Question

How did I do?
Use a power series to approximate the definite integral to six decimal places.
\[\int_{0}^{0.2}\frac{1}{1+x^5}dx\]
\[\int_{0}^{0.2}\frac{1}{1-(-x^5}dx\]
\[\int_{0}^{0.2}\sum_{n=0}^{\infty}(-x^5)^n dx\]
\[\int_{0}^{0.2}\sum_{n=0}^{\infty}(-1)^nx^{5n} dx\]
\[\sum_{n=0}^{\infty}\left[(-1)^n\frac{x^{5n+1}}{5n+1} \right]_0^{0.2}\]

Answer 1

\[ \Large \frac{1}{1-x} = \sum_{n=0}^\infty x^n\] You used this right?

Answer 2

Seems pretty correct to me.

Answer 3

Yes

Answer 4

I have a dumb question, so do I sub those limits in for n or x?

Answer 5

x of course because the limit is from x=0 to x=0.2... so what would I do with the n?

Answer 6

yes, but there is your problem, what you did here is correct. But usually you write the terms out, understand what I mean? So if you want to make a rough approximation you have to carry out the steps.

Answer 7

I mean the integral is from 0 to 0.2

Answer 8

Let me try to write it down, I better post a cheat sheet before I make some careless mistakes.

Answer 9

so when n=0 and n=1 and n=2 and so on and so forth?

Answer 10

http://mathworld.wolfram.com/MaclaurinSeries.html

Answer 11

I did post that for myself too by the way (-:

Answer 12

but it's power series isn't it?

Answer 13

Sooo

\[ \Large \frac{1}{1-x}=1+x+x^2+x^3+x^4+x^5 + \cdots \] The series is correct you have choosen, but you have found the most general answer, and as far as I know, I can't work with that to find a approximate solution.

My idea:

Make the same substitution for the Series above (not the general form) and then evaluate each term individually by taking the integral.

Answer 14

I feel soo unintelligent right now...why 1+x+x^2 etc.?

Answer 15

well that is the series. http://mathworld.wolfram.com/MaclaurinSeries.html I hope I didn't choose the wrong one.

Answer 16

\[ \Large \frac{1}{1-x}=1+x+x^2+x^3+ \cdots = \sum_{n=0}^\infty x^n \]

Answer 17

See, one is exact, right hand side, one is approximate, left hand side (middle)

Answer 18

So if you want to approximate, you don't want to use the 'exact' general case. I hope that makes some sense.

Answer 19

oh it makes sense now

Answer 20

put the values of x
alternating series can be approximated this way
http://en.wikipedia.org/wiki/Alternating_series#Approximating_sums

Answer 21

nevertheless @MathSofiya , just so you know, what you did above is not for nothing, you can still see how your new series looks like by running it through the way you have defined it, in other words substitute a few of the n terms like n=0, n=1 ... and then you can integrate that term by term

Answer 22

how far should I go? till n=5 maybe?

Answer 23

Erm usually I do this via try and error, maybe there is a more elegant way *grins* let me see real quick before I say something that I will later have to pull back heh.

Answer 24

I would say just take a few of them, and then when you evaluate your integral at zero, all terms will go to zero, so evaluate each of it at higher bound and then see how many decimals each successive integral will give.

Answer 25

use this
... the tn must be of 6th order ... all (converging) alternating series can be approximated this way.

Answer 26

ok I'll try it and I'll post what i get

Answer 27

Man this is difficult :\
\[(-1)^n\frac{(0.2)^{5n+1}}{5n+1}-(-1)^n\frac{(0)^{5n+1}}{5n+1}\]
and now when I do the series I use
\[\left[(-1)^3\frac{(0.2)^{5*3+1}}{5*3+1}+(-1)^2\frac{(0.2)^{5*2+1}}{5*2+1}+(-1)^1\frac{(0.2)^{5*1+1}}{5*1+1}+(-1)^0\frac{(0.2)^{5*0+1}}{5*0+1}\right]\]-
\[\left[(-1)^3\frac{(0)^{5*3+1}}{5*3+1}+(-1)^2\frac{(0)^{5*2+1}}{5*2+1}+(-1)^1\frac{(0)^{5*1+1}}{5*1+1}+(-1)^0\frac{(0)^{5*0+1}}{5*0+1}\right]\]

Answer 28

something like that?

Answer 29

well, it looks correct to me at first glance @MathSofiya , although you could have saved the second line (the subtraction) because it's all evaluated at x=0, therefore all the terms become zero.

