anonymous
  • anonymous
what is the derivative of tan(1)?
Mathematics
chestercat
  • chestercat
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yuki
  • yuki
0
yuki
  • yuki
because tan(1) is a constant
anonymous
  • anonymous
okay then i have to post the actual question bc i am having trouble

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yuki
  • yuki
okay, shoot it
anonymous
  • 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
yuki
  • yuki
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 ?
anonymous
  • anonymous
i have to find the error bound using the lagrange error bound for P(x)
anonymous
  • anonymous
that the formula that i typed
yuki
  • yuki
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) ?
anonymous
  • anonymous
no not yet
yuki
  • yuki
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.
anonymous
  • anonymous
okay
yuki
  • yuki
are you comfortable finding it ?
anonymous
  • anonymous
i found it i got -6( 1+ x^2)^-4
anonymous
  • anonymous
oh i did it wrong i wrote down the wrong derivative
anonymous
  • anonymous
i need helpp
yuki
  • yuki
the derivative of tan(x) = sec^2(x) which makes it really hard to find te 4th term. take your time
anonymous
  • anonymous
i did find it i got -24 sinx/(cosx)^5
yuki
  • yuki
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.
anonymous
  • anonymous
so what do i do??
yuki
  • yuki
give me one second
anonymous
  • anonymous
ok
yuki
  • yuki
were you given any interval ?
anonymous
  • anonymous
it just says x=0 and tan (1) so 0
yuki
  • yuki
nath lol that info was actually very important lmao
yuki
  • yuki
since x is between 0 and 1, we just need to find the upper bound of -24 sin/cos^5 between those intervals
yuki
  • yuki
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
anonymous
  • 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
yuki
  • yuki
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)
anonymous
  • anonymous
k ty
yuki
  • yuki
\[f = tanx\] \[f' = \sec^2x\] \[f" = 2\sec^2x tanx\] \[f^{(3)} = 4secx tanx + 2 \sec^4x\] \[f^{(4)} = 4secx \tan^2x + 4 \sec^3x + 8\sec^4tanx\]
yuki
  • yuki
let me see if I can simplify it
anonymous
  • anonymous
kk
yuki
  • yuki
I think this one looks best among others \[f^{(4)} = {2\sin^2(2x) +4\cos^2(x) +8\sin(x) \over \cos^5(x)}\]
yuki
  • yuki
so M would be 2sin^2(2)+4cos^2(1) + 8sin(1) over cos(1)
anonymous
  • anonymous
the thing is i said the derivative of tan x is 1/cos^2 x and i got something completely different from you
anonymous
  • anonymous
okay lets move on so how do i know what to plug into x
yuki
  • yuki
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 ...
yuki
  • yuki
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
anonymous
  • anonymous
so for this problem what my max varies for cos and and sin
anonymous
  • anonymous
bc their max is different
yuki
  • yuki
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
yuki
  • yuki
ops nvm
yuki
  • yuki
i couldn't get a nice number ...
yuki
  • yuki
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.
anonymous
  • anonymous
alright ty u very much

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