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MAT334--Lectures & Home Assignments / Re: What's the answer of FE q2 d.e.f?
« on: December 11, 2018, 08:31:32 PM »
Can you explain how you figured it out?
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MAT334--Misc / Will the marks for booklets 150-200 be posted today?
« on: December 03, 2018, 02:17:33 PM »
Will the marks for booklets 150-200 be posted today?
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Quiz-4 / Re: Q4 TUT 0202
« on: December 02, 2018, 01:45:54 PM »
Never got any feedback on this answer, is it correct?
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Quiz-7 / Re: Q7 TUT 5101
« on: December 01, 2018, 08:49:16 PM »
Because
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \ >\ 0$
(Added this to my previous answer)
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \ >\ 0$
(Added this to my previous answer)
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Quiz-7 / Re: Q7 TUT 5101
« on: December 01, 2018, 04:54:19 PM »
You're right, $\displaystyle f( iy)$ is totally irrelevant here. My mistake. Do you mean the fact that f([-R, R]) lies completely in the upper half plane. Because
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \ >\ 0$
When $\displaystyle z\ \rightarrow \ \pm \infty $, Re(f(z)) $\displaystyle \rightarrow \ \infty $ and Im(f(z)) $\displaystyle \rightarrow \ \infty \ $and the endpoints of $\displaystyle f([ -R,\ R]) \ $lie in the first quadrant for R really big. So the change in argument is going to be 0.
This can be coupled with what I said about the curve.
$\displaystyle f\left( Re^{it}\right)$ = $\displaystyle e^{4it} \ $from [0, $\displaystyle \pi $] because the 4 term dominates. So the change in argument is 4$\displaystyle \pi $.
So in total we get 2 zeros as the total change in argument is 4$\displaystyle \pi $.
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \ >\ 0$
When $\displaystyle z\ \rightarrow \ \pm \infty $, Re(f(z)) $\displaystyle \rightarrow \ \infty $ and Im(f(z)) $\displaystyle \rightarrow \ \infty \ $and the endpoints of $\displaystyle f([ -R,\ R]) \ $lie in the first quadrant for R really big. So the change in argument is going to be 0.
This can be coupled with what I said about the curve.
$\displaystyle f\left( Re^{it}\right)$ = $\displaystyle e^{4it} \ $from [0, $\displaystyle \pi $] because the 4 term dominates. So the change in argument is 4$\displaystyle \pi $.
So in total we get 2 zeros as the total change in argument is 4$\displaystyle \pi $.
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Quiz-7 / Re: Q7 TUT 5101
« on: November 30, 2018, 04:24:14 PM »
$\displaystyle \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ z^{4} \ +\ 3iz^{2} \ +\ z\ -\ 2\ +\ i\\
\\
f( iy) \ is\ always\ positive\\
\\
f( x) \ =\ x^{4} \ +\ 3ix^{2} \ +\ x\ -\ 2\ +\ i\\
\\
Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2
\end{array}$
This is positive at x = R, then negative when it hits x = 1 to x = -1 then it becomes positive again as x approaches -R
.
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$ This is always positive.
So, we are always in the upper half plane. Change in Arg = 0.
$\displaystyle f\left( Re^{it}\right) \ =\ e^{4it} \ $from [0, $\displaystyle \pi $] because the 4 term domindates. So the change in argument is 4$\displaystyle \pi $.
$\displaystyle \frac{0+4\pi }{2\pi } \ =\ 2$ zeros total
f( z) \ =\ z^{4} \ +\ 3iz^{2} \ +\ z\ -\ 2\ +\ i\\
\\
f( iy) \ is\ always\ positive\\
\\
f( x) \ =\ x^{4} \ +\ 3ix^{2} \ +\ x\ -\ 2\ +\ i\\
\\
Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2
\end{array}$
This is positive at x = R, then negative when it hits x = 1 to x = -1 then it becomes positive again as x approaches -R
.
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$ This is always positive.
So, we are always in the upper half plane. Change in Arg = 0.
$\displaystyle f\left( Re^{it}\right) \ =\ e^{4it} \ $from [0, $\displaystyle \pi $] because the 4 term domindates. So the change in argument is 4$\displaystyle \pi $.
$\displaystyle \frac{0+4\pi }{2\pi } \ =\ 2$ zeros total
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Quiz-7 / Re: Q7 TUT 0101
« on: November 30, 2018, 04:01:32 PM »
Using Rouche's Theorem, which is a corollary of the Argument Principle.
