anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\int\limits_{0}^{0.78}\sin(6x^2)dx\]
anonymous
  • anonymous
@mathstudent55
IrishBoy123
  • IrishBoy123
writing 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 \(x \ f'(0) = ...\)

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anonymous
  • anonymous
Just plug it in directly?
anonymous
  • anonymous
What do you think @sourwing
IrishBoy123
  • IrishBoy123
"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
  • anonymous
So basically \[\sum_{n = 0}^{\infty} \frac{ -1^n }{ (2n+1)! } * (6x^2)^{2n+1}\], and solve for n = 0 and n=1?
IrishBoy123
  • IrishBoy123
sounds good
anonymous
  • anonymous
Well what do I do to estimate the value of the series?
anonymous
  • anonymous
I ask because I still have an x term in my result
IrishBoy123
  • IrishBoy123
if you have an 'x' term in your result, well that is a good thing. :p
anonymous
  • anonymous
Okay 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
  • IrishBoy123
now, 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
  • IrishBoy123
nothing to do with limits, either. just knock out the first 2 terms of the [bogus] expansion
anonymous
  • anonymous
So get the first two terms in the series, but how do I use that to estimate the value of the series?
IrishBoy123
  • IrishBoy123
|dw:1439852188969:dw|
IrishBoy123
  • IrishBoy123
|dw:1439852553035:dw|
anonymous
  • anonymous
\[6x^2-x^5\]??
IrishBoy123
  • IrishBoy123
|dw:1439852806038:dw|
anonymous
  • anonymous
-1/6 * 6x^5
anonymous
  • anonymous
-6x^5/6
anonymous
  • anonymous
-x^5
IrishBoy123
  • IrishBoy123
you 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
  • anonymous
brb I have to go feed goats
anonymous
  • anonymous
@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
  • anonymous
Maclaurin 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
  • anonymous
oh, your problem asks for the first 2 terms, not 3 as I suggested. That is easier.!!
anonymous
  • anonymous
I'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
  • anonymous
OH, 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
  • anonymous
\(sin x =\sum_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}\) ok?
anonymous
  • anonymous
Okay
anonymous
  • anonymous
so, just replace x by 6x^2 to get Maclaurin series of sin(6x^2)
anonymous
  • anonymous
ok?
anonymous
  • anonymous
Alright
anonymous
  • anonymous
now, you want to take int of sin (6x^2) from 0 to 0.78 , right? Just take integral both sides.
anonymous
  • anonymous
The left hand side is \(\int_0^{0.78} sin(6x^2)dx\)
anonymous
  • anonymous
The right hand side is integral of the sum (...) right?
anonymous
  • anonymous
Yes or NO? where don't you get?
anonymous
  • anonymous
I'm just writing this in my notes as we go, sorry
anonymous
  • anonymous
Yes
anonymous
  • anonymous
And we know that integral of sum = sum of integral, right?
anonymous
  • anonymous
Okay
anonymous
  • anonymous
hence the right hand side is \[\sum_{n=0}^{\infty} \int_0^{0.78}(-1)^n\dfrac{(6x^2)^{2n+1}}{(2n+1)!}\] ok?
anonymous
  • anonymous
Okay, I'm with you
anonymous
  • anonymous
Now, calculate the integral. Can you?
anonymous
  • anonymous
I can try.
anonymous
  • anonymous
But what do I do with the fact that this has both n and x in it?
anonymous
  • anonymous
Don't forget \((6x^2)^{2n+1}= 6^{2n+1}*x^{4n+2}\)
anonymous
  • anonymous
hey, just x is the variable, n is a number!! nothing to do with n, take it as constant
anonymous
  • anonymous
you take integral w.r.t.x (dx) right? how n works as variable?
anonymous
  • anonymous
We will use n for the far left summation, not for integral.
anonymous
  • anonymous
Okay
anonymous
  • anonymous
It turns to \[\sum_{n=0}^{\infty}(-1)^n\dfrac{6^{2n+1}}{(2n+1)!}\int_0^{0.78} x^{4n+2}dx\]
anonymous
  • anonymous
ok?? take integral, what do you get?
