more calculus help

[Zain]

just starting this time



im not so fresh on this taylor stuff and representing functions as power series

[Zain]

bluntbill you beautiful genius where are you

[theredspecial]

help him or I'll kill you all

[Fluffy]

kill him or I'll help you all

[dp4perchas3r]

lold

[Follower]

1. without centered at π/3, the taylor series is
Σ k:[0,∞] {[(-1)^k] * [x^(2k+1)]} / {(2k+1)!}
That is, an alternator raised to the k power multiplied by x raised to the 2k+1 power
this entire quantity divided by (2k+1) factorial, which is also equal to (2k+1) * (2k)! because a factorial follows the sequence of -1 each next term.

When it is centered at π/3, you simply subtract that by x, so:
Σ k:[0,∞] {[(-1)^k] * [(x - π/3)^(2k+1)]} / {(2k+1)!}

Then write out the first four terms, so plug in 0 to 3 and add them all up. The alternator will change the sign.

2. Maclaurin is the same as Taylor, except the center, c = 0.

There is a known Maclaurin expansion for 1/(1+x)
so in this case, we'll use that to interpret x/(1+x)
1/(1+x) = Σ k:[0,∞] {[(-1)^k] * (x^k)}
x/(1+x) = x * Σ k:[0,∞] {[(-1)^k] * (x^k)}
x/(1+x) = Σ k:[0,∞] {[(-1)^k] * [x^(k + 1)]}

Notice we kept the x outside of the series first, then multiplied it into the series because of order of operations. This added one to the power counter of the x term and changed it to k+1

Now plug in 0 to 4.

3. Finding the interval of convergence relies on first taking the ratio test as the limit approaches infinity of the An+1 term divided by the An term.

lim n->infinity |An+1 / An|
set it up so you take the recipicol of An, then multiply that by the next term in the sequence (so add 1 to anywhere n is)

as you take the limit, there are various things you might get. you might be left with an x term, this will produce a valid interval. However, if you have n terms under the x term such as lim n->infinity |x/(n+1)] after you've simplified the original multiplication, than this will produce a limit of 0. If you get 0 for your limit, the radius is infinity, and the interval of convergence is what you centered the function at, in this case it was a Maclaurin series it would be {0}. **(Not the answer)

If you get infinity for your limit, produced by something like lim n->infinity x * (n+1), then your radius is infinity and your interval is (-infinity, infinity).

Now, going back to if you get just an x term, you do -1 < whatever x term here < 1

simplify the inequality so x is just in the middle, or if it was just x, then your interval would be (-1, 1)

you have to test convergence of these two points though, to see if they lie in the functions graph. Plug in each point, -1, and 1, into the series representation for the X term, not the K terms. Test convergence, and change the (-1, 1) to [-1, 1] or (-1, 1] vice versa.
Σ k:[0,∞] {[(-1)^k] * [x^(k + 1)]}

6.
f(x) = e^x the series representation is as follows
Σ k:[0,∞] (x^k) / (k)!

centered at 1
Σ k:[0,∞] [(x-1)^k] / (k)!

7. Binomial series, you modify the original term you're given 1/{[(1+x^2)]^3/2}
to
{[(1+x^2)]^-(3/2)}

So the binomal is -3/2, your x term is squared, but it is in the form of 1+x, so this is now solvable.

Binomial series = 1 + p + p(p-1) + p(p-1)(p-2) + ... = (?) you have to find out the pattern and plug into the series representation.

The series representation takes on the form of a geometric series
Σ k:[0,∞] (?) x^k, in this case x was squared so
Σ k:[0,∞] (?) x^2k

8. Approximating an integral is easy

First you have a known series for cos(x), its
Σ k:[0,∞] {[(-1)^k] * [x^(2k)]}/[(2k)!]

plugging in 1/x^2 into that series will produce
Σ k:[0,∞] {[(-1)^k] * [x^(2k - 2)]}/[(2k)!]

when you integrate
∫ Σ k:[0,∞] {[(-1)^k] * [x^(2k - 2)]}/[(2k)!]

this produces

C (constant of integration)+ all things raised to k get pulled out as constants, the x term power needs to be added by one / divided by the x term power added by 1, thus follows:

∫ Σ k:[0,∞] {[(-1)^k] * [x^(2k - 2)]}/[(2k)!] = C + Σ k:[0,∞] {[(-1)^k] * [x^(2k - 1)]}/[(2k - 1)(2k)!]

write out a few terms of the series, then plug in the limits of integration and that is the approximation.

[TheJokerOCG]

jesus christ...no wonder i failed algebra in college...when the fuck did we start using '!' in math?

[Raven]

jesus christ...no wonder i failed algebra in college...when the fuck did we start using '!' in math?

If you fail algebra, you must be a fucking idiot

[2fast]

jesus christ...no wonder i failed algebra in college...when the fuck did we start using '!' in math?

me too bro fuck math

[TheJokerOCG]

in certain things, yes i am...math is one of them...and science...and foreign languages

[Raven]

Welp, im not good at math either, but I made my way through high school math and just finished pre cal. Just pay attention and read the books

[TheJokerOCG]

the highest math i did was algebra 2 (as a senior in high school), then algebra in college, which i was failing so i dropped, i havent taken a math class or done anything math intensive for almost 6 years

[Zain]

thank you follower i worked most of them out but i needed this before 12:30pm today

thanks alot though this really helped with the stuff i was confused about

[ChewY]

I crammed for all my calculus tests and barely passed on extra credit w/ a low C. Srry mane.

