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.
For complete information on how to solve this question 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, read and understand it carefully till the end.
https://allin4.de/post-sitemap.xml
https://allin4.de/sitemap_index.xml
Let us know how to solve this question
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}}+2.23D photo Real part

Imaginary part

Contour plots

Imaginary part

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}})+constantglobal 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]>1Series 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 ?
This article has been completely solved by tireless effort from our side, still if any error remains in it then definitely write us your opinion in the comment box. If you like or understand the methods of solving all the questions in this article, then send it to your friends who are in need.
Note: If you have any such question, then definitely send it by writing in our comment box to get the answer.
Your question will be answered from our side.
Thank you once again from our side for reading or understanding this article completely.
You yourself can also easily solve this type of question with the help given below.

Expansions on branch cuts
For the function itself

Expansions at z==1
For the function itself

Expansions of (1+z)1/2at z==0