$\displaystyle f( z) =\ z^{4} \ -\ 2z\ -\ 2$ in $\displaystyle \frac{1}{2} \ < \ |z|\ < \ \frac{3}{2}$
$\displaystyle Let\ g( z) \ =\ -2$
For the smaller curve we have $\displaystyle z=\frac{1}{2} e^{i\theta }$,
$\displaystyle |f( z) -( -2) |\ \leq \ z^{4} \ +\ 2z\ \ =\ \frac{17}{16} \ < \ |g( z) |\ $
So, f(z) and g(z) have the same number of zeros in the smaller circle, which is 0.
$\displaystyle Now,\ let\ g( z) \ =\ z^{4}$
For the larger curve we have $\displaystyle z=\frac{3}{2} e^{i\theta }$
$\displaystyle |f( z) \ -\ z^{4} |\ \leq |2z\ +\ 2\ |\ \ =\ 5\ < \ |g( z) |\ =\ \frac{81}{16} \ \ $
So, f(z) and g(z) have the same number of zeros in the larger circle, which is 4.
Thus, all four zeros of f(z) are in $\displaystyle \frac{1}{2} \ < \ |z|\ < \ \frac{3}{2}$
$\displaystyle f( z) =\ z^{4} \ -\ 2z\ -\ 2$ in $\displaystyle \frac{1}{2} \ < \ |z|\ < \ \frac{3}{2}$
$\displaystyle Let\ g( z) \ =\ -2$
For the smaller curve we have $\displaystyle z=\frac{1}{2} e^{i\theta }$,
$\displaystyle |f( z) -( -2) |\ \leq \ z^{4} \ +\ 2z\ \ =\ \frac{17}{16} \ < \ |g( z) |\ $
So, f(z) and g(z) have the same number of zeros in the smaller circle, which is 0.
$\displaystyle Now,\ let\ g( z) \ =\ z^{4}$
For the larger curve we have $\displaystyle z=\frac{3}{2} e^{i\theta }$
$\displaystyle |f( z) \ -\ z^{4} |\ \leq |2z\ +\ 2\ |\ \ =\ 5\ < \ |g( z) |\ =\ \frac{81}{16} \ \ $
So, f(z) and g(z) have the same number of zeros in the larger circle, which is 4.
Thus, all four zeros of f(z) are in $\displaystyle \frac{1}{2} \ < \ |z|\ < \ \frac{3}{2}$
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Term Test 2 / Re: TT2A Problem 5
« on: November 29, 2018, 05:00:07 PM »
$\displaystyle f( z) \ =\ \frac{3z}{( z-2)( z+1)}$
By partial fractions:
$\displaystyle \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ \frac{A}{z-2} \ +\ \frac{B}{z+1}\\
\\
f( z) \ =\ \frac{2}{z-2} \ +\ \frac{1}{z+1}
\end{array}$
a) as $\displaystyle |z|\ < \ 1$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ -\frac{1}{1\ -\ \frac{z}{2}} \ +\ \frac{1}{1\ -\ ( -z)}\\
\\
=\ -\sum ^{\infty }_{n\ =\ 0}\left(\frac{z}{2}\right)^{n} \ +\ \sum ^{\infty }_{n\ =\ 0}( -z)^{n}\\
\\
=\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{2^{n}} \ +( -1)^{n}\right) z^{n}
\end{array}$
b) as $\displaystyle 1\ < \ |z|\ < \ 2$
$\displaystyle f( z) \ =\ $$\displaystyle -\frac{1}{1\ -\ \frac{z}{2}} \ +\ \frac{1}{z}\frac{1}{1\ +\ \frac{1}{z}}$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
=-\sum ^{\infty }_{n\ =\ 0}\left(\frac{z}{2}\right)^{n} \ +\ \frac{1}{z}\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{z}\right)^{n}\\
\\
=-\sum ^{\infty }_{n\ =\ 0}\frac{z^{n}}{2^{n}} \ +\ \sum ^{\infty }_{n\ =\ 0}\frac{( -1)}{z^{n+1}}^{n}\\
\\
=\sum ^{\infty }_{n\ =\ 0} -\frac{1}{2^{n}} \ z^{n} \ +\ \sum ^{1}_{n\ =\ -\infty }\frac{1}{( -1)^{n-1}} \ z^{n}
