Question: "I want to find the Taylor series for..."

Papierskiix5n

Answered question

2022-04-08

I want to find the Taylor series for the function:

f (x) = \ln (\tan (x)) - \ln (x)

Answer & Explanation

tutaonana223a

Beginner2022-04-09Added 15 answers

First, we have

\ln (\tan x) - \ln x = \ln (\frac{\tan x}{x}) = \ln (\frac{\sin x}{x} \frac{1}{\cos x})

So when

x \to 0

, the term inside the logarithm will go to 1, so no problems there. From this we can also already conclude that there will be no constant term in the Taylor series, as

\ln (1 \pm ϵ) \to 0

ϵ \to 0

Next, we can use the Taylor series for

\tan (x)

to get

\frac{\tan x}{x} = 1 + \frac{1}{3} x^{2} + \frac{2}{15} x^{4} + \frac{17}{315} x^{6} + \frac{62}{2835} x^{8} + \dots

Using the Taylor series for

\ln (1 + x)

then gives us

\ln (\frac{\tan x}{x}) = \ln (1 + (\frac{1}{3} x^{2} + \frac{2}{15} x^{4} + \frac{17}{315} x^{6} + \frac{62}{2835} x^{8} + \dots))

= y - \frac{1}{2} y^{2} + \frac{1}{3} y^{3} - \frac{1}{4} y^{4} + \dots

where

y = \frac{1}{3} x^{2} + \frac{2}{15} x^{4} + \frac{17}{315} x^{6} + \frac{62}{2835} x^{8} + \dots

\frac{1}{2} y^{2} = \frac{1}{18} x^{4} + \frac{2}{45} x^{6} + (\frac{17}{945} + \frac{2}{225}) x^{8} + \dots

\frac{1}{3} y^{3} = \frac{1}{81} x^{6} + \frac{2}{135} x^{8} + \dots

\frac{1}{4} y^{4} = \frac{1}{324} x^{8} + \dots

Adding it all up, we get

\ln (\tan x) - \ln x = y - \frac{1}{2} y^{2} + \frac{1}{3} y^{3} - \frac{1}{4} y^{4} + \dots

= \frac{1}{3} x^{2} + (\frac{2}{15} - \frac{1}{18}) x^{4} + (\frac{17}{315} - \frac{2}{45} + \frac{1}{81}) x^{6} + (\frac{62}{2835} - \frac{17}{945} - \frac{2}{225} + \frac{2}{135} - \frac{1}{324}) x^{8} + \dots

= \frac{1}{3} x^{2} + \frac{7}{90} x^{4} + \frac{62}{2835} x^{6} + \frac{127}{18900} x^{8} + \dots

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