Question

When finding a formula for truncation error, why do you only take the absolute value of the top?

Answer 1

Ex: \[\frac{ 128lxl ^{7} }{ 1-2x }\]

Answer 2

Why don't I take the absolute value of the bottom part?

Answer 3

can you show me the process whereby we formulate a truncation error?

Answer 4

What do you mean?

Answer 5

They gave me a function 1/(1-2x) and they asked for the error associated with P6(x)

Answer 6

do we just magically have formulas? or are they developed thru more basic measures, like taking limits of this or that and working it out.

Answer 7

So I generated the series for the function using a/(1-r), and then I created a series for every term after the 6th term

Answer 8

using a/(1-r) again

Answer 9

so we want to create a polynomial that represents 1/(1-2x)

if we cut off (truncate) the poly at the 6th degree; then the error of truncation is the sum of the parts from 7 to infinity ... am i understanding this correctly?

Answer 10

Exactly

Answer 11

i havent tried to work it so ill ahve to review some stuff first to refresh my memory on this. if none else come by with an answer ill post what i can dig up.

Answer 12

Ok thanks. I got the answer, but I didn't understand the absolute value thing

Answer 13

one thing ive got in the back of my head has to do with an alternating error thrm; the size of the error of truncation is +- no more than, the first neglected term.

but i cant be sure that this applies here or not. so im reviewing :)

Answer 14

post your work if you can ...

Answer 15

Ok. the function was 1/(1-2x). So I said a=1 and r=2x, so I generated the series 1+2x+4x^2+8x^3+16x^4+32x^5+64x^6+128x^7+256x^8...
I cut it off after 64x^6, and I made a series a/(1-r) with 128x^7 as the first term. That series was 128x^7/(1-2x)

Answer 16

and the series converges of course if |2x| < 1, or |x|<1/2

its just my thought, but i think it has to do with this .... but i could be completely wrong at the moment :)

Answer 17

They did give me the interval (-1/2,1/2), so I dont know if that had anything to do with it

Answer 18

lagrange form of Rn
\[R_n=f^{(n+1)}(c)\frac{(x-a)^{n+1}}{(n+1)!}\le \left|f^{(n+1)}(c)\frac{(x-a)^{n+1}}{(n+1)!}\right|\]

Answer 19

Yes, so why do I only have to put absolute value bars around the numerator?

Answer 20

1+2x+4x^2+....
--------------------
1-2x | 1
1-2x
2x
2x-4x^2
4x^2
----------------------------
\[\frac{1}{1-2x}=\sum_{n=0}^{\infty}~2^nx^n\]
\[\sum_{n=7}^{\infty}~2^nx^n\le\left|\frac{2^7x^7}{1-2x}\right|\]

Answer 21

1-2x < 0
2x > 1
x > 1/2

becasue the bottom is never negative

1-2(-1/2) = 2
1-2(1/2) = 0

the bottom is from 0 to 2

Answer 22

2^7 is never negative ... so ... whats that leave?

Answer 23

spose we have x=-1/3 the top goes negative right?

Answer 24

for x = -1/2 to 0, the top goes negative, the bottom stays positive ...

Answer 25

-1/2 < x < 1/2

-1 < -2x < 1

0 < 1-2x < 2

notice the bottom is always positive in this interval

but x^7 is negative from -1/2 < x < 0
-------------------------------

-1/2 < x < 1/2


-1/2^7 < x^7 < 1/2^7

|x|^7 <= 1/2

Answer 26

I'm really confused

Answer 27

tellme what about this explanation is not making sense

Answer 28

I just don't understand how you are proving that the bottom doesn't need abs val bars

Answer 29

when is 1-2x less than zero? when is it negative?

Answer 30

When x>2

Answer 31

x>1/2

Answer 32

sorry

Answer 33

x is between -1/2 and 1/2 right? now work x into 1-2x

given: -1/2 < x < 1/2 , multiply by -2

-2/2 < -2x < 2/2 , add 1

1-1 < 1-2x < 1+1

0 < 1-2x < 2

since the bottom can only be values from 0 to 2 .... its ALWAYS positive for a given x value in its domain. right?

Answer 34

Ooooh ok.

Answer 35

if A and B are always postive then
\[\left|\frac{Ax}{B}\right|\implies \frac{A|x|}{B}\]

Answer 36

if k is even: x^k is always positive, but 7 is not an even number :)

Answer 37

Oooh ok. That makes much more sense! So I technically could put the bars around the whole thing?

Answer 38

yes, it wouldnt make a difference either way

Answer 39

IF the stated domain made the bottom negative at any point ... then we would keep it covered in | |

but theres no real need for it in this case

Answer 40

I see! Thanks so much!

Answer 41

youre welcome