What is expand (e^{jt}-e^{-jt})^5 ?

The solution to expand (e^{jt}-e^{-jt})^5 is e^{5jt}-5e^{3jt}+10e^{jt}-10e^{-jt}+5e^{-3jt}-e^{-5jt}

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expand (e^{jt}-e^{-jt})^5

expand

(e^{j t} - e^{- j t})^{5}

e^{5 j t} - 5 e^{3 j t} + 10 e^{j t} - 10 e^{- j t} + 5 e^{- 3 j t} - e^{- 5 j t}

Solution steps

(e^{j t} - e^{- j t})^{5}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = e^{j t}, b = - e^{- j t}

= i = 0 \sum 5 (i 5) (e^{j t})^{(5 - i)} (- e^{- j t})^{i}

Expand summation

= \frac{5 !}{0 ! ( 5 - 0 ) !} (e^{j t})^{5} (- e^{- j t})^{0} + \frac{5 !}{1 ! ( 5 - 1 ) !} (e^{j t})^{4} (- e^{- j t})^{1} + \frac{5 !}{2 ! ( 5 - 2 ) !} (e^{j t})^{3} (- e^{- j t})^{2} + \frac{5 !}{3 ! ( 5 - 3 ) !} (e^{j t})^{2} (- e^{- j t})^{3} + \frac{5 !}{4 ! ( 5 - 4 ) !} (e^{j t})^{1} (- e^{- j t})^{4} + \frac{5 !}{5 ! ( 5 - 5 ) !} (e^{j t})^{0} (- e^{- j t})^{5}

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} (e^{j t})^{5} (- e^{- j t})^{0} : e^{5 j t}

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} (e^{j t})^{4} (- e^{- j t})^{1} : - 5 e^{3 j t}

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} (e^{j t})^{3} (- e^{- j t})^{2} : 10 e^{j t}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} (e^{j t})^{2} (- e^{- j t})^{3} : - 10 e^{- j t}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} (e^{j t})^{1} (- e^{- j t})^{4} : 5 e^{- 3 j t}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} (e^{j t})^{0} (- e^{- j t})^{5} : - e^{- 5 j t}

= e^{5 j t} - 5 e^{3 j t} + 10 e^{j t} - 10 e^{- j t} + 5 e^{- 3 j t} - e^{- 5 j t}

(y^{3} + k x y^{4} - 10 x) d x + (3 x y^{2} + 29 x^{2} y^{3})

\frac{d}{d x} ((2 - x^{2}) cos (x^{2})) + 2 x sin (x^{2})

(1 + tan (x))^{3}

\frac{( ( 3 s - 1 ) ( 2 s - 1 ) )}{( s ( s ^{2} - 2 s + 14 ) )}

\frac{s - 1}{( s + 1 ) ( s + 2 ) ( s + 3 ) ( s + 4 )}

What is expand (e^{jt}-e^{-jt})^5 ?
The solution to expand (e^{jt}-e^{-jt})^5 is e^{5jt}-5e^{3jt}+10e^{jt}-10e^{-jt}+5e^{-3jt}-e^{-5jt}