MATH1251 Questions HELP (2 Viewers)

1008 · Jul 30, 2016

Hey everyone

First of all I'd like to thank everyone, especially IG for helping me with MATH1151 last sem.

I am doing MATH1251 this semester, and I have an inaugural question lol:

$\\\text{Use }z \overline{z}=|z|^2\text{ to show that:}\\\left|1-\overline{z}w\right|^2-\left|z-w\right|^2=\left(1-\left|z\right|^2\right)\left(1-\left|w\right|^2\right)\\\text{and deduce that }\left|1-\overline{z}w\right|^2=\left|z-w\right|^2\text{ if either }z\text{ or }w\text{ lies on the unit circle.}$

InteGrand · Jul 30, 2016

1008 said:
Hey everyone

First of all I'd like to thank everyone, especially IG for helping me with MATH1151 last sem.

I am doing MATH1251 this semester, and I have an inaugural question lol:

$\\\text{Use }z \overline{z}=|z|^2\text{ to show that:}\\\left|1-\overline{z}w\right|^2-\left|z-w\right|^2=\left(1-\left|z\right|^2\right)\left(1-\left|w\right|^2\right)\\\text{and deduce that }\left|1-\overline{z}w\right|^2=\left|z-w\right|^2\text{ if either }z\text{ or }w\text{ lies on the unit circle.}$

$\noindent We have$

$\begin{align*} |1-\overline{z}w|^2 -|z-w|^2 &= \left(1-\overline{z}w\right)\left(1-z\overline{w}\right) -(z-w)(\overline{z}-\overline{w}) \\ &= 1 -z\overline{w} -\overline{z}w +|\overline{z}w|^2 -|z|^2 +z\overline{w} +w\overline{z} -|w|^2 \\ &= 1 + |z|^2 |w|^2 -|z|^2 -|w|^2 \\ &= \left(1-|z|^2\right)\left(1-|w|^2\right).\end{align*}$

$\noindent The `deduce' part follows easily from this and is left as an exercise for the reader.$

1008 · Jul 31, 2016

InteGrand said:
$\noindent We have$

$\begin{align*} |1-\overline{z}w|^2 -|z-w|^2 &= \left(1-\overline{z}w\right)\left(1-z\overline{w}\right) -(z-w)(\overline{z}-\overline{w}) \\ &= 1 -z\overline{w} -\overline{z}w +|\overline{z}w|^2 -|z|^2 +z\overline{w} +w\overline{z} -|w|^2 \\ &= 1 + |z|^2 |w|^2 -|z|^2 -|w|^2 \\ &= \left(1-|z|^2\right)\left(1-|w|^2\right).\end{align*}$

$\noindent The `deduce' part follows easily from this and is left as an exercise for the reader.$

Yeah, figured the second part of the question is pretty obvious. If $z$ or $w$ lie on the unit circle then $|z|$ or $|w|$ =1 and RHS of the equation derived =0 and so the statement is true.

Got another one, though:

$\\\text{Show that }\int_{0}^{1}x^m(1-x)^ndx=\frac{m!n!}{(m+n+1)!}\text{ for all integers }m,n\geq 0$

HeroicPandas · Jul 31, 2016

1008 said:
$\\\text{Show that }\int_{0}^{1}x^m(1-x)^ndx=\frac{m!n!}{(m+n+1)!}\text{ for all integers }m,n\geq 0$

$$Let $I = \int_0^1 x^m (1-x)^n dx $ and keep integrating by parts to reduce $x^m$ down to $x^0$,$ \\ \begin{align*} I &= \left[ -\frac{x^n(1-x)^{n+1}}{n+1} \right]_0^1 + \frac{m}{n+1} \int_0^1 x^{m-1}(1-x)^{n+1} dx \\&= \frac{m}{n+\color{red}{1}}\int_0^1 x^{m-\color{red}{1}}(1-x)^{n+\color{red}{1}}dx \\ &\stackrel{IBP}{=} \frac{m}{n+1} \left( \text{some junk that equals 0} + \frac{m-1}{n+\color{red}{2}} \int_0^1 x^{m-\color{red}{2}}(1-x)^{n+\color{red}{2}}dx\right ) \\ &= \frac{m(m-1)}{(n+1)(n+2)} \int_0^1 x^{m-2}(1-x)^{n+2}dx \\ &\vdots \\ &= \frac{m!}{(n+1)(n+2)\dots (n+\color{red}{m})} \int_0^1 x^{m-\color{red}{m}}(1-x)^{n+\color{red}{m}}dx \\ &= \frac{m!}{(n+1)(n+2)\dots (n+m)} \frac{1}{m+n+1} \\ &= \frac{n!}{n!}\frac{m! }{(n+1)(n+2)\dots (m+n+1)} \\ &= \frac{m!n!}{(m+n+1)!}\end{align*}$

InteGrand · Jul 31, 2016

1008 said:
Yeah, figured the second part of the question is pretty obvious. If $z$ or $w$ lie on the unit circle then $|z|$ or $|w|$ =1 and RHS of the equation derived =0 and so the statement is true.

Got another one, though:

$\\\text{Show that }\int_{0}^{1}x^m(1-x)^ndx=\frac{m!n!}{(m+n+1)!}\text{ for all integers }m,n\geq 0$

$\noindent There's several ways to do this. You can do it using a multi-parameter reduction as HeroicPandas has done above. Alternatively, recall that for any $k\in \left\{0,1,\ldots,N\right\}$, we have $\int_0 ^1 \binom{N}{k}x^{k}\left(1-x\right)^{N-k}\, \mathrm{d}x=\frac{1}{N+1}$, as can be shown using Bayes' Billiard Ball argument (see link below) or induction or however else you like.$

(Bayes' Billiard Ball argument mentioned here: http://community.boredofstudies.org...ng-mathematical-statements-5.html#post7091302 .)

$\noindent Rearranging the above, we obtain$

$\begin{align*}\int_0 ^1 x^{k}\left(1-x\right)^{N-k}\, \mathrm{d}x &= \frac{1}{N+1}\cdot \frac{1}{\binom{N}{k}}\\ &=\frac{1}{N+1} \cdot \frac{k! \left(n-k\right)!}{N!} \\ &= \frac{k! \left(N-k\right)!}{\left(N+1\right)!}.\end{align*}$

$\noindent This identity actually holds for any non-negative integer $N$ and integer $k$ with $0\leq k\leq N$ (the $N=0$ case can be confirmed directly).$

$\noindent Now, let $m,n$ be non-negative integers, and write $k=m \geq 0$, and let $m+n = N\geq 0$, so $N-k = n$. So our identity above becomes $\int _0 ^1 x^{m}\left(1-x\right)^{n}\, \mathrm{d}x = \frac{m!n!}{\left(m+n+1\right)!}$, as required. (In fact this identity (equivalent to the one with the $N$ and $k$) can be extended to cases where $m$ and $n$ aren't just integers but arbitrary numbers greater than $-1$, using the \textsl{gamma function} in place of factorials. You can read more about this on the Wikipedia page for the \textsl{beta function}:$

https://en.wikipedia.org/wiki/Beta_function ).

The gamma function page is here: https://en.wikipedia.org/wiki/Gamma_function .

1008 · Jul 31, 2016

HeroicPandas said:
$$Let $I = \int_0^1 x^m (1-x)^n dx $ and keep integrating by parts to reduce $x^m$ down to $x^0$,$ \\ \begin{align*} I &= \left[ -\frac{x^n(1-x)^{n+1}}{n+1} \right]_0^1 + \frac{m}{n+1} \int_0^1 x^{m-1}(1-x)^{n+1} dx \\&= \frac{m}{n+1}\int_0^1 x^{m-1}(1-x)^{n+1}dx \\ &\stackrel{IBP}{=} \frac{m}{n+1} \left( \text{some junk that equals 0} + \frac{m-1}{n+2} \int_0^1 x^{m-2}(1-x)^{n+2}dx\right ) \\ &= \frac{m(m-1)}{(n+1)(n+2)} \int_0^1 x^{m-2}(1-x)^{n+2}dx \\ &\vdots \\ &= \frac{m!}{(n+1)(n+2)\dots (n+m)} \int_0^1 x^{m-m}(1-x)^{n+m}dx \\ &= \frac{m!}{(n+1)(n+2)\dots (n+m)} \frac{1}{m+n+1} \\ &= \frac{n!}{n!}\frac{m! }{(n+1)(n+2)\dots (m+n+1)} \\ &= \frac{m!n!}{(m+n+1)!}\end{align*}$

Thanks, yeah I was applying the same approach with repeated IBP but trying to reduce (1-x) to the power of 0

1008 · Jul 31, 2016

InteGrand said:
$\noindent There's several ways to do this. You can do it using a multi-parameter reduction as HeroicPandas has done above. Alternatively, recall that for any $k\in \left\{0,1,\ldots,N\right\}$, we have $\int_0 ^1 \binom{N}{k}x^{k}\left(1-x\right)^{N-k}\, \mathrm{d}x=\frac{1}{N+1}$, as can be shown using Bayes' Billiard Ball argument (see link below) or induction or however else you like.$

(Bayes' Billiard Ball argument mentioned here: http://community.boredofstudies.org...ng-mathematical-statements-5.html#post7091302 .)

