感知機收斂定理：
X%5Cleft&space;(&space;n&space;%5Cright&space;)%5Cleq&space;0 "W^{T}\left ( n \right )X\left ( n \right )\leq 0")
，且輸入向量屬於子集 
，顯然，迭代過程中，感知機不能正確對X(1)，X(2)...進行正確分類，在η(n)=1的情況下，權值每次迭代都需要修正：=W%5Cleft&space;(&space;n&space;%5Cright&space;)+X%5Cleft&space;(&space;n&space;%5Cright&space;)&space;%5Cquad&space;X%5Cleft&space;(&space;n&space;%5Cright&space;)&space;%5Cin&space;%5Cmathbb%7BC%7D_%7B1%7D "W\left ( n+1 \right )=W\left ( n \right )+X\left ( n \right ) \quad X\left ( n \right ) \in \mathbb{C}_{1}")
（1.1）又由於W(0)=0，故
=X%5Cleft&space;(&space;1&space;%5Cright&space;)+X%5Cleft&space;(&space;2&space;%5Cright&space;)+%5Ccdots&space;+X%5Cleft&space;(&space;n&space;%5Cright&space;) "W\left ( n+1 \right )=X\left ( 1 \right )+X\left ( 2 \right )+\cdots +X\left ( n \right )")
假設 
 "W^{T}X\left ( n \right )>0")
,X%5Cleft&space;(&space;n&space;%5Cright&space;)%5Cin&space;%5Cmathbb%7BH%7D_%7B1%7D "\alpha =min\ W_{0}^{T}X\left ( n \right ),X\left ( n \right )\in \mathbb{H}_{1}")
在（1.2）式左右兩邊同乘以
=W_%7B0%7D%5E%7BT%7DX%5Cleft&space;(&space;1&space;%5Cright&space;)+W_%7B0%7D%5E%7BT%7DX%5Cleft&space;(&space;2&space;%5Cright&space;)+%5Ccdots&space;+W_%7B0%7D%5E%7BT%7DX%5Cleft&space;(&space;n&space;%5Cright&space;) "W_{0}^{T}W\left ( n+1 \right )=W_{0}^{T}X\left ( 1 \right )+W_{0}^{T}X\left ( 2 \right )+\cdots +W_{0}^{T}X\left ( n \right )")
根據（1.3）得
%5Cgeq&space;n%5Calpha "W_{0}^{T}W\left ( n+1 \right )\geq n\alpha")
根據Cauchy-Schwarz不等式，得
&space;%5Cright&space;%5C%7C%5E%7B2%7D%5Cgeq&space;%5Cleft&space;%5B&space;W_%7B0%7D%5E%7BT%7DW%5Cleft&space;(&space;n+1&space;%5Cright&space;)&space;%5Cright&space;%5D%5E%7B2%7D "\left \| W_{0}^{T} \right \|^{2}\left \| W\left ( n+1 \right ) \right \|^{2}\geq \left [ W_{0}^{T}W\left ( n+1 \right ) \right ]^{2}")
結合（1.5）得
&space;%5Cright&space;%5C%7C%5E%7B2%7D%5Cgeq&space;n%5E%7B2%7D%5Calpha&space;%5E%7B2%7D "\left \| W_{0}^{T} \right \|^{2}\left \| W\left ( n+1 \right ) \right \|^{2}\geq n^{2}\alpha ^{2}")
即
&space;%5Cright&space;%5C%7C%5E%7B2%7D%5Cgeq&space;%5Cfrac%7Bn%5E%7B2%7D%5Calpha&space;%5E%7B2%7D%7D%7B%5Cleft&space;%5C%7C&space;W_%7B0%7D%5E%7BT%7D&space;%5Cright&space;%5C%7C%5E%7B2%7D%7D "\left \| W\left ( n+1 \right ) \right \|^{2}\geq \frac{n^{2}\alpha ^{2}}{\left \| W_{0}^{T} \right \|^{2}}")
對（1.1）式兩邊同取歐幾里得範數的平方得：
