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anonymous
 one year ago
Assume that \sin(x) equals its Maclaurin series for all x.
Use the Maclaurin series for sin(6 x^2) to evaluate the integral
(integral [0, 0.78]) sin(6 x^2)dx
. Your answer will be an infinite series. Use the first two terms to estimate its value.
anonymous
 one year ago
Assume that \sin(x) equals its Maclaurin series for all x. Use the Maclaurin series for sin(6 x^2) to evaluate the integral (integral [0, 0.78]) sin(6 x^2)dx . Your answer will be an infinite series. Use the first two terms to estimate its value.

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{0.78}\sin(6x^2)dx\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4writing the Maclaurin sin expansion in terms of t: \(\large sin(t) = \Sigma_{0}^{\infty} \frac{(1)^n }{(2n+1)! }\ t^{2n+1}\) it seems they are happy for you to stuff \(t = 6 x^2\) in and go from there to use an actual Mac series for \(sin(6 x^2)\), you'd need to go \(f(0) = 0\) \(f'(x) = (sin(6 x^2))' = ...\) and \(f'(0) = ....\) so \(x \ f'(0) = ...\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just plug it in directly?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What do you think @sourwing

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4"Just plug it in directly?" yes, that is what the question mandates for #1, plug it in term by terms and come up with a nice summation with a \(\Sigma\) in it! for #2, generate a few terms and hope you get the first two quickly. [if you have a crack at the proper Mac expansion for that function, you will be busy for a while :p]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So basically \[\sum_{n = 0}^{\infty} \frac{ 1^n }{ (2n+1)! } * (6x^2)^{2n+1}\], and solve for n = 0 and n=1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well what do I do to estimate the value of the series?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I ask because I still have an x term in my result

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4if you have an 'x' term in your result, well that is a good thing. :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay lol. Sorry I'm clueless on the procedure here. Should I be evaluating the limit at both of the solutions? They are N_0 = 6x^2 and N_1 = 216x^2

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4now, forget about the \(\Sigma\) and just evaluate the first 2 terms you are not summing, you are just using the first 2 terms of the series make sense?!?! you need x terms because you have to integrate

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4nothing to do with limits, either. just knock out the first 2 terms of the [bogus] expansion

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So get the first two terms in the series, but how do I use that to estimate the value of the series?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4dw:1439852188969:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4dw:1439852553035:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4dw:1439852806038:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4you are *not* summing a series, or *valuing* a series you are hacking an integral; by approximating the integrand to a hack of its Mac expansion

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0brb I have to go feed goats

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Kainui @Hero I'm sorry, but I still don't understand how to "hack the integral"? I'm still not quite getting what IrishBoy is trying to have me do here. Could you or one of the other moderators help here?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Maclaurin series of \(sin(6x^2) \) is \(sin(6x^2) =\sum_{n=0}^{\infty}\dfrac{(1)^n (6x^2)^{2n+1}}{(2n+1)!}\) \[\int_0^{0.78} sin(6x^2)dx=\int_0^{0.78}\sum_{n=0}^{\infty}\dfrac{(1)^n(6x^2)^{2n+1}}{(2n+1)!}dx\\=\sum_{n=0}^{\infty}\int_0^{0.78}\dfrac{(1)^n(6x^2)^{2n+1}}{(2n+1)!} dx\] This integral is easy to take, right? After take out the integral, you just calculate the first 3 term of it (approximate theorem) and you get the answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, your problem asks for the first 2 terms, not 3 as I suggested. That is easier.!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry but nothing about this seems easy to me. I hate to keep bothering you but is there any way you can force me to understand this? @Crazyandbeautiful

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OH, I assumed you know the formula of Maclaurin series of sin (6x^2) If you don't know it, you can derive it from Maclaurin series of sinx, ok?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(sin x =\sum_{n=0}^{\infty} (1)^n \dfrac{x^{2n+1}}{(2n+1)!}\) ok?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, just replace x by 6x^2 to get Maclaurin series of sin(6x^2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now, you want to take int of sin (6x^2) from 0 to 0.78 , right? Just take integral both sides.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The left hand side is \(\int_0^{0.78} sin(6x^2)dx\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The right hand side is integral of the sum (...) right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes or NO? where don't you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm just writing this in my notes as we go, sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And we know that integral of sum = sum of integral, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hence the right hand side is \[\sum_{n=0}^{\infty} \int_0^{0.78}(1)^n\dfrac{(6x^2)^{2n+1}}{(2n+1)!}\] ok?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now, calculate the integral. Can you?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But what do I do with the fact that this has both n and x in it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Don't forget \((6x^2)^{2n+1}= 6^{2n+1}*x^{4n+2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hey, just x is the variable, n is a number!! nothing to do with n, take it as constant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you take integral w.r.t.x (dx) right? how n works as variable?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We will use n for the far left summation, not for integral.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It turns to \[\sum_{n=0}^{\infty}(1)^n\dfrac{6^{2n+1}}{(2n+1)!}\int_0^{0.78} x^{4n+2}dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok?? take integral, what do you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439861021631:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ x ^{4n+3} }{ 4n+3 }\]???

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where are your limits?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, evaluated from 0 to 0.78

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just sub .78 and 0 for x and subtract?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.00 for x =0, just plug x =0.8 in, ok?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So it's just \[\frac{ .78^{4n+3} }{ 4n+3 }\] ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, now you have \[\sum_{n=0}^{\infty}(1)^n\dfrac{6^{2n+1}}{(2n+1)!} *\dfrac{0.78^{4n+3}}{4n+3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You need just 2 terms!! easy!! n =0, plug in, n =1, plug in add them together, done.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's it? Just add the resulting two terms together?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, dat sit. If you want, you can add you and me to get 4 terms. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0show me your result, please.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, let's do together

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm absolutely sure that you're right, I just did mine mostly in my head...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=0}^{\infty}(1)^n\dfrac{6^{2n+1}}{(2n+1)!} *\dfrac{0.78^{4n+3}}{4n+3}\] \[(1)^0\dfrac{6^{2*0+1}}{(2*0+1)!} *\dfrac{0.78^{4*0+3}}{4*0+3}+(1)^1\dfrac{6^{2*1+1}}{(2*1+1)!} *\dfrac{0.78^{4*1+3}}{4*1+3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[1^{0}*\frac{ 6^{2(0)+1} }{ (2(0)+1)! }*\frac{ 0.78^{4(0)+3} }{ 4(0)+3 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh you're much faster at this than I am

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hahaha... I just copy and paste the previous post, then change the numbers. hehehe

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well for n = 0, term by term, the first one would equal 1 * 6/1* .78^3/3 which I said totals 0.949104

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and your first term = \(\dfrac{6}{1!}\dfrac{0.78^3}{3}\) right? = \(2*0.78^3=0.949104\) Yes,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The second term is ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For n = 1 I said 1* 6^3/6 * .78^7/7 which I thought should be 0.90337211

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\dfrac{6^3}{3!}\dfrac{0.78^7}{7}=0.9033721125\) Yes, again,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0add them together , you get??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I did just mess something up in my head the first time.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much Crazyandbeautiful, you are literally the most beautiful person in the world to me right now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haahaha... I'm not beautiful, I am handsome. hahaha.... jk

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol you're probably going to be the deciding factor in how well I do in my test next week actually, because I clearly had no idea what I was doing.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Handsome, beautiful, the actual envisionment of God, whatever you are dude you saved my life here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You got the method, right? don't scare me. I spent more than 1h to explain and you said you don't know what you are doing. it is a real scaring.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I had no idea BEFORE this chat string

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Believe me, I was taking notes
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