anonymous
  • anonymous
Who wants a medal? Can you teach me about series and sequences?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Taking the Calc BC test tomorrow? Anyway, I can probably say a few things. Firstly, Taylor Series: You can generate a taylor series centered at x = a for a function by f(x) + f'(x)(x-a) + f''(x)(x-a)^2/2! +... I think you get the idea. Usually these use f(0), f'(0), etc. but sometimes they will specify a different place to do it.
anonymous
  • anonymous
Now, on the note of convergence: The ratio test is your friend. Most convergence things can be done with by dividing the n+1th term by the nth term of the series. The series converges where the absolute value of that ratio < 1. Conditional convergence can occur where it = 1, you have to check for that other ways.
anonymous
  • anonymous
Are you talking about Arithmetic Progression, Geometric Progression, Harmonic Progression or the other series that Jamal is giving here? @gandalfwiz

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
1/(1-x) = 1 + x + x^2 + x^3 + ... + x^n where x < 1. You can substitute in (-x) for X to get 1/(1+x) = 1 - x + x^2 - ... + (-x)^n You can also integrate that series to get the ln(1+x) or -ln(1-x) depending on which one you integrate.
anonymous
  • anonymous
The series for sin(x) = X- X^3/3! + X^5/5! - ... + (-1)^nx^2n+1/(2n+1)! You can get the cosine series by taking the derivative of that or by using the taylor series formula I poster earlier. That's how you can derive the sin series if you can forget it.
anonymous
  • anonymous
The e^x series is just 1 + x + x^2/2! + ... +x^n/n!. Easy. Again, you can use the taylor series formula to get this.
anonymous
  • anonymous
Oh, I forgot to mention that the ratio test is the limit as n goes to infinity of the ratio I mentioned earlier. Stupid me.
anonymous
  • anonymous
Wait, before you go on Jamal, I'm not quite that advanced.
anonymous
  • anonymous
Sorry to let you go on for so long, but the site just let me load my question-- this is just for precalc.
anonymous
  • anonymous
Oh, lol, sorry. I just have the calc BC test tomorrow and I was studying this. Anything in particular I can help with? I'm not great at precalc, though.
anonymous
  • anonymous
Haha, I've got a lot of friends taking that tomorrow too. I was watching Khan academy and he filled in the questions I had. But thanks anyways! You still get a medal :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.