Question

use four terms of a Taylor series to evaluate

Answer 1

\[\lim_{x \rightarrow 0} \frac{ x^2 + 2cosx - 2 }{ 3x^4 }\]

Answer 2

how do i do this?

Answer 3

@freckles

Answer 4

tay tay series is given by:
\[f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n\]

Answer 5

our thingy
(function if you want to be all weird about it)
\[f(x)=\frac{x^2+2\cos(x)-2}{3x^4}\]

Answer 6

yes

Answer 7

we can find the first 4 terms as that thingy in your problem suggests
so we are going to need the first 4 derivatives

Answer 8

3 actually

Answer 9

because the first term isn't really a derivative
i guess you can call it the zero derivative (which is just the original function)

Answer 10

\[f(x)=\frac{x^2+2 \cos(x)-2}{3x^4} \\ f(x)=\frac{1}{3}x^{-2}+\frac{2}{3} x^{-4} \cos(x)-\frac{2}{3} x^{-4}\]

Answer 11

so basically what we're doing is finding its derivative 4 times?

Answer 12

well we only need the first derivative
the second and the third

Answer 13

but it says use four terms

Answer 14

right f(x) will be used in the first term

Answer 15

that is 4 terms

Answer 16

okay \[f(x) = \frac{ x^2 + 2cosx - 2 }{ 3x^4 }\]

Answer 17

\[f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2+\frac{f'''(a)}{6}(x-a)^3\]

Answer 18

how do you do quotient derivative again?

Answer 19

you don't have to

Answer 20

you can separate the 3 fractions as I did above
and then use law of exponents to rewrite
then just use power rule and product rule

Answer 21

but

Answer 22

if I'm not mistaken there might be an easier way to do this

Answer 23

what is it?

Answer 24

no I'm too afraid
yeah let's just find the first 3 derivatives of f(x)

Answer 25

lol alright

Answer 26

hey hey

Answer 27

I think I'm going to take away my fear
I think we can look at the taylor series separate for each term

Answer 28

so you did this right x^2 / 3x^4 + 2cosx / 3x^4 - 2 /3x^4?

Answer 29

polynomials are unnecessary for rewrite when it comes to tay series

Answer 30

what?

Answer 31

\[f(x) = \frac{ x^2 + 2cosx - 2 }{ 3x^4 } \]
just find the first 4 terms of cos(x) using taylor series

Answer 32

so x^2 / 3x^4 = 3x^-2 right?

Answer 33

right don't do that
that is going to be ugly

Answer 34

but it would be 1/3*x^(-2)

Answer 35

but yeah still don't do that
that is going to be really really super duper ugly

Answer 36

but isnt that the way you did it

Answer 37

can we simplify the top if we already know the series for 2cos(x)?

Answer 38

I want you to only find the first 4 terms of cos(x) using taylor series

Answer 39

okay

Answer 40

\[f(x) = cosx \]

Answer 41

\[\lim_{x \rightarrow 0} \frac{ x^2 + 2cosx - 2 }{ 3x^4 } \\ \lim_{x \rightarrow 0}\frac{x^2+2( ...... )-2}{3x^4}\]
when are you done you are going to put that mess there where those ..... things are

Answer 42

f(x) = cosx f'(x) = sinx f''(x) = -cosx f'''(x) = sinx

Answer 43

you want to me to put all for terms between those parenthesis

Answer 44

four*

Answer 45

f(x)=cos(x)
f'(x)=-sin(x)
f''(x)=-cos(x)
f'''(x)=sin(x)

just one sign on f' is gross

Answer 46

now plug in mister 0

Answer 47

we are going to evaluate the tay series at x=0

Answer 48

a=0*

Answer 49

im assuming its already known ...
\[\frac{1}{3x^4}\left(x^2-2+2\sum_0\frac{(-1)^n}{(2n)!}x^{2n}\right)\]
\[\frac{1}{3x^4}\left(x^2-2+2(1)-2(\frac{x^2}{2 })+2\sum_2\frac{(-1)^n}{(2n)!}x^{2n}\right)\]

Answer 50

do you already know the tay series for cos(x) el_arrow ?