anonymous
  • anonymous
|dw:1439861021631:dw|
anonymous
  • anonymous
\[\frac{ x ^{4n+3} }{ 4n+3 }\]???
anonymous
  • anonymous
where are your limits?
anonymous
  • anonymous
Sorry, evaluated from 0 to 0.78
anonymous
  • anonymous
ok, =?
anonymous
  • anonymous
Just sub .78 and 0 for x and subtract?
anonymous
  • anonymous
0 for x =0, just plug x =0.8 in, ok?
anonymous
  • anonymous
oh, 0.78
anonymous
  • anonymous
So it's just \[\frac{ .78^{4n+3} }{ 4n+3 }\] ??
anonymous
  • anonymous
So, now you have \[\sum_{n=0}^{\infty}(-1)^n\dfrac{6^{2n+1}}{(2n+1)!} *\dfrac{0.78^{4n+3}}{4n+3}\]
anonymous
  • anonymous
ok ?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
You need just 2 terms!! easy!! n =0, plug in, n =1, plug in add them together, done.
anonymous
  • anonymous
That's it? Just add the resulting two terms together?
anonymous
  • anonymous
Yes, dat sit. If you want, you can add you and me to get 4 terms. :)
anonymous
  • anonymous
show me your result, please.
anonymous
  • anonymous
Working on it now
anonymous
  • anonymous
-0.745188?
anonymous
  • anonymous
I got 0.0457318875
anonymous
  • anonymous
ok, let's do together
anonymous
  • anonymous
I'm absolutely sure that you're right, I just did mine mostly in my head...
anonymous
  • anonymous
\[\sum_{n=0}^{\infty}(-1)^n\dfrac{6^{2n+1}}{(2n+1)!} *\dfrac{0.78^{4n+3}}{4n+3}\] \[(-1)^0\dfrac{6^{2*0+1}}{(2*0+1)!} *\dfrac{0.78^{4*0+3}}{4*0+3}+(-1)^1\dfrac{6^{2*1+1}}{(2*1+1)!} *\dfrac{0.78^{4*1+3}}{4*1+3}\]
anonymous
  • anonymous
\[-1^{0}*\frac{ 6^{2(0)+1} }{ (2(0)+1)! }*\frac{ 0.78^{4(0)+3} }{ 4(0)+3 }\]
anonymous
  • anonymous
Oh you're much faster at this than I am
anonymous
  • anonymous
hahaha... I just copy and paste the previous post, then change the numbers. hehehe
anonymous
  • anonymous
Well for n = 0, term by term, the first one would equal 1 * 6/1* .78^3/3 which I said totals 0.949104
anonymous
  • anonymous
and your first term = \(\dfrac{6}{1!}\dfrac{0.78^3}{3}\) right? = \(2*0.78^3=0.949104\) Yes,
anonymous
  • anonymous
The second term is ??
anonymous
  • anonymous
For n = 1 I said -1* 6^3/6 * .78^7/7 which I thought should be -0.90337211
anonymous
  • anonymous
\(-\dfrac{6^3}{3!}\dfrac{0.78^7}{7}=-0.9033721125\) Yes, again,
anonymous
  • anonymous
add them together , you get??
anonymous
  • anonymous
0.045732
anonymous
  • anonymous
yup
anonymous
  • anonymous
So I did just mess something up in my head the first time.
anonymous
  • anonymous
easy, right?
anonymous
  • anonymous
Thank you so much Crazyandbeautiful, you are literally the most beautiful person in the world to me right now
anonymous
  • anonymous
haahaha... I'm not beautiful, I am handsome. hahaha.... jk
anonymous
  • anonymous
lol 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
  • anonymous
Handsome, beautiful, the actual envisionment of God, whatever you are dude you saved my life here
anonymous
  • anonymous
You 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
  • anonymous
I had no idea BEFORE this chat string
anonymous
  • anonymous
Believe me, I was taking notes
anonymous
  • anonymous
ok, good luck
anonymous
  • anonymous
Thanks again

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