espandere (x+Δ)^4

Soluzione

espandere

(x + Δ)^{4}

x^{4} + 4Δ x^{3} + 6 Δ^{2} x^{2} + 4 Δ^{3} x + Δ^{4}

Fasi della soluzione

(x + Δ)^{4}

Applicare il teorema binomiale:

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

a = x, b = Δ

= i = 0 \sum 4 (i 4) x^{(4 - i)} Δ^{i}

Espandere la sommatoria

= \frac{4 !}{0 ! ( 4 - 0 ) !} x^{4} Δ^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} x^{3} Δ^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} x^{2} Δ^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} x^{1} Δ^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} x^{0} Δ^{4}

Semplifica

\frac{4 !}{0 ! ( 4 - 0 ) !} x^{4} Δ^{0} : x^{4}

Semplifica

\frac{4 !}{1 ! ( 4 - 1 ) !} x^{3} Δ^{1} : 4Δ x^{3}

Semplifica

\frac{4 !}{2 ! ( 4 - 2 ) !} x^{2} Δ^{2} : 6 Δ^{2} x^{2}

Semplifica

\frac{4 !}{3 ! ( 4 - 3 ) !} x^{1} Δ^{3} : 4 Δ^{3} x

Semplifica

\frac{4 !}{4 ! ( 4 - 4 ) !} x^{0} Δ^{4} : Δ^{4}

= x^{4} + 4Δ x^{3} + 6 Δ^{2} x^{2} + 4 Δ^{3} x + Δ^{4}

e x p an d (x^{4} - x^{3} + x^{2} - x + 1) (x^{4} + x^{3} + x^{2} + x + 1)

e x p an d 6 (12 + 7)

e x p an d \frac{s + 1}{s ^{2} ( s ^{2} - 2 s + 5 )}

e x p an d x^{4} 4 - x (e^{k} - e^{- x})

e x p an d 4 (x + Δ x)^{2} - 3 (x + Δ x) - 2