Answer 51

\[\lim_{x \rightarrow 0} \frac{ x^2 + 2(cosx - sinx - cosx + sinx) -2 }{ 3x^4 }\]

Answer 52

no

Answer 53

\[\text{ Let } f(x)=\cos(x) \\f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2+\frac{f'''(a)}{6}(x-a)^3 \\ \text{ so far you have } f(x)=\cos(0)-\sin(a)(x-a)-\frac{\cos(a)}{2}(x-a)^2+\frac{\sin(a)}{6}(x-a)^3\]
using the general thingy I wrote above for the first 4 terms of a Taylor series
now it would be easiest to apply this series to find the limit when a is 0

Answer 54

lol I already accidentally replaced one of the a's with 0

Answer 55

its hard to imagine that they would not have covered the taylor series for cos or sin ... and instead give you this thing to play with.

Answer 56

oh oh I should be approximately sign instead of equal sign

Answer 57

how did you get 2 and 6 as the denominator

Answer 58

2!=2(1)=2
3!=3(2)(1)=6

Answer 59

oh

Answer 60

http://math.bard.edu/belk/math142af09/ApplicationsTaylorSeries.pdf
also read this later

Answer 61

for a dbl chk ...
\[\frac{1}{3x^4}\left(2\sum_2\frac{(-1)^n}{(2n)!}x^{2n}\right)\]
\[\frac23\sum_2\frac{(-1)^n}{(2n)!}x^{2n-4}\]
\[\frac23\sum_{0}\frac{(-1)^n}{(2(n+2))!}x^{2n}\]

Answer 62

what are we looking for now?

Answer 63

first 4 taylor series terms of cos(x)

Answer 64

\[\text{ Let } f(x)=\cos(x) \\f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2+\frac{f'''(a)}{6}(x-a)^3 \\ \text{ so far you have } f(x)=\cos(a)-\sin(a)(x-a)-\frac{\cos(a)}{2}(x-a)^2 \\ +\frac{\sin(a)}{6}(x-a)^3 \]
with a being 0

Answer 65

replace a with 0's
and evaluate cos(0) and sin(0) as they occur

Answer 66

the weird thing is I think we are going to need more terms
and I guess they aren't counting the terms that end up as 0 as terms

Answer 67

okay \[f(x) = \cos(0) - \sin(0)(x-0) - \frac{ \cos(0) }{ 2 } (x-0)^2 + \frac{ \sin(0) }{ 6}(x-0)^3\]

Answer 68

anyways yeah I'm going to just do what @amistre64 did
you can show
\[\cos(x) \approx 1-\frac{1}{2}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6\]
there are the first non 0 terms of cos(x)

Answer 69

these are the first 4 none- zero terms of cos(x)*

Answer 70

f(x) = 1 - 1/2 x^2 + 1/4x^4 - 1/6x^6

Answer 71

don't forget your ! thingy

Answer 72

I didn't put it on the 2 because 2! just means 2

Answer 73

oh right so now how do we find the none-zero terms for the whole equation

Answer 74

\[\lim_{x \rightarrow 0} \frac{ x^2 + 2cosx - 2 }{ 3x^4 } \\ \lim_{x \rightarrow 0}\frac{x^2+2( 1-\frac{1}{2}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6 )-2}{3x^4}\]

Answer 75

you will get a little cancel action on top

Answer 76

distribute and add like terms on top

Answer 77

oh okay cancel everything out

Answer 78

well not everything

Answer 79

just the x's?

Answer 80

you see the 2-2?
and the x^2-x^2?

Answer 81

yes

Answer 82

ok so that is the only things that add up to zero on top

Answer 83

you should have two terms left on top now

Answer 84

you can separate the fraction and evaluate the limits separately for both

Answer 85

or just plug in 0

Answer 86

lol

Answer 87

well after canceling

Answer 88

but if i plug 0 for x everything will go to zero

Answer 89

after the cancellation part

Answer 90

\[\lim_{x \rightarrow 0}(\frac{\frac{2}{4!}x^4}{3x^4}-\frac{\frac{2}{6!}x^6}{3x^4})\] cancel a little more

Answer 91

\[\frac{ \frac{ 1 }{ 4! } x^4 - \frac{ 1 }{ 6! } x^6}{ 3x^4 }\]

Answer 92

okay so the 2/4! turns to 1/2! and the x^4/x^4 = 1

Answer 93

right?

Answer 94

4!=4*3*2*1
2/(4*3*2*1)
=1/(4*3*1)

Answer 95

but yes x^4/x^4=1

Answer 96

so 2/6! x^6 / 3x^4 = 1 /( 6*5*4*3*1) x^2

Answer 97

you don't need to do that for the other term
remember x^6/x^4=x^2
this is the part you plug 0 into

Answer 98

and you are left with the other number