What is expand (4e^{2-2t}+1)(e^{j^2pint}) ?

The solution to expand (4e^{2-2t}+1)(e^{j^2pint}) is 4e^{2-2t+pij^2nt}+e^{pij^2nt}

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expand (4e^{2-2t}+1)(e^{j^2pint})

expand

(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}

Solution steps

(4 e^{2 - 2 t} + 1) (e^{j^{2} πn t})

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

Apply the distributive law:

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})

Simplify

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}

\frac{x ^{2}}{( x + 2 ) ( x - 1 )}

(2 y - x^{3}) d x + x d y

sin (π y) + (x - 1)^{2}

ln (x) (x^{2} + 1)^{2}

\frac{( 4 x + 1 )}{( x + 2 ) ^{2}}

What is expand (4e^{2-2t}+1)(e^{j^2pint}) ?
The solution to expand (4e^{2-2t}+1)(e^{j^2pint}) is 4e^{2-2t+pij^2nt}+e^{pij^2nt}