Blog: Let `z= cos theta +i sin theta` .Then the value of `sum_(m=1)^15` `lm(Z^(2m-1))` at `theta=2` deg is

Saturday, April 2, 2016

Let $\displaystyle{z}={\cos{\theta}}+{i}{\sin{\theta}}$ .Then the value of $\displaystyle{\sum_{{{m}={1}}}^{{15}}}$ $\displaystyle{l}{m}{\left({{Z}}^{{{2}{m}-{1}}}\right)}$ at $\displaystyle\theta={2}$ deg is

You need to evaluate the sum of imaginaries of the given powers of complex number z, such that:

$\displaystyle{\sigma_{{{m}={1}}}^{{15}}}{I}{m}{\left({{z}}^{{{2}{m}-{1}}}\right)}={I}{m}{\left({{z}}^{{{2}\cdot{1}-{1}}}\right)}+{I}{m}{\left({{z}}^{{{2}\cdot{2}-{1}}}\right)}+{I}{m}{\left({{z}}^{{{2}\cdot{3}-{1}}}\right)}+\ldots+{I}{m}{\left({{z}}^{{{2}\cdot{15}-{1}}}\right)}$

$\displaystyle{\sigma_{{{m}={1}}}^{{15}}}{I}{m}{\left({{z}}^{{{2}{m}-{1}}}\right)}={I}{m}{\left({{z}}^{{1}}\right)}+{I}{m}{\left({{z}}^{{3}}\right)}+{I}{m}{\left({{z}}^{{5}}\right)}+\ldots+{I}{m}{\left({{z}}^{{29}}\right)}$

All the powers of the complex number z, may be evaluated with De Moivre formula, such that:

$\displaystyle{{z}}^{{n}}={{\left({\cos{\theta}}+{i}\cdot{\sin{\theta}}\right)}}^{{n}}\Rightarrow{{z}}^{{n}}={\left({\cos{{\left({n}\cdot\theta\right)}}}+{i}\cdot{\sin{{\left({n}\cdot\theta\right)}}}\right)}$

$\displaystyle{{z}}^{{3}}={\cos{{\left({3}\theta\right)}}}+{i}\cdot{\sin{{\left({3}\theta\right)}}}$

........

$\displaystyle{{z}}^{{29}}={\cos{{\left({29}\theta\right)}}}+{i}\cdot{\sin{{\left({29}\theta\right)}}}$

The imaginary part of the sum is $\displaystyle{I}{m}{\left({{z}}^{{1}}+{{z}}^{{3}}+\ldots{{z}}^{{29}}\right)}={\cos{\theta}}+{\cos{{\left({3}\theta\right)}}}+\ldots+{\cos{{\left({29}\theta\right)}}}$

You may group the terms such that:

$\displaystyle{\cos{\theta}}+{\cos{{\left({29}\theta\right)}}}={2}{\cos{{\left(\frac{{\theta+{29}\theta}}{{2}}\right)}}}\cdot{\cos{{\left(\frac{{\theta-{29}\theta}}{{2}}\right)}}}$

$\displaystyle{\cos{\theta}}+{\cos{{\left({29}\theta\right)}}}={2}{\cos{{\left({15}\theta\right)}}}{\cos{{\left(-{14}\theta\right)}}}$

Since $\displaystyle{\cos{{\left(-\theta\right)}}}={\cos{\theta}}$

$\displaystyle{\cos{\theta}}+{\cos{{\left({29}\theta\right)}}}={2}{\cos{{\left({15}\theta\right)}}}{\cos{{\left({14}\theta\right)}}}$

You may calculate the next sum $\displaystyle{\cos{{\left({3}\theta\right)}}}+{\cos{{\left({27}\theta\right)}}},$ such that:

$\displaystyle{\cos{{\left({3}\theta\right)}}}+{\cos{{\left({27}\theta\right)}}}={2}{\cos{{\left({15}\theta\right)}}}{\cos{{\left({12}\theta\right)}}}$

