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Jamal_Negrar
 one year ago
Best ResponseYou've already chosen the best response.1Taking 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)(xa) + f''(x)(xa)^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.

Jamal_Negrar
 one year ago
Best ResponseYou've already chosen the best response.1Now, 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.

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.0Are you talking about Arithmetic Progression, Geometric Progression, Harmonic Progression or the other series that Jamal is giving here? @gandalfwiz

Jamal_Negrar
 one year ago
Best ResponseYou've already chosen the best response.11/(1x) = 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(1x) depending on which one you integrate.

Jamal_Negrar
 one year ago
Best ResponseYou've already chosen the best response.1The 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.

Jamal_Negrar
 one year ago
Best ResponseYou've already chosen the best response.1The 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.

Jamal_Negrar
 one year ago
Best ResponseYou've already chosen the best response.1Oh, I forgot to mention that the ratio test is the limit as n goes to infinity of the ratio I mentioned earlier. Stupid me.

gandalfwiz
 one year ago
Best ResponseYou've already chosen the best response.0Wait, before you go on Jamal, I'm not quite that advanced.

gandalfwiz
 one year ago
Best ResponseYou've already chosen the best response.0Sorry to let you go on for so long, but the site just let me load my question this is just for precalc.

Jamal_Negrar
 one year ago
Best ResponseYou've already chosen the best response.1Oh, 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.

gandalfwiz
 one year ago
Best ResponseYou've already chosen the best response.0Haha, 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 :)
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