Question

Approximating sum of series

Answer 1

have done binomial expansion before?

Answer 2

taylor?

Answer 3

taylor series yes

Answer 4

how isit related?

Answer 5

http://www.wolframalpha.com/input/?i=expand+sqrt%281%2Bx%5E4%29+at+x%3D0

Answer 6

calculate the taylor series ... integrate term by term of the taylor series.

Answer 7

hm.. which formula is that?

Answer 8

that is taylor series of the sqrt(1+x^4) expanded at zero.

Answer 9

I am guessing that this integral is elliptik

Answer 10

The question is, can you find how many terms are necessary for a given number of decimal places? How can you predict how fast it will converge.

Answer 11

the upper limit is 1, the lower limit is 0 ... so just use first few terms. looks like this is alternating. there is one way of approximating for alternating series but forgot it. most likely it's something like squeezing.

Answer 12

so means i have to integrate it then?

Answer 13

You are integrating an approximation of the Taylor series expansion for the function. So you are integrating a few terms of the series.

Answer 14

\[\sum_{n=0}^{\infty}\left(\begin{matrix}k \\ n\end{matrix}\right)x^n=1+kx+\frac{ k(k-1)x^2 }{ 2! }+\frac{ k(k-1)(k-2)x^3 }{ 3! }+...\] can i use this formula?

Answer 15

i dont really get it,

Answer 16

i actually got the series 1 + x^4/2 - x^8/8 +... but i dont know to proceed after that

Answer 17

integrate term by term

Answer 18

but up to how much term do you need to evaluate?
http://tutorial.math.lamar.edu/Classes/CalcII/EstimatingSeries.aspx

Answer 19

okay, so it will be x + x^5/10 - x^9/72+x^13/208+...

Answer 20

it should be correct to 2 decimal places..

Answer 21

hmm ... you get alternating series.

Answer 22

yeah i think i shd make it into a geometric series form?

Answer 23

do you know how to estimate the value of alternating series?

Answer 24

no ... this is not geometric. if you want general term ... you should use generalized binomial theorem ... but last time you rejected the idea when I suggested.

Answer 25

well, i have no idea how to do it with an integral haha can u teach me? :3

Answer 26

Damn stupid wolf ... do you have Mathematica?

Answer 27

no, what is that? haha, anw if u dont want to we could just proceed to where i am now

Answer 28

http://www.wolframalpha.com/input/?i=Series%5BSqrt%5B1%2Bx%5E4%5D%2C+%7Bx%2C+0%2C+20%7D%5D

all right ... first find Taylor expansion of this.

Answer 29

then integrate it term by term.

Answer 30

http://www.wolframalpha.com/input/?i=Integrate%5B1%2Bx%5E4%2F2-x%5E8%2F8%2Bx%5E12%2F16-%285+x%5E16%29%2F128%2B%287+x%5E20%29%2F256-%2821+x%5E24%29%2F1024%2C+x%5D

Answer 31

then find the term \( t_n \) such than \( t_n < 0. 009\)

Answer 32

so means my integration is wrong just now?

Answer 33

where is your integration?

Answer 34

its x+ x^5/10 - x^9/72 + x^13/208+..

Answer 35

yes integrate it ... term by term.

Answer 36

huh, i already did, thats the answer, isnt it that we integrate 1+x^4/2-x^8/8+...?

Answer 37

you mean integrate again? why so?

Answer 38

of course ... your integral is definite integral, you need to put the values of 'x from 0 to 1

Answer 39

http://www.wolframalpha.com/input/?i=Integrate%5Bx%2B+x%5E5%2F10+-+x%5E9%2F72+%2B+x%5E13%2F208%2C+%7Bx%2C+0%2C+1%7D%5D

I was trying to tell you that you only need to integrate the first two terms.

Answer 40

woops!! did something go wrong?

Answer 41

i have no idea haha, im just puzzled coz we had to integrate it 2 times.. i got the series expansion of (1+x^4)^1/2 then from then i dont know already

Answer 42

you said integrate t so i did, and i got it, but u said u need to integrate it again and i dont know why it is related from having a definite integral from 0 to 1

Answer 43

http://www.wolframalpha.com/input/?i=Integrate%5BSqrt%5B1%2Bx%5E4%5D%2C+%7Bx%2C+0%2C+1%7D%5D

Answer 44

http://www.wolframalpha.com/input/?i=Integrate%5B1%2Bx%5E4%2F2-x%5E8%2F8%2Bx%5E12%2F16+%2C+%7Bx%2C+0%2C+1%7D%5D

Answer 45

Evaluate up to this much
1+x^4/2-x^8/8+x^12/16-(5 x^16)/128

Answer 46

the problem is how to get how many terms to get an answer to 2 dec places?

Answer 47

http://www.wolframalpha.com/input/?i=Integrate%5B1%2Bx%5E4%2F2-x%5E8%2F8%2Bx%5E12%2F16-%285+x%5E16%29%2F128%2C+%7Bx%2C+0%2C+1%7D%5D

Answer 48

yes ... that is the problem ...
when you integrate
x^n/k from 0 to 1, you get
1/(n*k)

Answer 49

hmm, so means i just have to like try out and estimate and see which will give me <0.009?

Answer 50

i mean for decimals.

Answer 51

no ... find the term, whose integral from 0 to 1 ... is less than 0.0001

Answer 52

or 0.001

Answer 53

just look at the series
http://www.wolframalpha.com/input/?i=Series%5BSqrt%5B1%2Bx%5E4%5D%2C+%7Bx%2C+0%2C+20%7D%5D

see if this is less than 0.001 or not ... or just you can take less than 0.009

Answer 54

http://www.wolframalpha.com/input/?i=Integrate%5B%285+x%5E16%29%2F128%2C+%7Bx%2C+0%2C+1%7D%5D

Answer 55

this term is not significant ... so just leave it, and all terms after it. only evaluate the terms before it.

Answer 56

read this article if you find tyme ...
http://tutorial.math.lamar.edu/Classes/CalcII/EstimatingSeries.aspx

Answer 57

i got 1.4375

Answer 58

add more terms then

Answer 59

you should get 1.09

Answer 60

okay... btw, i've been taught o take the taylo series but not the expansion but because in this case \[(1+k)^k=\sum_{n=0}^{\infty}\left(\begin{matrix}k \\ n\end{matrix}\right)x^n\] i dont know how to interpret the matrix so i dont know how to find the general term , that would be easier though

Answer 61

yeah i did i got 1.09 thanks!(:

Answer 62

that is not matrix ... it's k choose n.
although you hardly use it ... instead you use generalized binomial theorem.

Answer 63

hm, how to use it? can u make this an example?

Answer 64

\[ (1+x)^r = 1 + r x + r(r-1)/2! x^2 + r(r-1)(r-2) x^3/3! + ... \]

Answer 65

haha okay xD thats what i did just now, thank you so much for your help! :D

Answer 66

http://www.wolframalpha.com/input/?i=expand+%281%2Bx%29%5Er++at+x%3D0

Answer 67

put r=1/2 and x-> x^4 ... and see what you get,
you will get your Taylor series.

Answer 68

yup! figured that out just now, but you were saying that there was some other method u were suggesting but i didnt follow

Answer 69

that is for estimating the value of your integral ...

Answer 70

okay! :D thanks anyways!

Answer 71

sorry for going offline suddenly ... please read this article
you will find it very helpful
http://tutorial.math.lamar.edu/Classes/CalcII/EstimatingSeries.aspx

Answer 72

thank you! :D