It's strange how arbitrary the numerical representation of π is. Compare 3.14159265358979... (equally random in any other base) to the platonic perfection - and inevitability - of the circle and its diameter.
Surely there is a compelling series for π which more clearly demonstrates its geometrical elegance?
The Leibniz formula is very elegant:
The Leibniz series for π is expressed as:
π ≈ 4 × (1 - 1/3 + 1/5 - 1/7 + ...)
The first four terms of the series are:
π ≈ 4 × (1 - 1/3 + 1/5 - 1/7) = 4 × 0.72380952 ≈ 2.8952381
but it comes out of the Taylor series for arctan. There seems no intuitive geometrical picture of how this ends up as π. There are many other series which converge to π: many of them have striking patterns. But the term-by-term connection to geometry is never there.
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An elementary mathematical identity is that sin2(θ)
+ cos2(θ)
= 1. I asked ChatGPT (which can now do maths) to show me how this comes out of the Taylor series for sin2(θ)
and cos2(θ)
separately.
Question:
What is the Taylor series for
sin2(θ)
andcos2(θ)
separately? How does their sum turn out to be invariantly one?
Answer:
1. Taylor Series for
sin2(θ)
The Taylor series for
sin(θ)
is:
sin(θ) = θ - θ3/3! + θ5/5! - θ7/7! + ...
Squaring this series gives:
sin2(θ) = (θ - θ3/3! + θ5/5! - ... )2
Which results in:
sin2(θ) = θ2 - θ4/3 + 2θ6/45 - ...
2. Taylor Series for
cos2(θ)
The Taylor series for
cos(θ)
is:
cos(θ) = 1 - θ2/2! + θ4/4! - θ6/6! + ...
Squaring this series gives:
cos2(θ) = (1 - θ2/2! + θ4/4! - ... )2
Which results in:
cos2(θ) = 1 - θ2 + θ4/3 - 2θ6/45 + ...
3. Adding Them Together:
sin2(θ) + cos2(θ)
Now, summing the Taylor series for
sin2(θ)
andcos2(θ)
:
sin2(θ) + cos2(θ) = (θ2 - θ4/3 + 2θ6/45 - ...) + (1 - θ2 + θ4/3 - 2θ6/45 + ...)
When you add the terms:
- The
θ2
terms cancel out. - The
θ4
terms cancel out. - The
θ6
terms cancel out. - And so on for all higher-order terms.
Thus, all terms cancel out except the constant 1:
sin2(θ) + cos2(θ) = 1
This shows how the sum of the Taylor series results in the invariant identity
sin2(θ) + cos2(θ) = 1
, independent of the value ofθ
.Credit: ChatGPT
I wish I'd seen this when I was fifteen.
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