Mathematics Thread

[samehsameh]

He doesnt need to justify everything if its obvious. If people had to justify things that had already been proven, then proofs would be so much longer.
\( {\text{Need to show that }}\sum\limits_{i = 1}^n i = \frac{{n(n + 1)}}{2}. \)
\( {\text{Let }}n{\text{ be some small value, say 5}}{\text{.}} \)
\(
{\text{Then}} \hfill \\
\)
\(
2\sum\limits_{i = 1}^n i = 2\sum\limits_{i = 1}^5 i = \sum\limits_{i = 1}^5 {2i} = 2 + 4 + 6 + 8 + 10 \)\( = (n - 3) + (n - 1) + (n + 1) + (n + 3) + 2n = 6n \hfill \\
\)
\( {\text{We already said }}6 = (n + 1){\text{, in the expansion so}} \)
\( 6n = (n + 1)n. \)
\( {\text{So}} \)
\( 2\sum\limits_{i = 1}^n i = (n + 1)n, \)
\( \sum\limits_{i = 1}^n i = \frac{{n(n + 1)}}{2},{\text{for}} n = 5{\text{.}} \)
\( {\text{Claim }}\frac{{n(n + 1)}}{2}{\text{, is correct for all }}n. \)
\(
{\text{Proof by induction}}{\text{.}} \hfill \\
{\text{Base case }}n = 1 \hfill \\ \)
\( \sum\limits_{i = 1}^1 i = 1,{\text{ and }}\frac{{n(n + 1)}}{2} = 1.{\text{ Condition holds}}{\text{.}} \)
\(
{\text{Inductive step}}{\text{.}} \hfill \\
{\text{Assume that }}\sum\limits_{i = 1}^k i = \frac{{k(k + 1)}}{2}{\text{ holds}}{\text{.}} \hfill \\
\)
\(
{\text{Then need to prove }}\sum\limits_{i = 1}^{k + 1} i = \frac{{(k + 1)(k + 2)}}{2} \hfill \\
\)
\(
\sum\limits_{i = 1}^{k + 1} i = \sum\limits_{i = 1}^k i + (k + 1) \hfill \\
{\text{ }} = \frac{{k(k + 1)}}{2} + (k + 1) \hfill \\ \)
\(
{\text{ }} = \frac{{k(k + 1)}}{2} + \frac{{2(k + 1)}}{2} \hfill \\
{\text{ }} = \frac{{k(k + 1) + 2(k + 1)}}{2} = \frac{{(k + 1)(k + 2)}}{2} \hfill \\
\)
\( {\text{so }}\sum\limits_{i = 1}^n i = \frac{{n(n + 1)}}{2}{\text{ holds for all }}n. \)

But you knew that already, so why be unnecessarily pedantique?

 

[ben1996123]

Or you could just use \( S_n=\frac{n(2a+(n-1)d)}{2} \)

 

[Christopher Mowla]

I don't know if anyone would be interested in this, but I figured out how to use Pascal's Triangle to write the power of sums formula for any integer power (greater than i^0) from my recurrence relation.

1) Draw a Pascal's Triangle upside down:

​

2) Label as follows:
​

​

Here's a direct iterative process (you don't have to deal with a series anymore).

Note that the iterations really start for i^1 and greater.

Spoiler

Given that \( \sum\limits_{i=1}^{n}{\left( 1 \right)}=n \), from the adjusted Pascal's Triangle,

\( \sum\limits_{i=1}^{n}{i}=\frac{1}{2}\left( n \right)+\frac{1}{2}\left( n \right)\left[ n \right]=\frac{n\left( n+1 \right)}{2} \)

\( \sum\limits_{i=1}^{n}{i^{2}}=\frac{1}{3}\left( n^{2} \right)+\frac{1}{3}\left( \left( 2n+1 \right)\left[ \frac{n\left( n+1 \right)}{2} \right]-n\left[ n \right] \right)=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \)

\( \sum\limits_{i=1}^{n}{i^{3}}=\frac{1}{4}\left( n^{3} \right)+\frac{1}{4}\left( \left( 3n+3 \right)\left[ \frac{n\left( n+1 \right)\left( 2n+1 \right)}{2} \right]-\left( 3n+1 \right)\left[ \frac{n\left( n+1 \right)}{2} \right]+n\left[ n \right] \right) \)
\( =\frac{n^{2}\left( n+1 \right)^{2}}{4} \)

