Alphabeta Math
TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-01
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The exponential addition formula exp⁡(x+y)=exp⁡(x)exp⁡(y)

Statement

For all real x,y, exp⁡(x+y)=exp⁡(x)exp⁡(y).

Facts & Assumptions

Given: x,y∈R.

[L1]

For fixed x,y, the auxiliary power series ∑n≥0xnzn/ι(n!) and ∑n≥0ynzn/ι(n!) have infinite radius by The exponential series converges absolutely for every real argument. Inside their common radius, their product is the Cauchy product of their coefficients (Inside the common radius the product of two power-series sums is represented by the Cauchy product of their coefficients).

[L3]

For k≤n, ι(nk)=ι(n!)/(ι(k!)ι((n−k)!)) ((nk) k! (n−k)!=n! for k≤n; hence (nk) k!=nk‾, the quotient n!/(k!(n−k)!) is a natural number, and (nk)=(nn−k)). Therefore 1/(ι(k!)ι((n−k)!))=ι(nk)/ι(n!), with all naturals read in R through The canonical natural ι(n)=n⋅1F of a field.

Proof

technique · direct
1.1

Apply [L1] at the auxiliary value z=1. The coefficient of degree n in the resulting Cauchy product for exp⁡(x)exp⁡(y) is ∑k=0nxkyn−k/(ι(k!)ι((n−k)!)).

L1given
2.1

Apply [L3] and [L2] to identify this finite sum with (x+y)n/ι(n!).

step 1.1L2L3algebra
3.1

Summing over n gives the exponential series at x+y, hence the formula.

step 2.1L1∎

Depends on

Used by

Dependency tree · two levels

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Sources