本文主要是介绍数字信号处理数学基础，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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泰勒级数
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + f ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + ⋯ + f ( n ) ( x 0 ) n ! ( x − x 0 ) n + R n ( x ) f\left( x \right)=f\left( {{x}_{0}} \right)+{{f}^{'}}\left( {{x}_{0}} \right)\left( x-{{x}_{0}} \right)+\frac{{{f}^{''}}\left( {{x}_{0}} \right)}{2!}\left( x-{{x}_{0}} \right)^2+\cdots +\frac{{{f}^{\left( n \right)}}\left( {{x}_{0}} \right)}{n!}{{\left( x-{{x}_{0}} \right)}^{n}}+{{R}_{n}}\left( x \right) f(x)=f(x0)+f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+⋯+n!f(n)(x0)(x−x0)n+Rn(x)
其中 R n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − x 0 ) n + 1 {{R}_{n}}\left( x \right)=\frac{{{f}^{\left( n+1 \right)}}\left( \xi \right)}{\left( n+1 \right)!}{{\left( x-{{x}_{0}} \right)}^{n+1}} Rn(x)=(n+1)!f(n+1)(ξ)(x−x0)n+1 -
复指数函数
e z = lim n → ∞   ( 1 + z n ) n = lim n → ∞   ∑ i = 0 n 1 i ! z i {{e}^{z}}=\underset{n\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{z}{n} \right)}^{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=0}^{n}{\frac{1}{i!}}{{z}^{i}} ez=n→∞lim(1+nz)n=n→∞limi=0∑ni!1z
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