The Arctan or Arctangent is the inverse of Tangent, which is a ratio, generally referring to the degrees of angles, when of course, x is a Real number.
Therefore, arctan x can be many things depending upon the value of x. Let’s explore some examples,
If,
tan y = x
Then the arctangent of x is equal to the inverse tangent function of x, which is equal to y:
arctan x= tan-1 x = y
Arctan(x) can be represented in the form of a McLaurin series.
For Example,
Recall the formula for sum of a geometric series (with a common ratio rr such that |r|<1|r|<1) :
∑n = 0 ∞ φn = 11 −φ ∑n = 0 ∞ φn = 11−φ
Let φ=−t2φ=−t2.
⟹∑n = 0 ∞ (−t2) n = 11 + t2
⟹∑n = 0 ∞ (−t2) n = 11+ t2
(−t2) n (−t2) n can be rewritten as (−1) nt2n (−1) nt2n.
Hence,
∑n = 0 ∞ (−1) nt2n = 11 + t2∑n = 0 ∞ (−1) nt2n =11+ t2
Take the integral of both sides over [0, x] [0, x]:
∑n = 0 ∞ ∫0x (−1) nt2ndt = ∫0x11 + t2dt ∑n = 0 ∞ ∫0x (−1) nt2ndt = ∫0x11 + t2dt
The RHS is just a standard Integral of arctan(x)
arctan(x):
Therefore,
∑n = 0 ∞ ∫0x (−1) nt2ndt = arctan(x)
Therefore,
∑n= 0 ∞ ∫0x (−1) nt2ndt = arctan(x)
Integrating the LHS:
∑n = 0 ∞ (−1) n (t2n+12n+1∣∣∣x0) = arctan(x)
Therefore,
arctan(x) = ∑n= 0 ∞ (−1) nx2n + 12n + 1
Derivative of arctan(x):
Let’s use our formula for the derivative of an inverse function to find the derivative of the inverse of the tangent function:
y = tan−1
x = arctan x.
We simplify the equation by taking the tangent of both sides:
y = tan−1 x
tan y = tan (tan−1 x)
Therefore,
tan y = x
Graphical Presentation of Arctan(x)
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