$\displaystyle{\cos{{\left({5}\theta\right)}}}+{\cos{{\left({25}\theta\right)}}}={2}{\cos{{\left({15}\theta\right)}}}{\cos{{\left({10}\theta\right)}}}$

$\displaystyle{\cos{{\left({7}\theta\right)}}}+{\cos{{\left({23}\theta\right)}}}={2}{\cos{{\left({15}\theta\right)}}}{\cos{{\left({8}\theta\right)}}}$

$\displaystyle{\cos{{\left({9}\theta\right)}}}+{\cos{{\left({21}\theta\right)}}}={2}{\cos{{\left({15}\theta\right)}}}{\cos{{\left({6}\theta\right)}}}$

$\displaystyle{\cos{{\left({11}\theta\right)}}}+{\cos{{\left({19}\theta\right)}}}={2}{\cos{{\left({15}\theta\right)}}}{\cos{{\left({4}\theta\right)}}}$

$\displaystyle{\cos{{\left({13}\theta\right)}}}+{\cos{{\left({17}\theta\right)}}}={2}{\cos{{\left({15}\theta\right)}}}{\cos{{\left({2}\theta\right)}}}$

Notice that the term $\displaystyle{\cos{{\left({15}\theta\right)}}}$ remains. You may take out the common factor $\displaystyle{\cos{{\left({15}\theta\right)}}}$ such that:

$\displaystyle{I}{m}{\left({{z}}^{{1}}+{{z}}^{{3}}+\ldots{{z}}^{{29}}\right)}={\cos{{\left({15}\theta\right)}}}{\left({2}{\cos{{\left({12}\theta\right)}}}+{2}{\cos{{\left({10}\theta\right)}}}+{2}{\cos{{\left({8}\theta\right)}}}+{2}{\cos{{\left({6}\theta\right)}}}+{2}{\cos{{\left({4}\theta\right)}}}+{2}{\cos{{\left({2}\theta\right)}}}+{1}\right)}$

You may put now $\displaystyle\theta={{2}}^{{o}}$ , such that:

$\displaystyle{\cos{{\left({15}\cdot{{2}}^{{o}}\right)}}}={{\cos{{30}}}}^{{o}}=\frac{\sqrt{{3}}}{{2}}$

You may group again the terms, such that:

$\displaystyle{2}{\cos{{\left({12}\theta\right)}}}+{2}{\cos{{\left({2}\theta\right)}}}={4}{\cos{{\left({7}\theta\right)}}}\cdot{\cos{{\left({5}\theta\right)}}}$

$\displaystyle{2}{\cos{{\left({10}\theta\right)}}}+{2}{\cos{{\left({4}\theta\right)}}}={4}{\cos{{\left({7}\theta\right)}}}\cdot{\cos{{\left({3}\theta\right)}}}$

$\displaystyle{2}{\cos{{\left({8}\theta\right)}}}+{2}{\cos{{\left({6}\theta\right)}}}={4}{\cos{{\left({7}\theta\right)}}}\cdot{\cos{{\left(\theta\right)}}}$

Factoring out 4 $\displaystyle{\cos{{\left({7}\theta\right)}}}$ yields:

$\displaystyle{2}{\cos{{\left({12}\theta\right)}}}+{2}{\cos{{\left({10}\theta\right)}}}+{2}{\cos{{\left({8}\theta\right)}}}+{2}{\cos{{\left({6}\theta\right)}}}+{2}{\cos{{\left({4}\theta\right)}}}+{2}{\cos{{\left({2}\theta\right)}}}={4}{\cos{{\left({7}\theta\right)}}}\cdot{\left({\cos{{\left({5}\theta\right)}}}+{\cos{{\left({3}\theta\right)}}}+{\cos{{\left(\theta\right)}}}\right)}$

You may group again the terms, such that:

$\displaystyle{\cos{{\left({5}\theta\right)}}}+{\cos{{\left(\theta\right)}}}={2}{\cos{{\left({3}\theta\right)}}}\cdot{\cos{{\left({2}\theta\right)}}}$

$\displaystyle{2}{\cos{{\left({3}\theta\right)}}}\cdot{\cos{{\left({2}\theta\right)}}}+{\cos{{\left({3}\theta\right)}}}={\cos{{\left({3}\theta\right)}}}{\left({2}{\cos{{\left({2}\theta\right)}}}+{1}\right)}{\)}$

Using the formula of double angle yields:

$\displaystyle{\cos{{\left({2}\theta\right)}}}={2}{{\cos}}^{{2}}\theta-{1}$

$\displaystyle{\cos{{\left({3}\theta\right)}}}{\left({2}{\cos{{\left({2}\theta\right)}}}+{1}\right)}{\)}={\cos{{\left({3}\theta\right)}}}{\left({4}{{\cos}}^{{2}}\theta-{2}+{1}\right)}{\)}$

$\displaystyle{\cos{{\left({3}\theta\right)}}}{\left({2}{\cos{{\left({2}\theta\right)}}}+{1}\right)}{\)}={\cos{{\left({3}\theta\right)}}}{\left({4}{{\cos}}^{{2}}\theta-{1}\right)}{\)}$

$\displaystyle{2}{\cos{{\left({12}\theta\right)}}}+{2}{\cos{{\left({10}\theta\right)}}}+{2}{\cos{{\left({8}\theta\right)}}}+{2}{\cos{{\left({6}\theta\right)}}}+{2}{\cos{{\left({4}\theta\right)}}}+{2}{\cos{{\left({2}\theta\right)}}}={4}{\cos{{\left({7}\theta\right)}}}\cdot{\cos{{\left({3}\theta\right)}}}\cdot{\left({4}{{\cos}}^{{2}}\theta-{1}\right)}{\)}$

$\displaystyle{I}{m}{\left({{z}}^{{1}}+{{z}}^{{3}}+\ldots{{z}}^{{29}}\right)}={\cos{{\left({15}\theta\right)}}}{\left({4}{\cos{{\left({7}\theta\right)}}}\cdot{\cos{{\left({3}\theta\right)}}}\cdot{\left({4}{{\cos}}^{{2}}\theta-{1}\right)}\right)}+{1}{\)}$

Hence, evaluating $\displaystyle{\sigma_{{{m}={1}}}^{{15}}}{I}{m}{\left({{z}}^{{{2}{m}-{1}}}\right)}={\left(\frac{\sqrt{{3}}}{{2}}\right)}\cdot{\)}{\left({4}{\cos{{\left({{14}}^{{o}}\right)}}}\cdot{\cos{{\left({{6}}^{{o}}\right)}}}\cdot{\left({4}{{\cos}}^{{2}}{\left({{2}}^{{o}}\right)}-{1}\right)}\right)}+{1}{\)}.$

Blog

Saturday, April 2, 2016

Let $\displaystyle{z}={\cos{\theta}}+{i}{\sin{\theta}}$ .Then the value of $\displaystyle{\sum_{{{m}={1}}}^{{15}}}$ $\displaystyle{l}{m}{\left({{Z}}^{{{2}{m}-{1}}}\right)}$ at $\displaystyle\theta={2}$ deg is

No comments:

Post a Comment

How does author Elie Wiesel use symbolism to contribute to the meaning of Night?

Saturday, April 2, 2016

Let .Then the value of at deg is

No comments:

Post a Comment

How does author Elie Wiesel use symbolism to contribute to the meaning of Night?

Let $\displaystyle{z}={\cos{\theta}}+{i}{\sin{\theta}}$ .Then the value of $\displaystyle{\sum_{{{m}={1}}}^{{15}}}$ $\displaystyle{l}{m}{\left({{Z}}^{{{2}{m}-{1}}}\right)}$ at $\displaystyle\theta={2}$ deg is