} = x\underbrace{\sum_{n=0}^{\infty}n x^{n-1}}_{t}$

    $\displaystyle T =\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$

    Now rewind it to f by differentiating and multiplying by x.


    See this page for some related info:
    Polylogarithm - Wikipedia, the free encyclopedia
    Insipid and not beautiful eh?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Thanks
    1
    Quote Originally Posted by galactus View Post
    $\displaystyle \sum_{n=0}^{\infty}n^{3}x^{n}$

    Let's work with the third derivative:

    (...)

    $\displaystyle =\frac{x^{3}+4x^{2}+x}{(x-1)^{4}}+\frac{a}{1-x}$
    Instead of differentiating 3 times, integrate once and then divide by x, integrate, then divide by x again. Do this until you get $\displaystyle \sum_{n=0}^{\infty}x^{n}$. It will be easier to deal with. I showed this approach in my second post.

    Quote Originally Posted by Mathstud28
    Insipid and not beautiful eh?
    It's not as creative as the other solution, but it works for any $\displaystyle \sum_{n=0}^{\infty}n^k x^n$
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by wingless View Post
    Instead of differentiating 3 times, integrate once and then divide by x, integrate, then divide by x again. Do this until you get $\displaystyle \sum_{n=0}^{\infty}x^{n}$. It will be easier to deal with. I showed this approach in my second post.



    It's not as creative as the other solution, but it works for any $\displaystyle \sum_{n=0}^{\infty}n^k x^n$
    I think yours was much more work though, creatively disguised under

    Now integrate that sum two times, then apply the geometric progression formula, then differentiate two times.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Thanks
    1
    Quote Originally Posted by Mathstud28 View Post
    I think yours was much more work though, creatively disguised under
    Now integrate that sum two times, then apply the geometric progression formula, then differentiate two times.
    What? Even reading this sentence is harder than doing the calculation ;p

    It's just,
    $\displaystyle \sum_{n=0}^{\infty}n(n-1)x^{n-2}~\underset{\text{int}}{\to}~\sum_{n=0}^{\infty}n (n-1)x^{n}\frac{1}{n(n-1)}~\to~\sum_{n=0}^{\infty}x^{n}~\to~\frac{1}{1-x}~\underset{\text{diff}}{\to}~\frac{2}{(1-x)^3}$
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by wingless View Post
    What? Even reading this sentence is harder than doing the calculation ;p

    It's just,
    $\displaystyle \sum_{n=0}^{\infty}n(n-1)x^{n-2}~\underset{\text{int}}{\to}~\sum_{n=0}^{\infty}n (n-1)x^{n}\frac{1}{n(n-1)}~\to{\color{red}{\cdots}}~\sum_{n=0}^{\infty}x^ {n}~\to~\frac{1}{1-x}~\underset{\text{diff}}{\to}{\color{red}{\cdots} }~\frac{2}{(1-x)^3}$
    ...

    But anways, even if your solution was more beautiful (which it is ), don't say another person's solution isn't as beautiful..even though I know you had no intention of it coming off that way...all math is equally beautiful!....except statistics
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Thanks
    1
    I said "not beautiful" for my own solution. Come on, do you really think I'm that rude?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by wingless View Post
    I said "not beautiful" for my own solution. Come on, do you really think I'm that rude?
    No ...I'm just a wuss
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum