sum from n=1 to infinity of (cos(1.4))^n

Lösung

n = 1 \sum \infty (cos (1.4))^{n}

0.20477 \dots

Schritte zur Lösung

n = 1 \sum \infty (cos (1.4))^{n}

Vereinfache

cos (1.4) : 0.16996 \dots

= n = 1 \sum \infty 0.16996 \dots^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty 0.16996 \dots^{n} - 0.16996 \dots^{0}

n = 0 \sum \infty 0.16996 \dots^{n} = 1.20477 \dots

= 1.20477 \dots - 0.16996 \dots^{0}

Vereinfache

= 0.20477 \dots

\int_{0}^{t} t^{2} e^{- 5 t} d t

\frac{d}{d t} (e^{- 6 t})

y^{^{'}} - 6 + \frac{y}{( 4 ( x + 250 ) )} = 0, y (0) = 70

\int_{0}^{\frac{1}{2}} arcsin (2 y) d y

\frac{d}{d x} (x ln (x) - 1)