I found out that: \(\displaystyle{\tan{{\left({x}\right)}}}={x}+{\frac{{{x}^{{3}}}}{{{3}}}}+{\frac{{{2}{x}^{{5}}}}{{{15}}}}+{O}{\left({x}^{{7}}\right)}\) If I wanted to

fsrlaihq55

fsrlaihq55

Answered question

2022-03-30

I found out that:
tan(x)=x+x33+2x515+O(x7)
If I wanted to find a value of tan(x), how would I use it to find a value for x?

Answer & Explanation

Alannah Campos

Alannah Campos

Beginner2022-03-31Added 10 answers

I supppose that what you mean is the following : knowing that, for small values of x we can approximate the tangent by the truncated Taylor expansion
tan(x)=x+x33+2x515+O(x7)
how could we find the value of x knowing the value of tanx=a
This is called series reversion. The idea is to write
x=n=1kAnan
and to replace. Using k=5 we then have
a=(n=1kAnan)+13(n=1kAnan)3+215(n=1kAnan)5
and expand, ignoring the terms of order higher than 5. This gives
a=aA1+a2A2+a3(A133+A3)+a4(A2A12+A4)+a5(2A1515+A3A12+A22A1+A5)+
Now, by identification, we then have
A1=1A2=0A133+A3=0A2A12+A4=02A1515+A3A12+A22A1+A5=0
from which
A1=1A2=0A3=13A4=0A5=15
meaning, by the end
x=tan(x)13tan3(x)+15tan5(x)+O(tan7(x))
Try using tanx=13 ; this will give
x=414530.52603
while π60.523599

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?