\end{array}$
c) as $\displaystyle |z|\ >\ 2$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ \frac{2}{z} \ \frac{1}{1\ -\ \frac{2}{z}} \ +\frac{1}{z}\frac{1}{1\ +\ \frac{1}{z}} \ \\
=\ \frac{2}{z} \ \sum ^{\infty }_{n\ =\ 0}\left(\frac{2}{z}\right)^{n} +\ \frac{1}{z}\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{z}\right)^{n}\\
\\
=\sum ^{\infty }_{n\ =\ 0}\left(\frac{2^{n+1}}{z^{n+1}}\right) \ +\ \sum ^{\infty }_{n\ =\ 0}\frac{( -1)}{z^{n+1}}^{n}\\
\\
=\sum ^{-1}_{n\ =\ -\infty }\left(\frac{1}{2^{n-2}} \ +\ \frac{1}{( -1)^{n-1}}\right) z^{n}\\
\\
\end{array}$
By partial fractions:
$\displaystyle \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ \frac{A}{z-2} \ +\ \frac{B}{z+1}\\
\\
f( z) \ =\ \frac{2}{z-2} \ +\ \frac{1}{z+1}
\end{array}$
a) as $\displaystyle |z|\ < \ 1$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ -\frac{1}{1\ -\ \frac{z}{2}} \ +\ \frac{1}{1\ -\ ( -z)}\\
\\
=\ -\sum ^{\infty }_{n\ =\ 0}\left(\frac{z}{2}\right)^{n} \ +\ \sum ^{\infty }_{n\ =\ 0}( -z)^{n}\\
\\
=\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{2^{n}} \ +( -1)^{n}\right) z^{n}
\end{array}$
b) as $\displaystyle 1\ < \ |z|\ < \ 2$
$\displaystyle f( z) \ =\ $$\displaystyle -\frac{1}{1\ -\ \frac{z}{2}} \ +\ \frac{1}{z}\frac{1}{1\ +\ \frac{1}{z}}$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
=-\sum ^{\infty }_{n\ =\ 0}\left(\frac{z}{2}\right)^{n} \ +\ \frac{1}{z}\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{z}\right)^{n}\\
\\
=-\sum ^{\infty }_{n\ =\ 0}\frac{z^{n}}{2^{n}} \ +\ \sum ^{\infty }_{n\ =\ 0}\frac{( -1)}{z^{n+1}}^{n}\\
\\
=\sum ^{\infty }_{n\ =\ 0} -\frac{1}{2^{n}} \ z^{n} \ +\ \sum ^{1}_{n\ =\ -\infty }\frac{1}{( -1)^{n-1}} \ z^{n}
\end{array}$
c) as $\displaystyle |z|\ >\ 2$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ \frac{2}{z} \ \frac{1}{1\ -\ \frac{2}{z}} \ +\frac{1}{z}\frac{1}{1\ +\ \frac{1}{z}} \ \\
=\ \frac{2}{z} \ \sum ^{\infty }_{n\ =\ 0}\left(\frac{2}{z}\right)^{n} +\ \frac{1}{z}\sum ^{\infty }_{n\ =\ 0}\left( -\frac{1}{z}\right)^{n}\\
\\
=\sum ^{\infty }_{n\ =\ 0}\left(\frac{2^{n+1}}{z^{n+1}}\right) \ +\ \sum ^{\infty }_{n\ =\ 0}\frac{( -1)}{z^{n+1}}^{n}\\
\\
=\sum ^{-1}_{n\ =\ -\infty }\left(\frac{1}{2^{n-2}} \ +\ \frac{1}{( -1)^{n-1}}\right) z^{n}\\
\\
\end{array}$
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MAT334--Lectures & Home Assignments / 2.5 Example 5
« on: November 18, 2018, 01:04:15 AM »
Can someone please help me with example 5? I understand the Res(r ; -2i) is F(z)/G'(z) but when I try that I am not getting the same calculation as the book.
**Solved, had to multiply by the conjugate to get the answer.
**Solved, had to multiply by the conjugate to get the answer.
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Quiz-6 / Re: Q6 TUT 5301
« on: November 17, 2018, 05:24:26 PM »
Why 2k? It should just be k.
