Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21
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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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An entire function of polynomial growth is a polynomial

Statement

Let f:C→C be entire. Suppose there are real numbers C,N≥0 such that

∣f(z)∣≤C(1+∣z∣)N(z∈C),

where real powers have the convention of Real powers for positive bases, with the zero-base positive-exponent convention. Put m=⌊N⌋. Then there are complex coefficients c0,…,cm such that

f(z)=∑k=0mckzk(z∈C).

Thus f is a polynomial, and if it is nonzero its degree is at most ⌊N⌋.

Facts & Assumptions

Given: An entire function f and real constants C,N≥0 satisfying the displayed growth bound.

[L1]

If f is holomorphic on D(a,R0), 0<r<R0, and ∣f(ζ)∣≤M on ∣ζ−a∣=r, then the nth Taylor coefficient cn satisfies ∣cn∣≤M/rn (Cauchy's inequalities bound the Taylor coefficients by the circle supremum).

[L2]

For a>0 and real x, the real power is ax=exp⁡(xlog⁡a); zero-base powers are defined only for positive exponents (Real powers for positive bases, with the zero-base positive-exponent convention).

[L3]

Positive-base real powers satisfy ar+s=aras and (ab)r=arbr (The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents).

[L4]

The natural logarithm is the inverse of the exponential on the positive reals (The natural logarithm as the inverse of the exponential function).

[L5]

The exponential tends to 0 at −∞ and to +∞ at +∞ (The exponential tends to +∞ at +∞ and to 0 at −∞).

[L6]

Every entire function equals its Taylor series at the origin on the whole complex plane (A holomorphic function equals its Taylor series throughout the largest centred disc in its domain).

[L7]

For every real x there is a unique integer ⌊x⌋ satisfying ⌊x⌋≤x<⌊x⌋+1 (Integer part: for every real x there is exactly one integer m with m≤x<m+1).

[L8]

The exponential function is strictly increasing on R (The exponential function is strictly increasing).

Proof

technique · direct
1.1givenL1

Let f(z)=∑n≥0cnzn be its Taylor series at 0, fix a natural n>N, and take any real R≥1; the growth hypothesis bounds ∣f∣ on ∣ζ∣=R by C(1+R)N, so [L1], applied with outer radius R+1, gives ∣cn∣≤C(1+R)N/Rn.

2.1step 1.1L2L3L4L5L8algebra

Since R≥1 gives 1+R≤2R, [L2], [L3], [L4], and [L8] give ∣cn∣≤C2NRN−n=C2Nexp⁡(−(n−N)log⁡R); by [L4], [L5], and [L8], log⁡R→+∞ as R→+∞, so the right side tends to 0, forcing cn=0.

3.1step 2.1L6

Step 2.1 applies to every natural n>N, and [L6] represents f globally by its Taylor series, so all terms with index exceeding N vanish and the series truncates.

4.1step 3.1L7

Put m=⌊N⌋. By [L7], m≤N<m+1, so every natural n≥m+1 satisfies n>N and has cn=0 by step 3.1; because N≥0, the integer m is a natural number, and the displayed finite polynomial has no term above m.

5.1step 3.1step 4.1algebra∎

If C=0, the hypothesis gives f=0 directly; if N=0, step 4.1 gives m=0 and f is constant. In every case step 4.1 proves the stated polynomial representation, with the degree qualification interpreted only for a nonzero polynomial.

Depends on

Used by

Dependency tree · two levels

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Sources