Question

Approximate tan inverse (0.1) to three decimal place accuracy using the Maclaurin series tan inverse x

Answer 1

@PlasmaFuzer

Answer 2

@ganeshie8

Answer 3

@pooja195

Answer 4

@Shalante

Answer 5

I did this so far

The book says that the error to 3 decimal places is

\[(0.1)- \frac{ 0.1^3 }{ 3 }\]

Answer 6

however when i put (0.1^3)/3 in my calculator the answer is 3.3333 * 10^-4

that is less that 0.5 * 10^-3

Answer 7

so why could that be the answer? wouldn't it just be 0.1 ?

Answer 8

@Hero

Answer 9

the question asks you to do it using the maclaurin for cos x

is that what you did?

Answer 10

yes

Answer 11

terrible graphics i know but it seems you plugged your numbers into the expansion for\(\tan^{-1}x\).

Answer 12

I wonder why it says to approximate arctan(0.1) using the maclaurin series for cos(x)
and here it says use maclaurin series for arctan(x)
I think we would want to use the series for arctan(x) since we are apporximating arctan( something)

Answer 13

yeah i did.

Answer 14

but it really does say cos(x)?

Answer 15

tan inverse (0.1)

Answer 16

You know I'm trying to tell you that there is conflicting information here right?

Answer 17

ohhh I see

Answer 18

my bad the one i copied was wrong

Answer 19

but the question here is right

Answer 20

ok this question makes more sense to me :)
Approximate tan inverse (0.1) to three decimal place accuracy using the Maclaurin series tan inverse x

\[0.1-\frac{0.1^3}{3} \text{ is a pretty good approximation for } \arctan(0.1) \\ 0.1-\frac{0.1^3}{3} \approx 0.099\overline{6} \\ \text{ while } \arctan(0.1) \text{ to 6 decimal spots is } 0.09966\]

Answer 21

Why is the the answer they have on the back of the book is

(0.1)−(0.1^3/3)

Answer 22

but the way I learned is by taking each monomial of the series and substituting the value of x

Answer 23

and (0.1^3)/3 is less that 5*10^-4

Answer 24

yep but then you have to add the terms in the series

Answer 25

where does 5*10^(-4) come from ?

Answer 26

from a rule that says that the decimal places they are asking you for is equal to

0.5*10^-n

where n is the number of decimal places

Answer 27

so i just simplified to 5*10^-4

Answer 28

and i need to find a term that is smaller than 5*10^-4 in the series

Answer 29

@PlasmaFuzer

Answer 30

oh I don't remember that
I would have honestly found arctan(0.1) first which is approximately 0.0996
from the calculator then added the terms in the series until I got with in digits of that number
\[\arctan(x)=\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{2k+1} \\ \arctan(x)=(-1)^0 \frac{x^{2(0)+1}}{2(0)+1}+(-1)^{1}\frac{x^{2(1)+1}}{2(1)+1}+(-1)^{2}\frac{x^{2(2)+1}}{2(2)+1} \\ + \cdots +(-1)^k \frac{x^{2k+1}}{2k+1}+ \cdots \\ \text{ so let's see if one term is enough } \\ \arctan(x) \approx x \\ \arctan(0.1) \approx 0.1 \text{ this is a really bad approximation } \\ \text{ how about two terms } \\ \arctan(x) \approx x- \frac{x^3}{3}\]
\[\arctan(0.1) \approx 0.1-\frac{0.1^3}{3} =.099\overline{6} \text{ this is an awesome approximation }\]

Answer 31

the approximation only gets better if we add more terms onto that

Answer 32

@blackstreet23 Hello... Freckles and irishboy have it right here... That formula you have listed (0.5*10^-n)I dont recall (though I think I can guess where it comes from), but if your issue is you want more accuracy simply add one more term to the approximating series no?

Answer 33

well i guess i will ask this one to my professor. Thanks guys for your help!

Answer 34

Sorry I couldn't help Im not sure what you are hung up on as far as the problem is concerned

Answer 35

ohh no problem! i know you did your best :)