[ReDNeCk GuRL]

I don't know who Follower is, but I'm pretty fucking turned on right now.

[BLUNTBILL]

my bad zain, next time

i didnt read through it but i think follower has you

[ganj]

I don't know who Follower is, but I'm pretty fucking turned on right now.

[qbert]

follower is mx lol

[ganj]

also wasnt there some site for this ?

[ReDNeCk GuRL]

follower is mx lol

Oh perfect, now I'm turned on and confused.

[qbert]

aren't you always?

[Follower]

actually zain I think I found a mistake

lim n->infinity |x/(n+1)] after you've simplified the original multiplication, than this will produce a limit of 0. If you get 0 for your limit, the radius is infinity

I forget if the radius is infinity or 0 if the limit is 0. Ask your teacher, I haven't taken calculus since 2007

[rmk]

1. without centered at π/3, the taylor series is
Σ k:[0,∞] {[(-1)^k] * [x^(2k+1)]} / {(2k+1)!}
That is, an alternator raised to the k power multiplied by x raised to the 2k+1 power
this entire quantity divided by (2k+1) factorial, which is also equal to (2k+1) * (2k)! because a factorial follows the sequence of -1 each next term.

When it is centered at π/3, you simply subtract that by x, so:
Σ k:[0,∞] {[(-1)^k] * [(x - π/3)^(2k+1)]} / {(2k+1)!}

Then write out the first four terms, so plug in 0 to 3 and add them all up. The alternator will change the sign.

2. Maclaurin is the same as Taylor, except the center, c = 0.

There is a known Maclaurin expansion for 1/(1+x)
so in this case, we'll use that to interpret x/(1+x)
1/(1+x) = Σ k:[0,∞] {[(-1)^k] * (x^k)}
x/(1+x) = x * Σ k:[0,∞] {[(-1)^k] * (x^k)}
x/(1+x) = Σ k:[0,∞] {[(-1)^k] * [x^(k + 1)]}

Notice we kept the x outside of the series first, then multiplied it into the series because of order of operations. This added one to the power counter of the x term and changed it to k+1

Now plug in 0 to 4.

3. Finding the interval of convergence relies on first taking the ratio test as the limit approaches infinity of the An+1 term divided by the An term.

lim n->infinity |An+1 / An|
set it up so you take the recipicol of An, then multiply that by the next term in the sequence (so add 1 to anywhere n is)

as you take the limit, there are various things you might get. you might be left with an x term, this will produce a valid interval. However, if you have n terms under the x term such as lim n->infinity |x/(n+1)] after you've simplified the original multiplication, than this will produce a limit of 0. If you get 0 for your limit, the radius is infinity, and the interval of convergence is what you centered the function at, in this case it was a Maclaurin series it would be {0}. **(Not the answer)

If you get infinity for your limit, produced by something like lim n->infinity x * (n+1), then your radius is infinity and your interval is (-infinity, infinity).

Now, going back to if you get just an x term, you do -1 < whatever x term here < 1

simplify the inequality so x is just in the middle, or if it was just x, then your interval would be (-1, 1)

you have to test convergence of these two points though, to see if they lie in the functions graph. Plug in each point, -1, and 1, into the series representation for the X term, not the K terms. Test convergence, and change the (-1, 1) to [-1, 1] or (-1, 1] vice versa.
Σ k:[0,∞] {[(-1)^k] * [x^(k + 1)]}

6.
f(x) = e^x the series representation is as follows
Σ k:[0,∞] (x^k) / (k)!

centered at 1
Σ k:[0,∞] [(x-1)^k] / (k)!

7. Binomial series, you modify the original term you're given 1/{[(1+x^2)]^3/2}
to
{[(1+x^2)]^-(3/2)}

So the binomal is -3/2, your x term is squared, but it is in the form of 1+x, so this is now solvable.

Binomial series = 1 + p + p(p-1) + p(p-1)(p-2) + ... = (?) you have to find out the pattern and plug into the series representation.

The series representation takes on the form of a geometric series
Σ k:[0,∞] (?) x^k, in this case x was squared so
Σ k:[0,∞] (?) x^2k

8. Approximating an integral is easy

First you have a known series for cos(x), its
Σ k:[0,∞] {[(-1)^k] * [x^(2k)]}/[(2k)!]

plugging in 1/x^2 into that series will produce
Σ k:[0,∞] {[(-1)^k] * [x^(2k - 2)]}/[(2k)!]

when you integrate
∫ Σ k:[0,∞] {[(-1)^k] * [x^(2k - 2)]}/[(2k)!]

this produces

C (constant of integration)+ all things raised to k get pulled out as constants, the x term power needs to be added by one / divided by the x term power added by 1, thus follows:

∫ Σ k:[0,∞] {[(-1)^k] * [x^(2k - 2)]}/[(2k)!] = C + Σ k:[0,∞] {[(-1)^k] * [x^(2k - 1)]}/[(2k - 1)(2k)!]

write out a few terms of the series, then plug in the limits of integration and that is the approximation.

[Fluffy]

well duh

[RPC]

fuck man im glad i knew that shit 5 years ago and have never needed it since.

Zain gl sir.

[morreo]

jesus christ...no wonder i failed algebra in college...when the fuck did we start using '!' in math?

5! = 5 * 4 * 3 * 2 * 1

3! = 3 * 2 * 1

[BLUNTBILL]

jesus christ...no wonder i failed algebra in college...when the fuck did we start using '!' in math?

thats factorial.... so like 7th grade