$\displaystyle \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ \pi \ cot\ ( \pi z)\\
\\
=\frac{\pi cos( \pi z)}{sin( \pi z)}
\end{array}$
So clearly, f(z) has a pole when sin($\displaystyle \pi z) \ =\ 0$ which happens whenever z is any integer
This pole is order 1.
$\displaystyle \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ \pi \ cot\ ( \pi z)\\
\\
=\frac{\pi cos( \pi z)}{sin( \pi z)}
\end{array}$
So clearly, f(z) has a pole when sin($\displaystyle \pi z) \ =\ 0$ which happens whenever z is any integer
This pole is order 1.
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Quiz-6 / Re: Q6 TUT 0203
« on: November 17, 2018, 04:32:23 PM »
$\displaystyle f( z) \ =\ \frac{g( z)}{( z-z_{0})^{l}}$
where $\displaystyle g( z)$ is analytic in $\displaystyle |z\ -\ z_{0} |\ < \ R\ and\ g( z_{0}) \ \neq 0$
Then,
$\displaystyle \frac{f'( z)}{f( z)} \ =\ \frac{g'( z)}{g( z)} \ -\ \frac{l}{z\ -\ z_{0}}$
Since $\displaystyle g( z_{0}) \ \neq \ 0,\ then\ \frac{g'( z)}{g( z)}$ is analytic at $\displaystyle z_{0}$ , so the residue of that first term is 0. So for the residue of the whole thing, we read the coefficent of the second term which is -l. So we get
$\displaystyle Res\left(\frac{f'( z)}{f( z)} ;l\right) \ =-l$
where $\displaystyle g( z)$ is analytic in $\displaystyle |z\ -\ z_{0} |\ < \ R\ and\ g( z_{0}) \ \neq 0$
Then,
$\displaystyle \frac{f'( z)}{f( z)} \ =\ \frac{g'( z)}{g( z)} \ -\ \frac{l}{z\ -\ z_{0}}$
Since $\displaystyle g( z_{0}) \ \neq \ 0,\ then\ \frac{g'( z)}{g( z)}$ is analytic at $\displaystyle z_{0}$ , so the residue of that first term is 0. So for the residue of the whole thing, we read the coefficent of the second term which is -l. So we get
$\displaystyle Res\left(\frac{f'( z)}{f( z)} ;l\right) \ =-l$
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Reading Week Bonus--sample problems for TT2 / Re: Term Test 2 sample P4M
« on: November 10, 2018, 04:53:13 PM »\begin{equation*}
I( b) \ =\int ^{\infty }_{0}\frac{e^{-bx} sin( x)}{x} \ dx\ for\ b\ \geq 0.\\
\\
I( 0) \ =\ \int ^{\infty }_{0}\frac{sin( x)}{x} \ ( This\ is\ the\ integral\ of\ interest) .\\
\\
I'( b) \ =\ -\ \int ^{\infty }_{0} \ e^{-bx} sinx\ dx\\
\\
=\ -\ \frac{1}{1\ +\ b^{2}}\\
\end{equation*}
Integrate this with respect to b and we get:
\begin{equation*}
I( b) \ =\ \pi /2\ -\ arctan( b) ,\ setting\ b\ =\ 0,\ gives\ us\ \pi /2\\
so,\\
\\
\int ^{\infty }_{0}\frac{sin( x)}{x} \ \ =\ \pi /2\ \\
\end{equation*}
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Quiz-4 / Re: Q4 TUT 0202
« on: November 07, 2018, 06:24:36 PM »
By Cauchy's theorem, we get:
\begin{equation*}
0\ =\ \int _{\gamma } e^{iz^{2}} dz\ =\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} ie^{i\left( Re^{i\theta }\right)^{2}} Re^{i\theta } d\theta \ +\ \int ^{0}_{R} e^{ir^{2} \ e^{i\pi /2}} e^{\pi /4} \ dr\\
\\
=\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{ir^{2}( cos\pi /2\ _{\ } +\ isin\pi /2)} dr\\
\\
=\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{-r^{2}} dr\\
\\
\\
\end{equation*}
Now, just handling the middle integral, we get:
\begin{equation*}
\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \\
=R\int ^{\pi /4}_{0} e^{iR^{2} cos2\theta \ -\ R^{2} sin( 2\theta ) \ } ie^{i\theta } d\theta \\
