blackstreet23
  • blackstreet23
Approximate tan inverse (0.1) to three decimal place accuracy using the Maclaurin series tan inverse x
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
blackstreet23
  • blackstreet23
@PlasmaFuzer
blackstreet23
  • blackstreet23
@ganeshie8
blackstreet23
  • blackstreet23
@pooja195

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

blackstreet23
  • blackstreet23
@Shalante
blackstreet23
  • blackstreet23
1 Attachment
blackstreet23
  • blackstreet23
I did this so far The book says that the error to 3 decimal places is \[(0.1)- \frac{ 0.1^3 }{ 3 }\]
blackstreet23
  • blackstreet23
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
blackstreet23
  • blackstreet23
so why could that be the answer? wouldn't it just be 0.1 ?
blackstreet23
  • blackstreet23
@Hero
IrishBoy123
  • IrishBoy123
the question asks you to do it using the maclaurin for cos x is that what you did?
blackstreet23
  • blackstreet23
yes
IrishBoy123
  • IrishBoy123
|dw:1442179486992:dw| terrible graphics i know but it seems you plugged your numbers into the expansion for\(\tan^{-1}x\).
freckles
  • freckles
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)
blackstreet23
  • blackstreet23
yeah i did.
freckles
  • freckles
but it really does say cos(x)?
blackstreet23
  • blackstreet23
tan inverse (0.1)
freckles
  • freckles
You know I'm trying to tell you that there is conflicting information here right?
blackstreet23
  • blackstreet23
ohhh I see
blackstreet23
  • blackstreet23
my bad the one i copied was wrong
blackstreet23
  • blackstreet23
but the question here is right
freckles
  • freckles
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 \[0.1-\frac{0.1^3}{3} \text{ is a pretty good approximation for } \arctan(0.1) \\ 0.1-\frac{0.1^3}{3} \approx 0.099\overline{6} \\ \text{ while } \arctan(0.1) \text{ to 6 decimal spots is } 0.09966\]
blackstreet23
  • blackstreet23
Why is the the answer they have on the back of the book is (0.1)−(0.1^3/3)
blackstreet23
  • blackstreet23
but the way I learned is by taking each monomial of the series and substituting the value of x
blackstreet23
  • blackstreet23
and (0.1^3)/3 is less that 5*10^-4
freckles
  • freckles
yep but then you have to add the terms in the series
freckles
  • freckles
where does 5*10^(-4) come from ?
blackstreet23
  • blackstreet23
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
blackstreet23
  • blackstreet23
so i just simplified to 5*10^-4
blackstreet23
  • blackstreet23
and i need to find a term that is smaller than 5*10^-4 in the series
blackstreet23
  • blackstreet23
@PlasmaFuzer
freckles
  • freckles
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 \[\arctan(x)=\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{2k+1} \\ \arctan(x)=(-1)^0 \frac{x^{2(0)+1}}{2(0)+1}+(-1)^{1}\frac{x^{2(1)+1}}{2(1)+1}+(-1)^{2}\frac{x^{2(2)+1}}{2(2)+1} \\ + \cdots +(-1)^k \frac{x^{2k+1}}{2k+1}+ \cdots \\ \text{ so let's see if one term is enough } \\ \arctan(x) \approx x \\ \arctan(0.1) \approx 0.1 \text{ this is a really bad approximation } \\ \text{ how about two terms } \\ \arctan(x) \approx x- \frac{x^3}{3}\] \[\arctan(0.1) \approx 0.1-\frac{0.1^3}{3} =.099\overline{6} \text{ this is an awesome approximation }\]
freckles
  • freckles
the approximation only gets better if we add more terms onto that
IrishBoy123
  • IrishBoy123
|dw:1442182723650:dw|
anonymous
  • anonymous
@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?
blackstreet23
  • blackstreet23
well i guess i will ask this one to my professor. Thanks guys for your help!
anonymous
  • anonymous
Sorry I couldn't help Im not sure what you are hung up on as far as the problem is concerned
blackstreet23
  • blackstreet23
ohh no problem! i know you did your best :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.