# MATH1251 Questions HELP (1 Viewer)

#### 1008

##### Active Member
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:

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

#### InteGrand

##### Well-Known Member
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:

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

\bg_white \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*}

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

#### 1008

##### Active Member
$\bg_white \noindent We have$

\bg_white \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*}

$\bg_white \noindent The deduce' part follows easily from this and is left as an exercise for the reader.$
Yeah, figured the second part of the question is pretty obvious. If $\bg_white z$ or $\bg_white w$ lie on the unit circle then $\bg_white |z|$ or $\bg_white |w|$=1 and RHS of the equation derived =0 and so the statement is true.

Got another one, though:

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

#### HeroicPandas

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

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#### InteGrand

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

Got another one, though:

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

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

$\bg_white \noindent Rearranging the above, we obtain$

\bg_white \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*}

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

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

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

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

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#### 1008

##### Active Member
\bg_white 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

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

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

$\bg_white \noindent Rearranging the above, we obtain$

\bg_white \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*}

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

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

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

#### 1008

##### Active Member
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...

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

##### Kosovo is Serbian
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...

$\bg_white \int\frac{dx}{x^2\sqrt{x^2+16}}$
trig substitution, think it was x=tan, for this case atleast

#### HeroicPandas

##### Kosovo is Serbian
yeah mb its been 2 years since ive done this :<

#### 1008

##### Active Member
trig substitution, think it was x=tan, for this case atleast
Yeah thanks, figured it out.

#### InteGrand

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

#### leehuan

##### Well-Known Member
Tan is becoming boring

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

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

Nice to know that \csch is not supported.

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...

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

##### Active Member
yeah mb its been 2 years since ive done this :<
$\bg_white \noindent And if you want to use a hyperbolic substitution, it is x = 4\sinh t. It may lead to a neater integral (hyperbolic substitutions often do), but you'll need to be familiar with hyperbolic identities and integrating hyperbolic analogs of the circular trig. functions.$
Yeah both methods work out very nicely. With the first one, you end up with $\bg_white \int \frac{1}{16}cot(t)cosec(t)dt$ which gives $\bg_white \frac{1}{16}cosec(t) +C$ while with the other approach you end up with $\bg_white \int \frac{1}{16}cosech^2(t)dt$ which gives $\bg_white \frac{-1}{16}coth(t) +C$

Then you sub. back in the x and voila!

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##### -insert title here-
Nice to know that \csch is not supported.
$\bg_white \text{csch}(x)$

Code:
\text{csch}(x)

#### leehuan

##### Well-Known Member
$\bg_white \text{csch}(x)$

Code:
\text{csch}(x)
Laaaaaaaaaaaaaame

But I guess it's the best possible case

##### -insert title here-
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

##### Well-Known Member
Lol I knew how to improvise. I was just being in my lazy + complainer modes simultaneously.