Welcome to my article 1.2+(sqrt(1-(sqrt(x^2+y^2))^2) + 1 – x^2-y^2) * (sin (10000 * (x*3+y/5+7))+1/4) from -1.6 to 1.6 . This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
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## 1.2+(sqrt(1-(sqrt(x^2+y^2))^2) + 1 – x^2-y^2) * (sin (10000 * (x*3+y/5+7))+1/4) from -1.6 to 1.6

Input

\displaystyle 1.2+(\sqrt{{1-\sqrt{{{{x}^{2}}+{{y}^{{{{2}^{2}}}}}}}}}+1-{{x}^{2}}-{{y}^{2}})

Result

\displaystyle \sqrt{{-{{x}^{2}}-}}{{y}^{2}}+1-{{x}^{2}}-{{y}^{2}}+2.2

3D photo Real part

Imaginary part

Contour plots

Imaginary part

\displaystyle \begin{array}{l}Alternate\text{ }form\0.2\left( {5\sqrt{{-{{x}^{2}}-{{y}^{2}}+1-}}5{{x}^{2}}-5{{y}^{2}}+11} \right)\end{array}

Pr opertiesasafunction

\displaystyle {(x,y)\in {{x}^{2}}+{{y}^{2}}\le 1}

Renge

\displaystyle \left( {z\in R:\frac{6}{5}\le z\le } \right)

\displaystyle \begin{array}{l}Even\series\exp ansionatx=0\end{array}

\displaystyle {-{{y}^{2}}+\sqrt{{1-{{y}^{2}}}}+2.2}+{{x}^{2}}{-\frac{1}{{2\sqrt{{1-{{x}^{2}}}}}}-1}-\frac{{{{x}^{4}}}}{{8{{{(1-{{y}^{2}})}}^{{\frac{3}{2}}}}}}+0({{x}^{5}})

\displaystyle series\exp eansion.at.x=\infty \displaystyle (-{{x}^{2}}+\frac{{\sqrt{{-{{x}^{2}}x}}}}{x}+(2.2-{{y}^{2}})-\frac{{x(y-1)(y+1)}}{{2\sqrt{{-{{x}^{2}}x}}}}+0{{{(\frac{1}{x})}^{2}}}

\displaystyle partial.derivatives

\displaystyle \frac{\partial }{{\partial x}}(\sqrt{{-{{x}^{2}}-{{y}^{2}}+1}}-{{x}^{2}}-{{y}^{2}}+2.2)=x(-\frac{1}{{\sqrt{{-{{x}^{2}}-{{y}^{2}}+1}}}}-2)

\displaystyle \frac{\partial }{{\partial x}}(\sqrt{{-{{x}^{2}}-{{y}^{2}}+1}}-{{x}^{2}}-{{y}^{2}}+2.2)=y(-\frac{1}{{\sqrt{{-{{x}^{2}}-{{y}^{2}}+1}}}}-2)

Indefinite integral

\displaystyle \int{{(2.2}}-{{x}^{2}}-{{y}^{2}}+\sqrt{{1-{{x}^{2}}-{{y}^{2}}}})dx=x(0.5\sqrt{{-{{x}^{2}}-{{y}^{2}}+1}}-0.333333{{x}^{2}}-{{y}^{2}}+2.2)+(0.5{{y}^{2}}-0.5){{\tan }^{{-1}}}(\frac{{x\sqrt{{-{{x}^{2}}-{{y}^{2}}+1}}}}{{{{x}^{2}}+{{y}^{2}}-1}})+constant

global maximum

\displaystyle \max {1.2+(\sqrt{{1-\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}+1-{{x}^{2}}-{{y}^{2}})}=\frac{{16}}{5}at(x,y)=(0,0)

global minima

\displaystyle \min {1.2+(\sqrt{{1-\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}+1-{{x}^{2}}-{{y}^{2}})}=\frac{6}{5}for({{x}^{2}}+{{y}^{2}}=1)

series representations

\displaystyle 1.2+(\sqrt{{1-{{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}^{2}}}}+1-{{x}^{2}}-{{y}^{2}})=2.2-{{x}^{2}}-{{y}^{2}}+\sqrt{{-{{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}^{2}}}}\sum\limits_{{k=0}}^{\infty }{{(\frac{{\frac{1}{2}}}{k})}}{{(-{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}^{2}})}^{{-k}}}for\left[ {{{x}^{2}}+{{y}^{2}}} \right]>1

\displaystyle 1.2+(\sqrt{{1-{{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}^{2}}}}+1-{{x}^{2}}-{{y}^{2}})=2.2-{{x}^{2}}-{{y}^{2}}+\sum\limits_{{k=0}}^{\infty }{{\frac{{{{{(-1)}}^{k}}(-\frac{1}{2})k{{{(-\sqrt{{{{x}^{2}}+{{y}^{2}}}})}}^{k}}}}{{k!}}}}for\left[ {{{x}^{2}}+{{y}^{2}}} \right]<1

\displaystyle 1.2+(\sqrt{{1-{{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}^{2}}}}+1-{{x}^{2}}-{{y}^{2}})=2.2-{{x}^{2}}-{{y}^{2}}\sqrt{{1-{{{\sqrt{{{{x}^{2}}+{{y}^{2}}}}}}^{2}}}}\sum\limits_{{k=0}}^{\infty }{{\frac{{{{{(-1)}}^{k}}(-\frac{1}{2})k{{{(-\sqrt{{{{x}^{2}}+{{y}^{2}}}})}}^{{-k}}}}}{{k!}}}}for\left[ {{{x}^{2}}+{{y}^{2}}} \right]>1

Series representations

Generalized power series

Expansions at generic point z==z0

For the function itself

1.2+(sqrt(1-(sqrt(x^2+y^2))^2) + 1 – x^2-y^2) * (sin (10000 * (x*3+y/5+7))+1/4) from -1.6 to 1.6 ?

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Expansions on branch cuts

For the function itself

Expansions at z==1

For the function itself

Expansions of (1+z)1/2at z==0

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