恆等式

恆等式 

若函數f(x)＝x(x+2)，其中x是正整數，則f(x+1)＝(x+1)(x+3)＝(x+2)²-1，因此1+f(x+1)＝(x+2)²，x+2＝1+f(x+1)−−−−−−−−−−√，所以f(x)＝x1+f(x+1)−−−−−−−−−−√。
使用函數f(x)＝x1+f(x+1)−−−−−−−−−−√迭代計算可得f(x+1)＝(x+1)1+f(x+2)−−−−−−−−−−√，因此f(x)＝x1+(x+1)1+f(x+2)−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−√。
因為f(x+2)＝(x+2)1+f(x+3)−−−−−−−−−−√，所以
f(x)＝x1+(x+1)1+(x+2)1+f(x+3)−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√。
因為f(x+3)＝(x+3)1+f(x+4)−−−−−−−−−−√，所以
f(x)＝x1+(x+1)1+(x+2)1+(x+3)1+f(x+4)−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√。
如此迭代計算下去，可得
f(x)＝x1+(x+1)1+(x+2)1+(x+3)1+(x+4)...1+(x+n−1)1+f(x+n)−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−⎷−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−⎷−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−⎷，其中n是趨向無限大的正整數。

當上列恆等式中的x＝1時，f(1)＝1×3＝3

3＝f(1)＝1×1+f(2)−−−−−−−√＝1×1+2×4−−−−−−−√

3＝f(1)＝1×1+21+f(3)−−−−−−−√−−−−−−−−−−−−−√＝1×1+21+3×5−−−−−−−√−−−−−−−−−−−−−√

3＝f(1)＝1×1+21+31+f(4)−−−−−−−√−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−√＝1×1+21+31+4×6−−−−−−−√−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−√

3＝f(1)＝1×1+21+31+41+f(5)−−−−−−−√−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−√＝1×1+21+31+41+5×7−−−−−−−√−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−−√

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20世紀印度數學英才拉馬努金(Ramanujan)，擁有非常人的數感，具有前瞻性的洞察力，在其數學成就中就包含約4000個恆等式，但是有一些恆等式雖有結論，卻沒有留下詳細的推論過程。
本文的恆等式就是由拉馬努金(Ramanujan)所提出的。