Question

Maclaurin series, see attachment

Answer 1

I was thinking to expand e^(-(1/2)((1-100)/16)^2

Answer 2

or integrate e with alpha being 100 and beta as 110.

Answer 3

Any ideas?

Answer 4

well I do think alpha would be 100 and beta would be 110


and I think we need to write a macluarin series for the integrand

Answer 5

and then integrate of course on the interval

Answer 6

Okay, thanks :)

Answer 7

=]

Answer 8

that's it I guess? your integrand is ugly and hopefully you don't have to find too many terms for your macularin series thing to get what they call a good approximation

Answer 9

The integrand does indeed look nasty, the process of how to do it is familiar but it takes awhile ;-;
Thanks @freckles

Answer 10

it won't be too hard if you can use:
\[e^u= \sum_{n=0}^\infty \frac{u^n}{n!} \\ e^{\frac{-1}{2}(\frac{x-100}{16})^2}=\sum_{n=0}^\infty \frac{(-\frac{1}{2}(\frac{x-100}{16})^2)^n}{n!} \\ \text{ first few terms of that looks like: } \\ 1+(-\frac{1}{2}(\frac{x-100}{16})^2)+\frac{1}{2!}(\frac{-1}{2}(\frac{x-100}{16})^2)^2+\frac{1}{3!}(\frac{-1}{2}(\frac{x-100}{16})^2)^3 \\ +\frac{1}{4!}(\frac{-1}{2}(\frac{x-100}{16})^2)^4\]

so that is what I would integrate (well and multiply by that scalar outside the integral afterwards)


compare these:
http://www.wolframalpha.com/input/?i=1%2F%2816%29*1%2F%28sqrt%282*pi%29%29integrate%28exp%28-1%2F2%28%28x-100%29%2F16%29%5E2%29%2Cx%3D100..110%29


http://www.wolframalpha.com/input/?i=1%2F%2816%29*1%2F%28sqrt%282*pi%29%29integrate%281-1%2F2%28%28x-100%29%2F16%29%5E2%2B1%2F8*%28%28x-100%29%2F16%29%5E4-1%2F%2848%29*%28%28x-100%29%2F16%29%5E6%2B1%2F%2824%29*%28-1%2F2%29%5E4%28%28x-100%29%2F16%29%5E8%2Cx%3D100..110%29



these mac way isn't too bad of an approximation with 5 terms that is

Answer 11

correction
it won't be hard if you can use that
but it will just be a lot of work

Answer 12

those terms though can be integrated pretty easily (By algebraic substitution )
it is just work no one wants to do :p (or at least me anyways )

Answer 13

omg im so dumb
I didn't see them say use the first 3 terms

Answer 14

that makes this a little less work :p
and we don't have to see crazy big numbers

Answer 15

http://www.wolframalpha.com/input/?i=1%2F%2816%29*1%2F%28sqrt%282*pi%29%29integrate%281-1%2F2%28%28x-100%29%2F16%29%5E2%2B1%2F8*%28%28x-100%29%2F16%29%5E4%2Cx%3D100..110%29

still not too bad of an approximation

Answer 16

Thanks a lot! I was in the middle of my third term and the integration was kinda tough for me. I used wolfram for help. Oh, and yes, it says i only need three terms.
i didn't realize that you'll do everything haha.
You're very helpful, awesome work, thanks a lot <33

Answer 17

well I didn't actually do the work for myself
because it just seems too repetitive and a little un-interesting (no offense to you of course; I just don't find this type of math fun)

Answer 18

I dont find it fun either xD

Answer 19

great to know :)

Answer 20

did you find three terms doing the derivative over and over ?

Answer 21

or did you use the e^u power series expansion I wrote above

Answer 22

I did the derivative

Answer 23

that's what i normally use, i'm not familiar with the approach you used

Answer 24

I first, use the derivative and analyze the series from there

Answer 25

Though, my first three terms were exactly like yours

Answer 26

i just used that
\[e^u= \sum_{n=0}^\infty \frac{u^n}{n!} \]
and replaced u with the -1/2((x-100)/16)^2 thing

\[e^{\frac{-1}{2}(\frac{x-100}

{16})^2}=\sum_{n=0}^\infty \frac{(-\frac{1}{2}(\frac{x-100}{16})^2)^n}{n!} \]

Answer 27

.