Question

Find the partial sum of a series

Answer 1

give me sec to write the equation

Answer 2

\[s _{10} \sum_{n=1}^{\infty} 1/(n^4)\]

Answer 3

\[s _{10} of the series\]

Answer 4

http://www.wolframalpha.com/input/?i=sum+n+%3D+1+to+10+1%2Fn^4

Answer 5

\[ s _{10} + \int\limits_{?}^{\infty} 1/ x^4 dx \le s \le s _{10} + \int\limits_{?}^{\infty} 1/x^4 dx\]

Answer 6

according to an example in the book...don't really understand it though

Answer 7

i don't understand what you wrote

Answer 8

here is what's in the book:
the inequalities give a lower bound and an upper bound for s . They provide a more accurate approximation to the sum of the series than the partial sum sn does

Answer 9

maybe something like
\[\int_1^{\infty}\frac{dx}{x^4}\leq S\leq \int_0^{\infty}\frac{dx}{x^4}\]?

Answer 10

\[s_{10} +\int\limits_{11}^{\infty} 1/x4dx ≤ s ≤ s _{10}+\int\limits_{10}^{\infty} 1/x4 dx \]

Answer 11

oooooooooooooooh ok

Answer 12

i guess it is like upper sums vs lower sums or left hand endpoints vs right hand ones for the integrals

Answer 13

\[s_{10}+ \int\limits_{n+1}^{\infty} 1/x4dx ≤ s≤ s_{10}+ \int\limits_{n}^{}1/x4dx \]

Answer 14

I guess so... what do they mean by \[s_{n} + R_{n} = s\]

Answer 15

i have no idea, nor do i know why this should be true
\[\sum_{k=1}^{\infty}\frac{1}{k^4}<\sum_{k=1}^{10}\frac{1}{k^4}+\int_n^{\infty}\frac{dx}{x^4}\] in fact is seem rather unlikely

Answer 16

have you heard of Riemann zeta functions?

Answer 17

maybe
\[\sum_{k=1}^{\infty}\frac{1}{k^4}\leq\sum_{k=1}^{n}\frac{1}{k^4}+\int_n^{\infty}\frac{dx}{x^4}\]

Answer 18

yeah as a matter of fact i just bought a cheap book on amazon about it

Answer 19

I can write up the whole question because it's based on riemann zeta functions ...give me a second to write it...it should make more sense that way

Answer 20

so we know \[\sum_{k=1}^{\infty}\frac{1}{k^4}=\zeta(4)\]

Answer 21

and i guess you are getting a bound for this? or trying to find its value?

Answer 22

let me grab my book

Answer 23

What is the doman of zeta?
(a) find the partial sum \[s_{10}\] of the series. Estimate the error in using \[s_{10}\] as an approximation to the sum of the series.

Answer 24

i am not sure how to answer the first question as the reimann zeta function is the analytic continuation of \(f(s) =\sum_{n=1}^{\infty}\frac{1}{n^s}\) as a real valued function the domain of \(\zeta(s)\) is \(s>-1)\) because otherwise the sum will not converge \sum_{k=1}^{\infty}\frac{1}{k^4}<\sum_{k=1}^{10}\frac{1}{k^4}+\int_n^{\infty}\frac{dx}{x^4}

Answer 25

finding \(\sum_{k=1}^{10}\frac{1}{k^4}\) is an annoying but direct computation, i used wolfram

Answer 26

got \(1.0820365834937565467850774213178081250883564048689005...\)

Answer 27

wolfram is helpful in this regard, but unfortunately im not allowed to use it on an exam.
When I follow the example i'm able to get a solution...but in the example s10 =1.197532 how will I find the value of s10?

Answer 28

as for the last part, i think you are supposed to say that the error is between \[\int_{10}^{\infty}\frac{dx}{x^4}\] and
\[\int_{11}^{\infty}\frac{dx}{x^4}\]

Answer 29

\[\sum_{n=1}^{\infty} = s _{10}\]

Answer 30

that makes sense but the s10 is not quite clicking

Answer 31

hold on
my understanding is that \(S_{10}\) is the tenth partial sum, i.e. you sum from 1 to 10 right?

Answer 32

forgot to add f(x) to the equation before the equal sign

Answer 33

yes

Answer 34

ooooh i see

Answer 35

then \[S_{10}=\sum_{k=1}^{10}\frac{1}{n^4}\]

Answer 36

ok typo there, but you get the idea right? it i the tenth partial sum

Answer 37

instead of to infinity

Answer 38

exactly

Answer 39

how do i find the error?

Answer 40

so what they are telling you is that the error of the tenth partial to the actual sum is the integral from 10 to infinity on one side and the integral from 11 to infinity on the other

Answer 41

i see

Answer 42

you wrote
\[s_{10} +\int\limits_{11}^{\infty} \frac{1}{x^4}dx ≤ s ≤ s _{10}+\int\limits_{10}^{\infty} \frac{1}{x^4} dx\]

Answer 43

correct

Answer 44

and so \(S_{10}\) is a number known to you, as are those integrals (once you compute them) therefore giving you an upper and lower bound for \[S=\sum_{n=1}^{\infty}\frac{1}{n^4}\]

Answer 45

yes

Answer 46

oh...is it the difference between that and the answer i get from the inequalities

Answer 47

computing the integrals is easy. but without a calculator i have no idea how to compute
\[S_{10}=\sum_{n=1}^{10}\frac{1}{n^4}\]

Answer 48

yes exactly
you have an upper and lower bound on \(S\) by the inequalities you wrote

Answer 49

ok. It's all making sense now

Answer 50

integral on the left is \(\frac{1}{3993}\) and integral on the right is \(\frac{1}{3000}\) so basically you have
\[S_{10}+\frac{1}{3993}<S<S_{10}+\frac{1}{3000}\]

Answer 51

for S10 i get = 1/1 + 1/(2^4) + 1/( 3^4)..... =

Answer 52

yes up to ten

Answer 53

i don't think there is a snap way to do this, i think it requires a calculator

Answer 54

yep

Answer 55

think about how much this sucks with all the ugly denominators. you do not want to do this by hand
that is why i used wolfram

Answer 56

I will...but at least now I know why i'm plugging and chugging numbers

Answer 57

ok so that is done right? we have a lower bound on the left and an upper one on the right