展開する (4e^{2-2t}+1)(e^{j^2pint})

解

展開する

(4 e^{2 - 2 t} + 1) (e^{j^{2} πn t})

4 e^{2 - 2 t + π j^{2} n t} + e^{π j^{2} n t}

解答ステップ

(4 e^{2 - 2 t} + 1) (e^{j^{2} πn t})

= (e^{π j^{2} n t}) (4 e^{2 - 2 t} + 1)

分配法則を適用する:

a (b + c) = ab + a c

a = (e^{j^{2} πn t}), b = 4 e^{2 - 2 t}, c = 1

= (e^{j^{2} πn t}) \cdot 4 e^{2 - 2 t} + (e^{j^{2} πn t}) \cdot 1

= 4 e^{2 - 2 t} (e^{π j^{2} n t}) + 1 \cdot (e^{π j^{2} n t})

簡素化

4 e^{2 - 2 t} (e^{π j^{2} n t}) + 1 \cdot (e^{π j^{2} n t}) : 4 e^{2 - 2 t + π j^{2} n t} + e^{π j^{2} n t}

= 4 e^{2 - 2 t + π j^{2} n t} + e^{π j^{2} n t}