This\ is\ estimated\ above\ by:\\
\\
R\int ^{\pi /4}_{0} e^{-R^{2} sin( 2\theta ) \ } d\theta \ \leq R\int ^{\pi /4}_{0} e^{-2R^{2}\frac{\theta }{\pi } \ } d\theta \ < \ \frac{\pi }{2R}\\
\end{equation*}
\begin{equation*}
So\ the\ limit\ as\ R\ \rightarrow \ \infty ,\ this\ will\ be\ 0,\ and\ we\ will\ get:\\
0=\int ^{R}_{0} e^{ix^{2}} dx\ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{-r^{2}} dr\\
0=\int ^{\infty }_{0}\left( cos\left( x^{2}\right) +\ isin\left( x^{2}\right) dx\right) \ -\ \frac{1+i}{\sqrt{2}} \ \int ^{\infty }_{0} e^{-r^{2}} dr\\
\frac{1+i}{\sqrt{2}} \ \int ^{\infty }_{0} e^{-r^{2}} dr\ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\frac{1+i}{\sqrt{2}} \ \frac{\sqrt{\pi }}{2} \ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\frac{\sqrt{2\pi }}{4} \ +\ i\ \frac{\sqrt{2\pi }}{4} \ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\\
\end{equation*}
So, we separate imaginary and real parts and we get
\begin{equation*}
\int ^{\infty }_{0} cos\left( x^{2}\right) dx\ =\ \int ^{\infty }_{0} sin\left( x^{2}\right) dx\ =\frac{\sqrt{2\pi }}{4} \
\end{equation*}
\begin{equation*}
0\ =\ \int _{\gamma } e^{iz^{2}} dz\ =\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} ie^{i\left( Re^{i\theta }\right)^{2}} Re^{i\theta } d\theta \ +\ \int ^{0}_{R} e^{ir^{2} \ e^{i\pi /2}} e^{\pi /4} \ dr\\
\\
=\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{ir^{2}( cos\pi /2\ _{\ } +\ isin\pi /2)} dr\\
\\
=\int ^{R}_{0} e^{ix^{2}} dx\ +\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{-r^{2}} dr\\
\\
\\
\end{equation*}
Now, just handling the middle integral, we get:
\begin{equation*}
\int ^{\pi /4}_{0} e^{iR^{2}( cos2\theta \ +\ isin( 2\theta ) \ } iRe^{i\theta } d\theta \\
=R\int ^{\pi /4}_{0} e^{iR^{2} cos2\theta \ -\ R^{2} sin( 2\theta ) \ } ie^{i\theta } d\theta \\
This\ is\ estimated\ above\ by:\\
\\
R\int ^{\pi /4}_{0} e^{-R^{2} sin( 2\theta ) \ } d\theta \ \leq R\int ^{\pi /4}_{0} e^{-2R^{2}\frac{\theta }{\pi } \ } d\theta \ < \ \frac{\pi }{2R}\\
\end{equation*}
\begin{equation*}
So\ the\ limit\ as\ R\ \rightarrow \ \infty ,\ this\ will\ be\ 0,\ and\ we\ will\ get:\\
0=\int ^{R}_{0} e^{ix^{2}} dx\ -\ e^{i\pi /4} \ \int ^{R}_{0} e^{-r^{2}} dr\\
0=\int ^{\infty }_{0}\left( cos\left( x^{2}\right) +\ isin\left( x^{2}\right) dx\right) \ -\ \frac{1+i}{\sqrt{2}} \ \int ^{\infty }_{0} e^{-r^{2}} dr\\
\frac{1+i}{\sqrt{2}} \ \int ^{\infty }_{0} e^{-r^{2}} dr\ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\frac{1+i}{\sqrt{2}} \ \frac{\sqrt{\pi }}{2} \ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\frac{\sqrt{2\pi }}{4} \ +\ i\ \frac{\sqrt{2\pi }}{4} \ =\ \int ^{\infty }_{0} cos\left( x^{2}\right) dx+\ i\int ^{\infty }_{0} sin\left( x^{2}\right) dx\\
\\
\end{equation*}
So, we separate imaginary and real parts and we get
\begin{equation*}
\int ^{\infty }_{0} cos\left( x^{2}\right) dx\ =\ \int ^{\infty }_{0} sin\left( x^{2}\right) dx\ =\frac{\sqrt{2\pi }}{4} \
\end{equation*}
15
MAT334--Misc / Forum Bonus Marks
« on: October 28, 2018, 08:13:35 PM »
How are the bonus marks calculated. Is each "karma" point worth a bonus point?
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