Question

(Using error theorem) For approximately what values of x can you replace sin x by x-(x^3/6) with an error magnitude no greater than 5x10^-4?

Answer 1

@amistre64

Answer 2

im guessing youve truncated the taylor series?

Answer 3

Yes

Answer 4

what does your error thrm tell us that helps us out then

Answer 5

\[Rn(x)\le \frac{ f ^{n+1}(c)x ^{n+1} }{ (n+1)! }\]

Answer 6

I was pretty sure that I got the right answer, but apparently not :/

Answer 7

that seems to be an approximation for the 'at most' error. this is asking more for a specific value of x that will keep us in a valid range

Answer 8

Ok. Well it said that there was a way to use the theorem. There's another theorem that they gave us, but we never use it in class, and my teacher says to use this one

Answer 9

I got really close with my error formula

Answer 10

But I guess I messed part of it up

Answer 11

ok, its asking for a at most error

perhaps we are looking for the value of c? does that seem valid?

Answer 12

For c I used 1 because that is the most that the n+1th derivative will ever be

Answer 13

My formula ended up being (x^(2n+3))/((2n+3)!)

Answer 14

And if you plug in n=2, that spits out x^7/7!. The answer is apparently x^5/5!

Answer 15

well, we did truncate at x^3, then next term in the series is the x^5 so the answer at least makes sense

Answer 16

Oh shoot, I just realized that I am supposed to use n=4... Now that doesn't make any sense whatsoever

Answer 17

I am extremely confused

Answer 18

if we retain our terms 1 by 1, instead of only seeing the odd terms

\[sin(x)=0+x-0x^2/2!-x^3/3!+0x^4/4!+x^5/5! -0x^6/6!-x^7/7!+...\]
right?

Answer 19

Yes

Answer 20

But I thought that I was able to generate a rule strictly for the nonzero terms. The sine maclaurin series ignores the zero terms

Answer 21

it ignores them simply because it deals with (2n+1) .. it really isnt ignoring them is it?

at any rate, the next term that was left out was the x^5 term, and the error thrm says that the |error| is no more than the first neglected term. if memory serves

Answer 22

So my error formula has to account for the zero terms too?

Answer 23

can you post your workings? or a screenshot of them?

Answer 24

I mean all I did was\[R2(x)\le \frac{ 1(x ^{2(n+1)+1} }{ (2(n+1)+1)! }\]

Answer 25

I forgot to close the parentheses on top

Answer 26

But basically I just replaced n with n+1

Answer 27

And I plugged it into my calculator with n=2, and it looked nothing like what it should've

Answer 28

ok, n+1 is the next n value ... good and sin(x) or cos(x) is maxed at 1 good

Answer 29

Ok. I can't figure out where I went wrong

Answer 30

For approximately what values of x can you replace sin x by x-(x^3/6) with an error magnitude no greater than 5x10^-4

error thrm: the max error is no more than the first neglected term. the first neglected term is x^5/5!

|x^5/5!| <= 5x10^-4

|x^5| <= 5!(5)(10^-4)

|x| <= (5!(5)(10^-4))^(1/5)

|x| <= .5697

this is how im reading the problem

Answer 31

Well thats the answer

Answer 32

I don't understand what you did though

Answer 33

then im not sure why you did an n+1 and such

Answer 34

what is the first neglected term?

Answer 35

x^5/4!

Answer 36

5!

Answer 37

then, why not use it?

when is x value of the first neglected term less than or equal to 5(10^-4) ...

Answer 38

Why am I only concerned about the first neglected term?

Answer 39

because the error thrm says so; and that is what you are spose to use.

am i misstating the error thrm any?

Answer 40

The theorem was that equation that I posted a while back

Answer 41

yeah, at least the lagrange form of it.

Answer 42

Yes

Answer 43

That takes into account the infinite amount of terms that you cut off

Answer 44

the sin is an alternating series isnt it ... (-1)^n

Answer 45

Yes

Answer 46

then im prolly remembering something about an alternating series which states what i stated. The error in an alternating series is no more than the first neglected term.

\[sin(x)=\sum_{0}^{}\frac{(-1)^n}{(2n+1)!}x^{2n+1}\]
\[sin(x)=x-\frac{x^3}{3!}+\sum_{2}^{}\frac{(-1)^n}{(2n+1)!}x^{2n+1}\]

Answer 47

What do you mean that the error is no more than the first neglected term? We never learned anything like that

Answer 48

you havent, but i have :) let me do some review, but notice that the Lagrange form also uses the first neglected term ....

Answer 49

Ima try to explain without talking giberish i have solved this on another site a whole while back :)

Answer 50

Ok thanks!!

Answer 51

I know that it uses the first neglected term, but it accounts for all of the ones after that

Answer 52

My question isn't why you are using the first neglected term. Its why you aren't accounting for all of the ones after that

Answer 53

\[R_2(x)<=\left|(1)\frac{x^{2(2)+1}}{(2(2)+1)!}\right|\]
we are using your langrange setup, and f^(n)(c)=1 regarless
why is it you are not using this?

Answer 54

by your own admission, the lagrange form accounts for all the other terms in this package right?

Answer 55

the lagrange form utilizes the mean value thrm

Answer 56

I am using that. I was confused as to why you were only using x^5/5!

Answer 57

i was using it out of some repressed memory lol

Answer 58

And I am confused as to why my answer was wrong

Answer 59

lol

Answer 60

you used x^7/7! instead of using the lagrange form correctly, you plugged in n=3 instead of n=2

Answer 61

Is that really all I did.

Answer 62

I spent over an hour trying to rectify this and that is all I did T_T

Answer 63

I quit math

Answer 64

\[\underbrace{\frac{x^{2(0)+1}}{(2(0)+1)!}-\frac{x^{2(1)+1}}{(2(1)+1)!}}_{=x-\frac{x^3}{6}}+\frac{x^{2(2)+1}}{(2(2)+1)!}-\]

Answer 65

maybe you confused 6 as 6! and assumed the next term? i dunno ... its your psyche :)

Answer 66

I just realized that you are completely right. I feel like a moron now XD

Answer 67

Thank you so much for helping me!!