## blackstreet23 one year ago Approximate tan inverse (0.1) to three decimal place accuracy using the Maclaurin series tan inverse x

1. blackstreet23

@PlasmaFuzer

2. blackstreet23

@ganeshie8

3. blackstreet23

@pooja195

4. blackstreet23

@Shalante

5. blackstreet23

6. blackstreet23

I did this so far The book says that the error to 3 decimal places is $(0.1)- \frac{ 0.1^3 }{ 3 }$

7. 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

8. blackstreet23

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

9. blackstreet23

@Hero

10. IrishBoy123

the question asks you to do it using the maclaurin for cos x is that what you did?

11. blackstreet23

yes

12. IrishBoy123

|dw:1442179486992:dw| terrible graphics i know but it seems you plugged your numbers into the expansion for$$\tan^{-1}x$$.

13. 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)

14. blackstreet23

yeah i did.

15. freckles

but it really does say cos(x)?

16. blackstreet23

tan inverse (0.1)

17. freckles

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

18. blackstreet23

ohhh I see

19. blackstreet23

my bad the one i copied was wrong

20. blackstreet23

but the question here is right

21. 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 $0.1-\frac{0.1^3}{3} \text{ is a pretty good approximation for } \arctan(0.1) \\ 0.1-\frac{0.1^3}{3} \approx 0.099\overline{6} \\ \text{ while } \arctan(0.1) \text{ to 6 decimal spots is } 0.09966$

22. blackstreet23

Why is the the answer they have on the back of the book is (0.1)−(0.1^3/3)

23. blackstreet23

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

24. blackstreet23

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

25. freckles

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

26. freckles

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

27. 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

28. blackstreet23

so i just simplified to 5*10^-4

29. blackstreet23

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

30. blackstreet23

@PlasmaFuzer

31. 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 $\arctan(x)=\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{2k+1} \\ \arctan(x)=(-1)^0 \frac{x^{2(0)+1}}{2(0)+1}+(-1)^{1}\frac{x^{2(1)+1}}{2(1)+1}+(-1)^{2}\frac{x^{2(2)+1}}{2(2)+1} \\ + \cdots +(-1)^k \frac{x^{2k+1}}{2k+1}+ \cdots \\ \text{ so let's see if one term is enough } \\ \arctan(x) \approx x \\ \arctan(0.1) \approx 0.1 \text{ this is a really bad approximation } \\ \text{ how about two terms } \\ \arctan(x) \approx x- \frac{x^3}{3}$ $\arctan(0.1) \approx 0.1-\frac{0.1^3}{3} =.099\overline{6} \text{ this is an awesome approximation }$

32. freckles

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

33. IrishBoy123

|dw:1442182723650:dw|

34. 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?

35. blackstreet23

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

36. anonymous

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

37. blackstreet23

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