How do you use a Taylor polynomial to estimate #tan(x)# around #x=0# for #n=0, 1, 2#?

1 Answer
Truong-Son N.
Jun 5, 2017

Assuming you already know what the Maclaurin polynomial for #tanx# is (i.e. the Taylor series around #a = 0# truncated at a finite #n# number of terms):

#tanx = sum_(n=0)^(N) (f^((n))(0))/(n!) x^n#

#= x + 1/3x^3 + 2/15x^5 + 17/315 x^7 + . . . #

To estimate #tanx# for any #x# close enough to #bb(a = 0)#...

For #n = 0#, that corresponds to the #(n=0)#th order term, #f^((0))(0) = f(0)#, which is just #tan(0) = 0#. So, we say that the #0#th order term is small near #a = 0#.

To estimate it up to the #n = 1# term, we have:

#tanx ~~ cancel(f^((0))(x))^"small" + f'(x)#

#~~ x#

To estimate it up to the #n = 2# term, we have:

#tanx ~~ cancel(f^((0))(x))^"small" + f'(x) + f''(x)#

#~~ x + 1/3 x^3#

For example,

#tan("0.05 rad") ~~ color(blue)(0.0500417)#

whereas truncated at #n = 1# we have

#0.05 = color(blue)(0.0500000) < tanx#

and truncated at #n = 2# we have:

#0.05 + 1/3(0.05)^3 = color(blue)(0.050041bar(66)) ~~ tanx#

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