## anonymous 5 years ago what is the derivative of tan(1)?

1. anonymous

0

2. anonymous

because tan(1) is a constant

3. anonymous

okay then i have to post the actual question bc i am having trouble

4. anonymous

okay, shoot it

5. anonymous

use this formula |E_{n}(x)|\le\frac{M}{(n+1)!}|x-a|^{{n+1}} find the bound for the error in approximating the quantity with a third degree polynomial for tan 1, f(x) =tanx about x=0

6. anonymous

I have no idea what you plugged in lol are you asking us to find what tan(1) is using the McLaurin series with degree 3 ?

7. anonymous

i have to find the error bound using the lagrange error bound for P(x)

8. anonymous

that the formula that i typed

9. anonymous

okay, then let's do it step by step. did you find the Taylor (in our case McLaurin because the center is at x=0) Series of tan (x) ?

10. anonymous

no not yet

11. anonymous

to use the Lagrange error bound you need the (n+1)th term in the Taylor polynomial for tan(x). since we are approximating tan(x) by a 3rd degree, we need the 4th term of the Taylor polynomial.

12. anonymous

okay

13. anonymous

are you comfortable finding it ?

14. anonymous

i found it i got -6( 1+ x^2)^-4

15. anonymous

oh i did it wrong i wrote down the wrong derivative

16. anonymous

i need helpp

17. anonymous

the derivative of tan(x) = sec^2(x) which makes it really hard to find te 4th term. take your time

18. anonymous

i did find it i got -24 sinx/(cosx)^5

19. anonymous

now we need to find the upper bound of that term. the problem here is that -24sin(x)/(cosx)^5 has no bound because when cos(x) is close to 0, this number becomes arbitrarily big.

20. anonymous

so what do i do??

21. anonymous

give me one second

22. anonymous

ok

23. anonymous

were you given any interval ?

24. anonymous

it just says x=0 and tan (1) so 0<x<1

25. anonymous

nath lol that info was actually very important lmao

26. anonymous

since x is between 0 and 1, we just need to find the upper bound of -24 sin/cos^5 between those intervals

27. anonymous

cosine decreases as x increases from 0 to 1 (radians) and sine increases as x increases from 0 to 1 so the larges number M can be is found by plugging in x =1 into the fourth derivative of tan(x) just to let you know I am not really sure if it is -24sin(x)/cos^5(x), but if you are 100% sure we can proceed

28. anonymous

is there anyway you could do it really quickly just to make sure its correct cause we are make an effort to do the problem

29. anonymous

I will try to find it, then. try to trust me for the derivative. even if we don't get the right answer the concept is solid. give me a minute to find f(^4)(x)

30. anonymous

k ty

31. anonymous

$f = tanx$ $f' = \sec^2x$ $f" = 2\sec^2x tanx$ $f^{(3)} = 4secx tanx + 2 \sec^4x$ $f^{(4)} = 4secx \tan^2x + 4 \sec^3x + 8\sec^4tanx$

32. anonymous

let me see if I can simplify it

33. anonymous

kk

34. anonymous

I think this one looks best among others $f^{(4)} = {2\sin^2(2x) +4\cos^2(x) +8\sin(x) \over \cos^5(x)}$

35. anonymous

so M would be 2sin^2(2)+4cos^2(1) + 8sin(1) over cos(1)

36. anonymous

the thing is i said the derivative of tan x is 1/cos^2 x and i got something completely different from you

37. anonymous

okay lets move on so how do i know what to plug into x

38. anonymous

let's try that as well $f' = { 1 \over \cos^2(x)}$ $f" = {2\sin(x) \over \cos^4(x)}$ which simplyfies to 2sec^2(x)tan(x) $f^{(3)} = {2\cos^4(x) +6\cos^2(x)\sin^2(x) \over \cos^6(x)}$ I don't think I really want to do the next one ...

39. anonymous

basically your idea is "what is the maximum value of f^(n+1) in the given interval ?" if you are using a Taylor series with center a to apprx, tan(2), the interval is from a < x < 2 if they want you to evaluate tan(x) from -3 < x < 5 for example, you will pick the x that maximizes f^(n+1) between -3 and 5

40. anonymous

so for this problem what my max varies for cos and and sin

41. anonymous

bc their max is different

42. anonymous

so if we used the f^(4) I got, what you can say is I cannot calculate sin(1) or cos(1) , but I know that 1 radian is less than 60 degrees and greater than 0. to maximize the expression that I have, I need to make the numerator as small as possible and the denominator as large as possible. so since 0<sin(1)<sin(60), 0<cos(1)<cos(60) I will get the following

43. anonymous

ops nvm

44. anonymous

i couldn't get a nice number ...

45. anonymous

anyway, I don't feel like I was of much help, but the Idea is this. once you get a valid M, you start solving for n. since it will involve factorials, all you have to do is to plug in a couple of integers to see when it gives you a true statement. since tangent is not an easy function to calculate the derivatives, I am not 100% sure with my calculations either.

46. anonymous

alright ty u very much