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anonymous

  • 5 years ago

what is the derivative of tan(1)?

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  1. Yuki
    • 5 years ago
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    0

  2. Yuki
    • 5 years ago
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    because tan(1) is a constant

  3. anonymous
    • 5 years ago
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    okay then i have to post the actual question bc i am having trouble

  4. Yuki
    • 5 years ago
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    okay, shoot it

  5. anonymous
    • 5 years ago
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    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. Yuki
    • 5 years ago
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    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
    • 5 years ago
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    i have to find the error bound using the lagrange error bound for P(x)

  8. anonymous
    • 5 years ago
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    that the formula that i typed

  9. Yuki
    • 5 years ago
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    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
    • 5 years ago
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    no not yet

  11. Yuki
    • 5 years ago
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    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
    • 5 years ago
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    okay

  13. Yuki
    • 5 years ago
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    are you comfortable finding it ?

  14. anonymous
    • 5 years ago
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    i found it i got -6( 1+ x^2)^-4

  15. anonymous
    • 5 years ago
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    oh i did it wrong i wrote down the wrong derivative

  16. anonymous
    • 5 years ago
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    i need helpp

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

  18. anonymous
    • 5 years ago
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    i did find it i got -24 sinx/(cosx)^5

  19. Yuki
    • 5 years ago
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    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
    • 5 years ago
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    so what do i do??

  21. Yuki
    • 5 years ago
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    give me one second

  22. anonymous
    • 5 years ago
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    ok

  23. Yuki
    • 5 years ago
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    were you given any interval ?

  24. anonymous
    • 5 years ago
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    it just says x=0 and tan (1) so 0<x<1

  25. Yuki
    • 5 years ago
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    nath lol that info was actually very important lmao

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

  27. Yuki
    • 5 years ago
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    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
    • 5 years ago
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    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. Yuki
    • 5 years ago
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    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
    • 5 years ago
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    k ty

  31. Yuki
    • 5 years ago
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    \[f = tanx\] \[f' = \sec^2x\] \[f" = 2\sec^2x tanx\] \[f^{(3)} = 4secx tanx + 2 \sec^4x\] \[f^{(4)} = 4secx \tan^2x + 4 \sec^3x + 8\sec^4tanx\]

  32. Yuki
    • 5 years ago
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    let me see if I can simplify it

  33. anonymous
    • 5 years ago
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    kk

  34. Yuki
    • 5 years ago
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    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. Yuki
    • 5 years ago
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    so M would be 2sin^2(2)+4cos^2(1) + 8sin(1) over cos(1)

  36. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    okay lets move on so how do i know what to plug into x

  38. Yuki
    • 5 years ago
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    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. Yuki
    • 5 years ago
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    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
    • 5 years ago
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    so for this problem what my max varies for cos and and sin

  41. anonymous
    • 5 years ago
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    bc their max is different

  42. Yuki
    • 5 years ago
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    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. Yuki
    • 5 years ago
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    ops nvm

  44. Yuki
    • 5 years ago
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    i couldn't get a nice number ...

  45. Yuki
    • 5 years ago
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    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
    • 5 years ago
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    alright ty u very much

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