Whakaoti i te F=A(1+t)^N-1/t

$$F = A \frac { ( 1 + t ) ^ { N } - 1 } { t }$$

Whakaoti mō A (complex solution)

$$\left\{\begin{matrix}A=\frac{Ft}{\left(t+1\right)^{N}-1}\text{, }&\left(\left(Re(N)\right)^{2}+\left(Im(N)\right)^{2}=0\text{ or }\nexists n_{1}\in \mathrm{Z}\text{ : }t=e^{-\frac{2\pi n_{1}iRe(N)}{\left(Re(N)\right)^{2}+\left(Im(N)\right)^{2}}-\frac{2\pi n_{1}Im(N)}{\left(Re(N)\right)^{2}+\left(Im(N)\right)^{2}}}-1\right)\text{ and }t\neq 0\\A\in \mathrm{C}\text{, }&F=0\text{ and }\exists n_{2}\in \mathrm{Z}\text{ : }N=\frac{2\pi n_{2}i}{\ln(t+1)}\text{ and }t\neq 0\text{ and }t\neq -1\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakareatia ngā taha e rua o te whārite ki te t.

$$Ft=A\left(\left(1+t\right)^{N}-1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te A ki te \left(1+t\right)^{N}-1.

$$Ft=A\left(1+t\right)^{N}-A$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$A\left(1+t\right)^{N}-A=Ft$$

Pahekotia ngā kīanga tau katoa e whai ana i te A.

$$\left(\left(1+t\right)^{N}-1\right)A=Ft$$

He hanga arowhānui tō te whārite.

$$\left(\left(t+1\right)^{N}-1\right)A=Ft$$

Whakawehea ngā taha e rua ki te \left(1+t\right)^{N}-1.

$$\frac{\left(\left(t+1\right)^{N}-1\right)A}{\left(t+1\right)^{N}-1}=\frac{Ft}{\left(t+1\right)^{N}-1}$$

Mā te whakawehe ki te \left(1+t\right)^{N}-1 ka wetekia te whakareanga ki te \left(1+t\right)^{N}-1.

$$A=\frac{Ft}{\left(t+1\right)^{N}-1}$$

Whakaoti mō A

$$\left\{\begin{matrix}A=\frac{Ft}{\left(t+1\right)^{N}-1}\text{, }&\left(t=-1\text{ and }N>0\right)\text{ or }\left(N\neq 0\text{ and }t\neq -2\text{ and }Denominator(N)\text{bmod}2=1\text{ and }t<-1\right)\text{ or }\left(t<-1\text{ and }Numerator(N)\text{bmod}2=1\text{ and }Denominator(N)\text{bmod}2=1\right)\text{ or }\left(N\neq 0\text{ and }t\neq 0\text{ and }t>-1\right)\\A\in \mathrm{R}\text{, }&t\neq 0\text{ and }t\neq -1\text{ and }\left(N=0\text{ or }t<-1\right)\text{ and }\left(t=-2\text{ or }N=0\right)\text{ and }Numerator(N)\text{bmod}2=0\text{ and }Denominator(N)\text{bmod}2=1\text{ and }F=0\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakareatia ngā taha e rua o te whārite ki te t.

$$Ft=A\left(\left(1+t\right)^{N}-1\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te A ki te \left(1+t\right)^{N}-1.

$$Ft=A\left(1+t\right)^{N}-A$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$A\left(1+t\right)^{N}-A=Ft$$

Pahekotia ngā kīanga tau katoa e whai ana i te A.

$$\left(\left(1+t\right)^{N}-1\right)A=Ft$$

He hanga arowhānui tō te whārite.

$$\left(\left(t+1\right)^{N}-1\right)A=Ft$$

Whakawehea ngā taha e rua ki te \left(1+t\right)^{N}-1.

$$\frac{\left(\left(t+1\right)^{N}-1\right)A}{\left(t+1\right)^{N}-1}=\frac{Ft}{\left(t+1\right)^{N}-1}$$

Mā te whakawehe ki te \left(1+t\right)^{N}-1 ka wetekia te whakareanga ki te \left(1+t\right)^{N}-1.

$$A=\frac{Ft}{\left(t+1\right)^{N}-1}$$

Whakaoti mō F (complex solution)

$$F=-\frac{A\left(-\left(t+1\right)^{N}+1\right)}{t}$$

$$t\neq 0$$

Whakaoti mō F

$$F=-\frac{A\left(-\left(t+1\right)^{N}+1\right)}{t}$$

$$\left(t>-1\text{ and }t\neq 0\right)\text{ or }\left(N>0\text{ and }t=-1\right)\text{ or }\left(Denominator(N)\text{bmod}2=1\text{ and }t<-1\right)$$