&space;%5Cright&space;%5C%7C%5E%7B2%7D=%5Cleft&space;%5C%7C&space;W%5Cleft&space;(&space;k&space;%5Cright&space;)&space;%5Cright&space;%5C%7C%5E%7B2%7D+%5Cleft&space;%5C%7C&space;X%5Cleft&space;(&space;k&space;%5Cright&space;)&space;%5Cright&space;%5C%7C%5E%7B2%7D+2W%5E%7BT%7D%5Cleft&space;(&space;k&space;%5Cright&space;)X%5Cleft&space;(&space;k&space;%5Cright&space;)&space;%5Cquad&space;k=1,2,%5Ccdots&space;,n "\left \| W\left ( k+1 \right ) \right \|^{2}=\left \| W\left ( k \right ) \right \|^{2}+\left \| X\left ( k \right ) \right \|^{2}+2W^{T}\left ( k \right )X\left ( k \right ) \quad k=1,2,\cdots ,n")
但是
X%5Cleft&space;(&space;k&space;%5Cright&space;)&space;%5Cleq&space;0 "W^{T}\left ( k \right )X\left ( k \right ) \leq 0")
&space;%5Cright&space;%5C%7C%5E%7B2%7D%5Cleq&space;%5Cleft&space;%5C%7C&space;W%5Cleft&space;(&space;k&space;%5Cright&space;)&space;%5Cright&space;%5C%7C%5E%7B2%7D+%5Cleft&space;%5C%7C&space;X%5Cleft&space;(&space;k&space;%5Cright&space;)&space;%5Cright&space;%5C%7C%5E%7B2%7D "\left \| W\left ( k+1 \right ) \right \|^{2}\leq \left \| W\left ( k \right ) \right \|^{2}+\left \| X\left ( k \right ) \right \|^{2}")
等價於
&space;%5Cright&space;%5C%7C%5E%7B2%7D-&space;%5Cleft&space;%5C%7C&space;W%5Cleft&space;(&space;k&space;%5Cright&space;)&space;%5Cright&space;%5C%7C%5E%7B2%7D%5Cleq&space;%5Cleft&space;%5C%7C&space;X%5Cleft&space;(&space;k&space;%5Cright&space;)&space;%5Cright&space;%5C%7C%5E%7B2%7D "\left \| W\left ( k+1 \right ) \right \|^{2}- \left \| W\left ( k \right ) \right \|^{2}\leq \left \| X\left ( k \right ) \right \|^{2}")
把k=1,2,...,n的情況都相加得：
&space;%5Cright&space;%5C%7C%5E%7B2%7D%5Cleq&space;%5Csum_%7Bk=1%7D%5E%7Bn%7D%5Cleft&space;%5C%7C&space;X%5Cleft&space;(&space;k&space;%5Cright&space;)&space;%5Cright&space;%5C%7C%5E%7B2%7D%5Cleq&space;n%5Cbeta "\left \| W\left ( n+1 \right ) \right \|^{2}\leq \sum_{k=1}^{n}\left \| X\left ( k \right ) \right \|^{2}\leq n\beta")
其中
&space;%5Cright&space;%5C%7C%5E%7B2%7D,X%5Cleft&space;(&space;k&space;%5Cright&space;)%5Cin&space;%5Cmathbb%7BH%7D_%7B1%7D "\beta =max\ \left \| X\left ( k \right ) \right \|^{2},X\left ( k \right )\in \mathbb{H}_{1}")
綜合（1.8）（1.12）可得
不等式右邊是平方項，顯然這個不等式不可能永遠成立，n的最大取值即爲取等號的時候。
那麼感知機的權值迭代過程最多在nmax停止。
現在概述一下感知機的固定增量收斂定理：
設訓練向量子集是線性可分的，感知機的輸入來自這兩個子集。感知機在有限步迭代後收斂。
感知機經典習題：
參考文獻：神經網絡與機器學習（加）S.Haykin著 申富饒 徐燁 鄭俊 晁靜 翻譯
Neural Networks and Learning Machines S.Haykin