$\noindent Rearranging the above, we obtain$

$\begin{align*}\int_0 ^1 x^{k}\left(1-x\right)^{N-k}\, \mathrm{d}x &= \frac{1}{N+1}\cdot \frac{1}{\binom{N}{k}}\\ &=\frac{1}{N+1} \cdot \frac{k! \left(n-k\right)!}{N!} \\ &= \frac{k! \left(N-k\right)!}{\left(N+1\right)!}.\end{align*}$

$\noindent This identity actually holds for any non-negative integer $N$ and integer $k$ with $0\leq k\leq N$ (the $N=0$ case can be confirmed directly).$

$\noindent Now, let $m,n$ be non-negative integers, and write $k=m \geq 0$, and let $m+n = N\geq 0$, so $N-k = n$. So our identity above becomes $\int _0 ^1 x^{m}\left(1-x\right)^{n}\, \mathrm{d}x = \frac{m!n!}{\left(m+n+1\right)!}$, as required. (In fact this identity (equivalent to the one with the $N$ and $k$) can be extended to cases where $m$ and $n$ aren't just integers but arbitrary numbers greater than $-1$, using the \textsl{gamma function} in place of factorials. You can read more about this on the Wikipedia page for the \textsl{beta function}:$

https://en.wikipedia.org/wiki/Beta_function ).

The gamma function page is here: https://en.wikipedia.org/wiki/Gamma_function .

Wow thanks! The BoS link that you provided also has lots of other cool mathematical statements. I'll check those out too

leehuan · Jul 31, 2016

You know I made a thread for this stuff already

1008 · Jul 31, 2016

leehuan said:
You know I made a thread for this stuff already

That thread is for ALL 1231/1241 and 1251 subs. This one's exclusively for 1251 haha

However, seriously, if you want to post your questions here you're more than welcome. I prefer one thread per subject because there are various topics that are different between MATH1231/41/51.

Also, I know this may make me seem like a noob, but how would you do this one? I am surprised myself I can't do this...

$\int\frac{dx}{x^2\sqrt{x^2+16}}$

RenegadeMx · Jul 31, 2016

1008 said:
That thread is for ALL 1231/1241 and 1251 subs. This one's exclusively for 1251 haha

However, seriously, if you want to post your questions here you're more than welcome. I prefer one thread per subject because there are various topics that are different between MATH1231/41/51.

Also, I know this may make me seem like a noob, but how would you do this one? I am surprised myself I can't do this...

$\int\frac{dx}{x^2\sqrt{x^2+16}}$

trig substitution, think it was x=tan, for this case atleast

HeroicPandas · Jul 31, 2016

RenegadeMx said:
trig substitution, think it was x=4tan, for this case atleast

.

RenegadeMx · Jul 31, 2016

HeroicPandas said:
.

yeah mb its been 2 years since ive done this :<

1008 · Jul 31, 2016

RenegadeMx said:
trig substitution, think it was x=tan, for this case atleast

Yeah thanks, figured it out.

InteGrand · Jul 31, 2016

leehuan · Jul 31, 2016

Tan is becoming boring

$x=4\sinh t \implies dx = 4\cosh t \, dt$

$\int \frac{dx}{x^2\sqrt{x^2+16}}\\ = \int \frac{4\cosh t \, dt}{16\sinh^2 t \sqrt{16\cosh^2 t} }\\ = \frac{1}{16}\int \csc h^2 t\, dt\\ =-\frac{1}{16} \coth t + C \\ = -\frac{1}{16} \frac{\sqrt{16+x^2}}{x}+C$

Nice to know that \csch is not supported.

1008 said:
That thread is for ALL 1231/1241 and 1251 subs. This one's exclusively for 1251 haha

However, seriously, if you want to post your questions here you're more than welcome. I prefer one thread per subject because there are various topics that are different between MATH1231/41/51.

Also, I know this may make me seem like a noob, but how would you do this one? I am surprised myself I can't do this...

$\int\frac{dx}{x^2\sqrt{x^2+16}}$

A very small amount. They did complex numbers before probability and stats and we have that reversed. We also have a head start in calculus

1008 · Jul 31, 2016

RenegadeMx said:
yeah mb its been 2 years since ive done this :<

InteGrand said:
$\noindent And if you want to use a hyperbolic substitution, it is $x = 4\sinh t$. It may lead to a neater integral (hyperbolic substitutions often do), but you'll need to be familiar with hyperbolic identities and integrating hyperbolic analogs of the circular trig. functions.$

Yeah both methods work out very nicely. With the first one, you end up with $\int \frac{1}{16}cot(t)cosec(t)dt$ which gives $\frac{1}{16}cosec(t) +C$ while with the other approach you end up with $\int \frac{1}{16}cosech^2(t)dt$ which gives $\frac{-1}{16}coth(t) +C$

Then you sub. back in the x and voila!

Paradoxica · Jul 31, 2016

leehuan said:
Nice to know that \csch is not supported.

$\text{csch}(x)$

Code:

\text{csch}(x)

leehuan · Jul 31, 2016

Paradoxica said:
$\text{csch}(x)$

Code:

\text{csch}(x)

Laaaaaaaaaaaaaame

But I guess it's the best possible case

Paradoxica · Jul 31, 2016

leehuan said:
Laaaaaaaaaaaaaame

Not everything is supported on the codebase used by BOS, you should be an intelligent human being and learning how to improvise.

Oh wait you're supposed to be doing that all the time.

leehuan · Jul 31, 2016

Lol I knew how to improvise. I was just being in my lazy + complainer modes simultaneously.

MATH1251 Questions HELP (2 Viewers)

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