Resolver (z^2)^7 | Microsoften solucionagailu matematikoa

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antzeko 5 arazoen antzekoak:

Bilaketaren antzeko arazoak webgunean

https://socratic.org/questions/how-do-you-simplify-3z-2-2-and-write-it-using-only-positive-exponents

https://math.stackexchange.com/questions/1868205/on-real-part-of-the-complex-number-1iz2

https://math.stackexchange.com/questions/387629/does-pa-pb-rightarrow-a-b

https://math.stackexchange.com/q/2890453

From the definition of Cosine Integral : \begin{align} \text{Ci}(x)&=-\int_{x}^{\infty}\frac{\cos{t}}{t}\,dt=\gamma+\log{x}+\int_{0}^{x}\frac{\cos{t}-1}{t}\,dt \\[8mm] I_r&=\int r^2\,\sin\left(\frac{1-3z^2}{r^3}\right)\,dr \\[2mm] &\color{blue}{\small\left\{u=\sin\left(\frac{1-3z^2}{r^3}\right)\Rightarrow du=\frac{-3(1-3z^2)}{r^4}\cos\left(\frac{1-3z^2}{r^3}\right) dr,\quad dv=r^2 dr\Rightarrow v=\frac{r^3}{3}\right\}} \\[2mm] I_r&=\frac{r^3}{3}\,\sin\left(\frac{1-3z^2}{r^3}\right)+(1-3z^2)\int\frac{1}{r}\,\cos\left(\frac{1-3z^2}{r^3}\right)\,dr \\[2mm] &\color{blue}{\small\left\{\frac{d}{dx}\left[\text{Ci}(x)\right]=\frac{\cos{x}}{x}\Rightarrow\frac{d}{dx}\left[\text{Ci}\left(\frac{1-3z^2}{r^3}\right)\right]=\frac{\cos\left(\frac{1-3z^2}{r^3}\right)}{\frac{1-3z^2}{r^3}}\frac{-3(1-3z^2)}{r^4}=-\frac{3}{r}\cos\left(\frac{1-3z^2}{r^3}\right)\right\}} \\[2mm] I_r&=\frac{r^3}{3}\,\sin\left(\frac{1-3z^2}{r^3}\right)-\frac{1-3z^2}{3}\,\text{Ci}\left(\frac{1-3z^2}{r^3}\right)+const \\[8mm] \color{red}{I}&=\int_{-1}^{+1}\int_{0}^{\infty}r^2\,\sin\left(\frac{1-3z^2}{r^3}\right)\,dr\,dz=2\int_{0}^{1}\int_{0}^{\infty}r^2\,\sin\left(\frac{1-3z^2}{r^3}\right)\,dr\,dz\quad\color{blue}{\left\{\rightarrow z^2\right\}} \\[2mm] &=2\int_{0}^{1}\left[\frac{r^3}{3}\,\sin\left(\frac{1-3z^2}{r^3}\right)-\frac{1-3z^2}{3}\,\text{Ci}\left(\frac{1-3z^2}{r^3}\right)\right]_{r=0}^{r=\infty}\,dz \\[2mm] &=2\int_{0}^{1}\left[\frac{1-3z^2}{3}\left(1-\gamma-\log\left(1-3z^2\right)\right)\right]\,dz \quad\color{blue}{\left\{\rightarrow\text{Ci}(x)=\gamma+\log{x}-\text{Cin}(x)\right\}} \\[2mm] &=\frac{2}{3}(1-\gamma)\int_{0}^{1}\left(1-3z^2\right)\,dz\,-\,\frac{2}{3}\int_{0}^{1}\left(1-3z^2\right)\left(\log\left(1-3z^3\right)\right)\,dz\qquad\color{blue}{\left\{\rightarrow\text{IBP}\right\}} \\[2mm] &=\frac{2}{3}(1-\gamma)\left[z-z^3\right]_{0}^{1}\,+\,\frac{2}{3}\left[\frac{2}{3}z+\left(z-z^3\right)\left(\frac{2}{3}+\log\left(1-3z^2\right)\right)-\frac{4}{3\sqrt{3}}\text{arctanh}(\sqrt3z)\right]_{0}^{1} \\[2mm] &=\color{red}{\frac{4}{9}-\frac{8}{9\sqrt{3}}\,\text{arctanh}{\sqrt{3}}}\,\approx{0.106513} \end{align}

https://math.stackexchange.com/questions/3082989/suppose-z-in-mathbbc-is-an-nth-root-of-unity-that-is-suppose-that

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Kopiatu portapapeletan

\left(z^{2}\right)^{7}

Erabili berretzaileen arauak adierazpena sinplifikatzeko.

z^{2\times 7}

Berretura bat berretzeko, biderkatu berretzaileak haien artean.

z^{14}

Egin 2 bider 7.

7\left(z^{2}\right)^{7-1}\frac{\mathrm{d}}{\mathrm{d}z}(z^{2})

F bi funtzio diferentziagarrien (f\left(u\right) eta u=g\left(x\right) funtzioen) konposaketa bada, hau da, F\left(x\right)=f\left(g\left(x\right)\right) bada, F-ren deribatua hau izango da: f-ren deribatua u-rekiko, bider g-ren deribatua x-rekiko, hots, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

7\left(z^{2}\right)^{6}\times 2z^{2-1}

Polinomioaren deribatua haren deribatuen gaien batura da. Gai konstante guztien deribatua 0 da. ax^{n} ekuazioaren deribatua nax^{n-1} da.

14z^{1}\left(z^{2}\right)^{6}

Sinplifikatu.

14z\left(z^{2}\right)^{6}

t gaiei dagokienez, t^{1}=t.

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