Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-03
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Principal arctangent: derivative, integral, power series, and the Gregory–Leibniz series

Statement

For every xRx\in\mathbb R,

ddxarctanx=11+x2,arctanx=0xdt1+t2.\frac{d}{dx}\arctan x=\frac1{1+x^2},\qquad \arctan x=\int_0^x\frac{dt}{1+t^2}.

For x<1|x|<1,

arctanx=n=0(1)nx2n+12n+1.\arctan x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}.

At the endpoint, the ordinarily convergent alternating series satisfies

π4=113+1517+.\frac\pi4=1-\frac13+\frac15-\frac17+\cdots.

Facts & Assumptions

Given: No hypotheses beyond those quantified in the statement.

[L1]

Principal arctangent is the continuous increasing inverse of tangent on (π/2,π/2)(-\pi/2,\pi/2) (The principal inverse tangent arctan:R(π/2,π/2)\arctan:\mathbb R\to(-\pi/2,\pi/2)).

[L3]

(tanu)=sec2u=1+tan2u(\tan u)'=\sec^2u=1+\tan^2u on the tangent domain, and tan0=0\tan0=0 because sin0=0\sin0=0, cos0=1\cos0=1, and tan0=sin0/cos0\tan0=\sin0/\cos0 (Tangent, cotangent, secant, and cosecant on their exact natural domains, The derivatives of sine and cosine are cosine and minus sine, Derivatives and fundamental periods of tangent, cotangent, secant, and cosecant, Pythagorean and parity identities for all six trigonometric functions on their natural domains).

[L5]

For r<1|r|<1, n0rn=1/(1r)\sum_{n\ge0}r^n=1/(1-r), and a real power series may be integrated termwise on compact subintervals of its convergence interval (For r<1|r| < 1, k0rk=1/(1r)\sum_{k \ge 0} r^k = 1/(1-r), and for r1|r| \ge 1 the series diverges, Inside its radius a real power series may be integrated term by term on every closed subinterval).

Proof

technique · direct
1.1

For yRy\in\mathbb R, put u:=arctanyu:=\arctan y. Then tanu=y\tan u=y and [L3] gives (tan)(u)=1+y2>0(\tan)'(u)=1+y^2>0. Applying [L2] to the principal branch proves (arctany)=1/(1+y2)(\arctan y)'=1/(1+y^2).

L1L2L3
2.1

The function t1/(1+t2)t\mapsto1/(1+t^2) is continuous. By [L4], its oriented integral from 00 to xx has derivative 1/(1+x2)1/(1+x^2) and value 00 at x=0x=0. By [L3], tan0=0\tan0=0, so the inverse identity in [L1] gives arctan0=0\arctan0=0; step 1.1 gives its derivative and [L8] makes it continuous. Their difference is therefore continuous on R\mathbb R with zero derivative, so [L8] makes it zero.

step 1.1L1L3L4L8
3.1

If t<1|t|<1, [L5] with r=t2r=-t^2 gives 11+t2=n=0(1)nt2n.\frac1{1+t^2}=\sum_{n=0}^{\infty}(-1)^nt^{2n}. Termwise integration between 00 and xx (reversing endpoints when x<0x<0) and step 2.1 give the asserted arctangent series for x<1|x|<1.

step 2.1L5
4.1

Let S:=n0(1)n/(2n+1)S:=\sum_{n\ge0}(-1)^n/(2n+1), which exists by [L6]. Abel's theorem and step 3.1 yield S=limx1n0(1)nx2n+12n+1=limx1arctanx=arctan1.S=\lim_{x\uparrow1}\sum_{n\ge0}\frac{(-1)^nx^{2n+1}}{2n+1} =\lim_{x\uparrow1}\arctan x=\arctan1. By [L7] and the principal range, arctan1=π/4\arctan1=\pi/4.

step 3.1L1L6L7
5.1

Steps 1.1–4.1 establish all four displayed claims.

step 1.1step 2.1step 3.1step 4.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

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Sources