What is the simplest form of i^16 plus i^6 minus 2i^5 plus i^13?

1 answer:

6 0

$\bf \textit{recall that}\qquad i^4=1\qquad i^3=-i\qquad i^2=-1\\\\
-------------------------------\\\\
i^{16}+i^6-2i^5+i^{13}\implies i^{4\cdot 4}~+~i^{4+2}~-~2i^{3+2}~+~i^{4+4+4+1}
\\\\\\
(i^4)^4~+~i^4\cdot i^2~-~2\cdot i^3\cdot i^2~+~i^4\cdot i^4\cdot i^4\cdot i^1
\\\\\\
(1)^4+[1\cdot -1]-[2(-i)(-1)]+[(1)(1)(1)(i)]
\\\\\\
1-1-2i+i\implies -i$

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I have an expression

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floating around in my head; let's see if it makes sense.

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$E( (b-p)^2 ) = (-p)^2(1-p) + (1-p)^2p = p(1-p)$

That's for an individual sample. For the observed average we divide by n, and for the standard deviation we take the square root:

$\sigma = \sqrt{p(1-p)/n}$

Plugging in the numbers,

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Step-by-step explanation:

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