Question

How to calculate sum of alternating series whose terms are decreasing in magnitude summation n = 0 to infinity [ (-1)^n/(2n)!]

Answer 1

e^(-1)

Answer 2

i think it might be cosine

Answer 3

\[\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...\] so i think it is \(\cos(1)\)

Answer 4

How to compute the sum as accurately as possible without using cos(1). Any hints on steps ?

Answer 5

you can't exactly calculate the sum ... if you want to calculate the sum then you have to do it numerically. of course there are errors ... we make approximations.

Answer 6

this is pretty much the same question as "how do you compute \(\cos(1)\)
approximate it to any degree of accuracy you want by computing the sum you wrote above

Answer 7

\[\sum_{k=0}^n\frac{(-1)^k}{(2k)!}\]with whatever \(n\) you want for accuracy

Answer 8

Hi exactly i have to calculate the sum as accurately as possible so what value of n should i take ?

Answer 9

I can use four digit chopping arithmetic ( it is a numerical analysis question )

Answer 10

i would whip out a calculator, hit \(\cos(1)\) and copy down as many digits as i care to

Answer 11

i have never heard of four digit chopping method, maybe
@experimentX knows it

Answer 12

i've never heard it too .. but this method is useful in approximating alternating series.
http://en.wikipedia.org/wiki/Alternating_series#Approximating_sums
same as used in showing the convergence of alternating series
http://en.wikipedia.org/wiki/Alternating_series#Approximating_sums

Answer 13

up to what terms you want accurate ... you can calculate the number of terms you have to add up.

Answer 14

http://www.physicsforums.com/showthread.php?t=670526

Please see the last post on this page i did not understand how the value of n and sum was obtained.

Answer 15

sorry .. hold on a bit.

Answer 16

this is same as approximating the sum!!
you can represent the value of cos(1) as \( 0.a_1 a_2 a_3 a_4 a_5a_5 ...\)
your know that your value is given by \( \sum_{k=0}^n\frac{(-1)^k}{(2k)!} \) but you want the value correct up to 5 terms only. \( 0.a_1 a_2 a_3 a_4 a_5 \)

Answer 17

but the sum goes to infinity. In order to evaluate that ... up to how many terms you must add up in your calculator?

Your order is of \( 10^{-6 } \)

Answer 18

ok thx :)

Answer 19

just start with (1/(2n)! > 10^-6 ) .. find the value of n, and add up only to that term.

Answer 20

more can be found here
http://tutorial.math.lamar.edu/Classes/CalcII/EstimatingSeries.aspx
seems i also have forgotten.

Answer 21

thanks a lot @experimentX

Answer 22

no probs!! i was confused myself at first also thanks to Sat

Answer 23

thank you satellite73