#How do you calculate the sum of these series
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Question 11 is directly asking you about the Leibniz Pi Approximation
Question 15 is asking about the taylor expansion of e^x
We can infer that x is 3, looking at the possible answers
However, n = 2, so we know we should subtract something from e^3: the values of n for 0 and 1
If n=0, 3^0 / 0! = 1
If n=1, 3^1/1! = 3
So it is e^3 - 1 - 3
i dont think we learked about leibniz pi apprxomiation
is it anything to do with integral remainder estimation or something like that
You're familiar with integration?
yes
this is cal 2 homework
sorry was asleep
we've also learned partial sums if it has anything to do with that? I must have missed this unit and cant find it
Have you learned taylor series?
try the taylor expansion of pi*x, and relate that to q11
Okay, so. Are you familiar with the series sum(n = 0, inf) x^n and what it evaluates to?
im covering this today so i think ill look into thos
no
ik how to find convergence, radius, etc
with series
remainder too
but didnt know how to calculate sum at all other than partial sums
yes
Okay, so if we have a finite sum of sequential powers of x, that is, sum(i = 0, n) x^i, how can the difference of powers formula help us evaluate that?
@peak totem (sorry if pinging you is annoying, you just don't respond if I don't so I don't know if you're paying attention otherwise.)
no its totally okay i appreciate it so much
Im not sure, but i really appreciate you pointing me in the right direction
ill watch some videos on it
I mean, I can explain it all.
I just need you to be engaged in telling me what you do and don't know or can or can't figure out so that I know what I need to explain and what you can do on your own.
I feel like I just learn better from YT is the thing so I just wanted to know like what its called to solve these problems
Okay, look. The short version is that (1 - x) * (sum (i = 0, n) x^i) = 1 - x^(n + 1), therefore sum(i = 0, n) x^i = (1 - x^(n + 1))/(1 - x). Take the limit as n -> inf and what do you get?
$\sum_{i = 0}^n x^i = \frac{1 - x^{n + 1}}{1 - x}$
Techie Literate
@peak totem This help?
Not always.
How did you determine that?
hold on gtg need to take my mom to her appointment
Ping me when you're back.
bet
Right, that's exactly correct. And the cool thing about this is that we can use this and a bit of calculus to find series representations of functions that normally don't have them. For instance, ln(1 + x).
Actually, in hindsight it's seemingly just an easier way to derive the Taylor series. A Taylor series is a way to approximate any infinitely-differentiable function with a series: $f(x) = \sum_{n = 0}^\infty \frac{f^{(n)}(0)x^n}{n!}$
Techie Literate
Where $f^{(n)}(0)$ refers to the nth derivative of f evaluated at 0, and $f^{(0)}(x) = f(x)$.
Techie Literate
But honestly the sequential derivatives of some of these functions get annoyingly long, so I still prefer the shortcut derivation of the series.