Ooo. I see. But we haven't figured out the diffrentiated thing yet, so we shouldn't go back yet? If we go back and try and solve, we'd be nowhere, no?
Or have we figured it out and I completely missed it? Or do we integrate the simple thing and that'll give an answer? But what's the gurantee...
No, no, I really appreciate your persistance. But could you just explain it a little bit to me? You have to understand I know very little about series-- only that I am suppose to add numbers. The first example I did, my friend did it with limits, so I thought that ALL could be done with limits...
Oh alright that makes sense then. :rolleyes:
\sum_{k=1}^\infty \frac{x^{k-1}}{k+1}\allowbreak \allowbreak =\allowbreak \frac 1{x-1}\frac{1-x}x-\frac 1{x^2}\ln \left( 1-x\right)
Now if I plug in one, I get a lot of illegal values, i.e. 0 in denominator and ln(0)...so I diffrentiate...
Perhaps I should explain first that our professor did not cover a lot of this stuff, I am an ECON major, and have been fiddling through a book which does not cover series. (Everything I know is through research online. And none of it makes sense.)
What does this series have to do with anything:
\sum_{k=1}^\infty \frac{x^{k-1}}{k+1}\allowbreak \allowbreak =\allowbreak \frac 1{x-1}\frac{1-x}x-\frac 1{x^2}\ln \left( 1-x\right)
at least this one has the correct left hand side:
\sum_{k=1}^\infty \frac{x^k}{k(k+1)}=\allowbreak \frac...
I was thinking one thing:
The power series is as such:
e^x = \frac{x^n}{n!}
Could I use this? All that's different about what I am doing is that there is also (n+1)! in the denominator.
So would I just do:
e^x * \sum_{n=0}^\infty \frac{1}{(n+1)!}
So if I take sum from negative infinity to positive infinity, all the places where x is raised to an odd power will get canceled out. So only those which are left would be the one with the even power...which I take the sum of. However, how do I get the sum of the even power? i.e. how do I start...
Hello!
Here is the problem I am attempting:
Sum to infinity:
\frac{x}{1*2} + \frac{x^2}{2*3} + \frac{x^3}{3*4} + ...
Here is what I get:
S = \frac{x^n}{(n)(n+1)}
\lim_{n \to \infty} \frac{x^n}{(n)(n+1)}}
Now what? Should I do partial fractions to split the equation?
I thought that Chebychev's inequality is what would be used to solve this question, but the exceeding instead of the or more throws me off. Here is the question from Jim Pitman's book (I am studying for my final...)
Suppose the IQ scores of million individuals have a mean of 100 and a SD of...
Oh sorry, it is in fact,
F(x) = \int_{0}^{x}f(s)ds
xanthym, I have no idea what you did. Though I do understand that
\ \ \ \ g\left(x\right) \ \, = \, \ x^{\displaystyle \alpha} F\left(x\right) .
That was cool. I didn't realize that. How did you figure that out so fast? Was it obvious...