Answer 30

I don't quite understand what you mean by it's evaluated at x=0

Answer 31

By the way, this is the exact result http://www.wolframalpha.com/input/?i=integrate+1%2F%281%2Bx%5E5%29 , you can click on "Show steps", it's not less painful.

Answer 32

well, taylor/maclaurin series are polynomial approximations, polynomials without a constant c evaluated at zero are zero.

Answer 33

am i totally wrong, or for the first one would you not simply use \(1-x^5\) ?

Answer 34

where @satellite73 ?

Answer 35

for the first one

Answer 36

Like, lets say you have integrated some ominous function that I don't want to define any more carefully now, for a definite integral and you have obtained the following result:

\[ \Large F(x)= \left. x^7+3x^3+2x+\right|_0^3 \] If you plug in zero all the terms with an x will vanish.

Answer 37

you mean the second line...all the way at the top of the page?

Answer 38

oh I see @spacelimbus. So since we don't have any constants, whatever is to the right of the minus sign will disappear.

Answer 39

yeah i came here late
but \(\frac{1}{1+x^5}=1-x^5+x^{10}-...\) expanded at 0, and you are only going out to 6 decimal places, which means you can ignore the \((.2)^{10}\) part

Answer 40

yes, the minus will disappear.

Answer 41

how did you come to the conclusion that (.2)^10 is the "limit"

Answer 42

or i mean that's where i would stop

Answer 43

so i think (correct me if i am wrong) you just need to evaluate \(x-\frac{x^6}{6}\) at \(x=.2\)

Answer 44

shoot me. Really?

Answer 45

you are only looking for 6 decimal place accuracy right?

Answer 46

yes

Answer 47

\((.2)^{11}\) is some little tiny number beyond the sixth decimal place

Answer 48

oh. I see...
so back to the \[x-\frac{x^6}{6}\]
how and why?

Answer 49

i didn't mean to butt in so late and have not really read all the above posts, but the first one looked pretty straightforward

Answer 50

\(\frac{1}{1+x^5}\) expands as \(1-x^5+x^{10}-x^{15}+...\)

Answer 51

No problem @satellite73 , I believe the derivation we have done is correct, for the approximation of the given integral, I just don't know how to "bind" the problem to given decimals.

Answer 52

that is, take well known expansion for \(\frac{1}{1-x}\) and replace \(x\) by \(-x^5\)

Answer 53

now integrate term by term

Answer 54

for whatever reason \[1-x^5+x^{10}-x^{15}+...\] is still a magical occurrence

Answer 55

you get
\[x-\frac{x^6}{6}+\frac{x^{11}}{11}-...\]

Answer 56

I can do the integral, but the first step is still a mystery

Answer 57

I believe it's exactly what we derived isn't it? Sorry if I increase the confusion,

Answer 58

and since you only are looking for six decimal place accuracy, you can ignore everything beyond \(\frac{x^6}{6}\)

Answer 59

the gimmick is always to start with
\[\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k\]

Answer 60

make whatever you have look like that if you can

Answer 61

I did that and came up with
\[\sum_{n=0}^{\infty}(-1)^nx^{5n} dx\]

Answer 62

without the dx of course

Answer 63

oooohhhh. I think i see it now

Answer 64

yes that is right

Answer 65

oh you wrote them out before you integrated...

Answer 66

\[1-x^5+x^{10}-x^{15}-...\]

Answer 67

then integrate term by term

Answer 68

and evaluate at x=0.2

Answer 69

yes

Answer 70

and notice that \(\frac{.2^{11}}{11}\) is way beyond the sixth decimal place

Answer 71

but the way that @Spacelimbus and i did is correct too?

Answer 72

i believe so, yes

Answer 73

in fact, what @satellite73 wrote here, I have mentioned above too, I believe we just have had some misunderstanding.

*repost*
yes, but there is your problem, what you did here is correct. But usually you write the terms out, understand what I mean? So if you want to make a rough approximation you have to carry out the steps.

Answer 74

Mystery solved guys! Thank you so much Everyone!

Answer 75

I wasn't strict enough I believe which caused some confusion (-: So thanks for cleaning things up @satellite73

Answer 76

<---can kitten have medal too?

Answer 77

sure, but in the next post because i used mine up

Answer 78

lol that's ok :P

Answer 79

I will have to safe mine too it's already gone here (=

Answer 80

no prob. Thanks for helping