MHB View Attachment to Understanding Taxes

  • Thread starter Thread starter anil86
  • Start date Start date
  • Tags Tags
    Taxes
AI Thread Summary
The discussion revolves around mathematical series and their representations, particularly using binomial sums. The initial focus is on the series expansion of the function 1/sqrt(1 - x^2) and its implications for further calculations. The conversation includes deriving a complex expression involving the imaginary part of a function related to exponential terms. The participants are encouraged to proceed with the calculations based on the provided formulas. The exchange emphasizes the importance of mathematical rigor in understanding these series.
anil86
Messages
10
Reaction score
0
Please view attachment!
 

Attachments

  • Image0349.jpg
    Image0349.jpg
    87.8 KB · Views: 134
Mathematics news on Phys.org
anil86 said:
Please view attachment!

Wellcome on MHB anil86!... You can start from the binomial sum...

$\displaystyle \frac{1}{\sqrt{1 - x^{2}}} = 1 + \frac{1}{2}\ x^{2} + \frac{3}{8}\ x^{4} + ...\ (1)$

... and from (1) ...

$\displaystyle \frac{x}{\sqrt{1 - x^{2}}} = x + \frac{1}{2}\ x^{3} + \frac{3}{8}\ x^{5} + ... \ (2)$

... so that the sum of Your series is... $\displaystyle S = \text{Im} \{\frac{e^{i\ \theta}}{\sqrt{1 - e^{2\ i\ \theta}}}\}\ (3)$

Are You able to proceed?...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
Wellcome on MHB anil86!... You can start from the binomial sum...

$\displaystyle \frac{1}{\sqrt{1 - x^{2}}} = 1 + \frac{1}{2}\ x^{2} + \frac{3}{8}\ x^{4} + ...\ (1)$

... and from (1) ...

$\displaystyle \frac{x}{\sqrt{1 - x^{2}}} = x + \frac{1}{2}\ x^{3} + \frac{3}{8}\ x^{5} + ... \ (2)$

... so that the sum of Your series is... $\displaystyle S = \text{Im} \{\frac{e^{i\ \theta}}{\sqrt{1 - e^{2\ i\ \theta}}}\}\ (3)$

Are You able to proceed?...

Kind regards

$\chi$ $\sigma$

Please view attachment!View attachment 1687
 

Attachments

  • Image0352.jpg
    Image0352.jpg
    111.3 KB · Views: 103
chisigma said:
Wellcome on MHB anil86!... You can start from the binomial sum...

$\displaystyle \frac{1}{\sqrt{1 - x^{2}}} = 1 + \frac{1}{2}\ x^{2} + \frac{3}{8}\ x^{4} + ...\ (1)$

... and from (1) ...

$\displaystyle \frac{x}{\sqrt{1 - x^{2}}} = x + \frac{1}{2}\ x^{3} + \frac{3}{8}\ x^{5} + ... \ (2)$

... so that the sum of Your series is... $\displaystyle S = \text{Im} \{\frac{e^{i\ \theta}}{\sqrt{1 - e^{2\ i\ \theta}}}\}\ (3)$

Is...

$\displaystyle 1 - e^{2\ z} = - e^{z}\ (e^{z} - e^{- z}) -> \sqrt{1 - e^{2\ z}} = i\ e^{\frac{z}{2}}\ \sqrt {e^{z} - e^{- z}}\ (1)$

... so that for $\displaystyle z = i\ \theta $ is...

$\displaystyle \text{Im}\ \{\frac{e^{i\ \theta}}{\sqrt{1 - e^{2\ i\ \theta}}}\} = \text{Im}\ \{ \frac{e^{i\ \frac{\theta}{2}}}{i\ \sqrt{e^{i\ \theta} + e^{- i\ \theta}}} \} = \frac{\sin \frac{\theta}{2} + \cos {\frac{\theta}{2}}}{2\ \sqrt{\sin \theta}}\ (2)$

Kind regards

$\chi$ $\sigma$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top