sigma notation [series & sequences] - Tilted Forum Project Discussion Community

keyshawn · 03-29-2005, 08:01 AM

Hi,

I want to double check this math problem for the spring break take-home test:

q: use sigma notation to express the series: 0.2 - .8 + 3.2 - 12.8 + 51.2 - ..........

A: I realized that the powers of 2 are at odd integers, then divided by 10.

[i.e. 2^1=2
2^3=8
etc.....]

but I'm trying to figure out how to express this [the fact only odd integers are used as the powers] in notation....

so far i have:

Code:

   [infinity symbol]          [2^k] (-1)^(k+1)
 [Sigma symbol]                       _____________________
  K= 1                                                  10

thanks for the help,
will.

elsesomebody · 03-29-2005, 08:34 AM

Well my first suggestion would be to put the series {1,3,5,7,9...} in sigma notation. These are the exponents you're raising two to. This is what you're hung up on, and hopefully by breaking it down this way you can do it.

iccky · 03-29-2005, 09:14 AM

Quote:

Originally Posted by keyshawn

so far i have:

Code:

   [infinity symbol]          [2^k] (-1)^(k+1)
 [Sigma symbol]                       _____________________
  K= 1                                                  10

Almost there, but like you say the formula you have now will give you .2-.4+.8 ...

Try this:

Code:

   [infinity symbol]          [2^(2k+1)] (-1)^(k)
 [Sigma symbol]                       _____________________
  K= 0                                                  10

I'm a bit rusty and working quickly but that should work.

elsesomebody · 03-29-2005, 10:12 AM

Quote:

Originally Posted by iccky

Almost there, but like you say the formula you have now will give you .2-.4+.8 ...

Try this:

Code:

   [infinity symbol]          [2^(2k+1)] (-1)^(k)
 [Sigma symbol]                    _____________________
  K= 0                                  10

I'm a bit rusty and working quickly but that should work.

Actually I believe it should be:

Code:

   [infinity symbol]          [2^(2k-1)] (-1)^(k+1)
 [Sigma symbol]                _____________________
  K= 0                               10

keyshawn · 03-30-2005, 12:07 PM

aha - that's it !

thanks for your help ickky and elsesomebody !

catcha back on the flipside,
keyshawn

Hain · 03-31-2005, 09:43 PM

Well first of this cant be an infinite series since the sum is divergent. And also the equations presented before are somewhat difficult to work with when dealing with a sum.

Code:

 n
Sum   -((-4)^k)/20
k=1

Then the sum just becomes

Code:

remember: a1*(1-r^n)/(1-r) ; where r is -4 and a1 is the first term .2
          so:
(1-Tn)/25 ; Where Tn is the finishing term of your sum

iccky · 04-05-2005, 06:59 AM

Quote:

Originally Posted by keyshawn

aha - that's it !

thanks for your help ickky and elsesomebody !

catcha back on the flipside,
keyshawn

No problem

I always wanted to answer a math problem in tilted knowledge but all these smart TFPers always beat me to it.

03-29-2005, 08:01 AM	#1 (permalink)
keyshawn Psycho	sigma notation [series & sequences] Hi, I want to double check this math problem for the spring break take-home test: q: use sigma notation to express the series: 0.2 - .8 + 3.2 - 12.8 + 51.2 - .......... A: I realized that the powers of 2 are at odd integers, then divided by 10. [i.e. 2^1=2 2^3=8 etc.....] but I'm trying to figure out how to express this [the fact only odd integers are used as the powers] in notation.... so far i have: Code: [infinity symbol] [2^k] (-1)^(k+1) [Sigma symbol] _____________________ K= 1 10 thanks for the help, will. __________________ currently reading: currently playing :

03-30-2005, 12:07 PM	#5 (permalink)
keyshawn Psycho	aha - that's it ! thanks for your help ickky and elsesomebody ! catcha back on the flipside, keyshawn __________________ currently reading: currently playing :

03-31-2005, 09:43 PM	#6 (permalink)
Hain has a plan Location: middle of Whywouldanyonebethere	Well first of this cant be an infinite series since the sum is divergent. And also the equations presented before are somewhat difficult to work with when dealing with a sum. Code: n Sum -((-4)^k)/20 k=1 Then the sum just becomes Code: remember: a1*(1-r^n)/(1-r) ; where r is -4 and a1 is the first term .2 so: (1-Tn)/25 ; Where Tn is the finishing term of your sum __________________ ////\\// IRONSKY.NET \\//\\\\

03-29-2005, 08:34 AM	#2 (permalink)
elsesomebody Tilted Location: North Carolina	Well my first suggestion would be to put the series {1,3,5,7,9...} in sigma notation. These are the exponents you're raising two to. This is what you're hung up on, and hopefully by breaking it down this way you can do it.