Question

what is the derivative of tan(1)?

Answer 1

0

Answer 2

because tan(1) is a constant

Answer 3

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

Answer 4

okay, shoot it

Answer 5

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

Answer 6

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 ?

Answer 7

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

Answer 8

that the formula that i typed

Answer 9

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) ?

Answer 10

no not yet

Answer 11

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.

Answer 12

okay

Answer 13

are you comfortable finding it ?

Answer 14

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

Answer 15

oh i did it wrong i wrote down the wrong derivative

Answer 16

i need helpp

Answer 17

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

Answer 18

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

Answer 19

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.

Answer 20

so what do i do??

Answer 21

give me one second

Answer 22

ok

Answer 23

were you given any interval ?

Answer 24

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

Answer 25

nath lol
that info was actually very important lmao

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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)

Answer 30

k ty

Answer 31

\[f = tanx\]

\[f' = \sec^2x\]

\[f" = 2\sec^2x tanx\]

\[f^{(3)} = 4secx tanx + 2 \sec^4x\]

\[f^{(4)} = 4secx \tan^2x + 4 \sec^3x + 8\sec^4tanx\]

Answer 32

let me see if I can simplify it

Answer 33

kk

Answer 34

I think this one looks best among others

\[f^{(4)} = {2\sin^2(2x) +4\cos^2(x) +8\sin(x) \over \cos^5(x)}\]

Answer 35

so M would be

2sin^2(2)+4cos^2(1) + 8sin(1) over cos(1)

Answer 36

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

Answer 37

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

Answer 38

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 ...

Answer 39

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

Answer 40

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

Answer 41

bc their max is different

Answer 42

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

Answer 43

ops nvm

Answer 44

i couldn't get a nice number ...

Answer 45

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.

Answer 46

alright ty u very much