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

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Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0because tan(1) is a constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay then i have to post the actual question bc i am having trouble

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use this formula E_{n}(x)\le\frac{M}{(n+1)!}xa^{{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

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0I 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 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have to find the error bound using the lagrange error bound for P(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that the formula that i typed

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0okay, 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) ?

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0to 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.

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0are you comfortable finding it ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i found it i got 6( 1+ x^2)^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i did it wrong i wrote down the wrong derivative

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0the derivative of tan(x) = sec^2(x) which makes it really hard to find te 4th term. take your time

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i did find it i got 24 sinx/(cosx)^5

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0now 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.

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0were you given any interval ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it just says x=0 and tan (1) so 0<x<1

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0nath lol that info was actually very important lmao

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0since x is between 0 and 1, we just need to find the upper bound of 24 sin/cos^5 between those intervals

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0cosine 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is there anyway you could do it really quickly just to make sure its correct cause we are make an effort to do the problem

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0I 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)

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0\[f = tanx\] \[f' = \sec^2x\] \[f" = 2\sec^2x tanx\] \[f^{(3)} = 4secx tanx + 2 \sec^4x\] \[f^{(4)} = 4secx \tan^2x + 4 \sec^3x + 8\sec^4tanx\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0let me see if I can simplify it

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0I think this one looks best among others \[f^{(4)} = {2\sin^2(2x) +4\cos^2(x) +8\sin(x) \over \cos^5(x)}\]

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0so M would be 2sin^2(2)+4cos^2(1) + 8sin(1) over cos(1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the thing is i said the derivative of tan x is 1/cos^2 x and i got something completely different from you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay lets move on so how do i know what to plug into x

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0let'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 ...

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0basically 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so for this problem what my max varies for cos and and sin

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bc their max is different

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0so 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

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0i couldn't get a nice number ...

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0anyway, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright ty u very much
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