\( \sum\limits_{i=1}^{n}{i^{4}}=\frac{1}{5}\left( n^{4} \right)+\frac{1}{5}\left( 4n+6 \right)\left[ \frac{n^{2}\left( n+1 \right)^{2}}{4} \right]-\frac{1}{5}\left( 6n+4 \right)\left[ \frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right] \)
\( +\frac{1}{5}\left( 4n+1 \right)\left[ \frac{n\left( n+1 \right)}{2} \right]-\frac{1}{5}n\left[ n \right]=\frac{n\left( n+1 \right)\left( 2n+1 \right)\left( 3n^{2}+3n-1 \right)}{30} \)

\( \sum\limits_{i=1}^{n}{i^{5}}= \)
\( \frac{1}{6}n^{5}+\frac{1}{6}\left( 5n+10 \right)\left[ \frac{n\left( n+1 \right)\left( 2n+1 \right)\left( 3n^{2}+3n-1 \right)}{30} \right]-\frac{1}{6}\left( 10n+10 \right)\left[ \frac{n^{2}\left( n+1 \right)^{2}}{4} \right] \)
\( +\frac{1}{6}\left( 10n+5 \right)\left[ \frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right]-\frac{1}{6}\left( 5n+1 \right)\left[ \frac{n\left( n+1 \right)}{2} \right]+\frac{1}{6}n\left[ n \right] \)
\( \frac{n^{2}\left( n+1 \right)^{2}\left( 2n^{2}+2n-1 \right)}{12} \)

1) The fractions correspond directly to the row numbers in the triangle.
2) For the minus signs, it is only negative when the first (left) number in the pair of numbers you choose from the triangle.
That is, for example, when we use two 4's and the 6 in the triangle to find \( \sum\limits_{i=1}^{n}{i^{4}} \),
\( \begin{matrix}
4 & -6 \\
\end{matrix}\ne \left( 4n-6 \right),\text{ but}\begin{matrix}
-6 & 4 \\
\end{matrix}=-\left( 6n+4 \right) \) (the sign in the binomials is always positive).
3) The numbers in the adjusted Pascal's Triangle are never negative in the left-most column, but the pattern follows the idea that they are the first negative numbers, and then every other number afterwords is negative.

You can also just do it directly (without iterations) by looking at the adjusted triangle. That is, view each of the following calculations as independent ones.

Spoiler

\( \sum\limits_{i=1}^{n}{i^{1}}=\frac{1}{2}\left( n+n\left( n \right) \right)=\frac{n\left( n+1 \right)}{2} \)

\( \sum\limits_{i=1}^{n}{i^{2}}=\frac{1}{3}\left( n^{2}+\left( 2n+1 \right)\frac{1}{2}\left[ n+n\left( n \right) \right]-n\left( n \right) \right)=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \)

\( \sum\limits_{i=1}^{n}{i^{3}}=\frac{1}{4}\left( \begin{matrix}
n^{3}+\left( 3n+3 \right)\frac{1}{3}\left[ n^{2}+\left( 2n+1 \right)\frac{1}{2}\left[ n+n\left( n \right) \right]-n\left( n \right) \right] \\
-\left( 3n+1 \right)\frac{1}{2}\left[ n+n\left( n \right) \right]+n\left( n \right) \\
\end{matrix} \right)=\frac{n^{2}\left( n+1 \right)^{2}}{4} \)

\( =\frac{n\left( n+1 \right)\left( 2n+1 \right)\left( 3n^{2}+3n-1 \right)}{30} \)

\( =\frac{n^{2}\left( n+1 \right)^{2}\left( 2n^{2}+2n-1 \right)}{12} \)

 

[ben1996123]

Have you found an equation for \( \sum_{i=1}^{n}i^{i} \)? I typed it in to Wolfram Alpha and it didn't give me one.

Edit: I find it pretty cool that \( \sum_{i=1}^{n}i^{3}=(\sum_{i=1}^{n}i)^{2} \)

 

[jonlin]

Could anyone give me an explanation why 0! is 1?

I might sound stupid...

 

[ben1996123]

Could anyone give me an explanation why 0! is 1?

I might sound stupid...

By definition.

 

[aronpm]

Could anyone give me an explanation why 0! is 1?

I might sound stupid...

http://en.wikipedia.org/wiki/Empty_product

if you multiply nothing together, you get the multiplicative identity, 1. if you add nothing together, you get the additive identity, 0.

@Ben: saw this on my tutorial question sheet, lol'd: http://i.imgur.com/B7Av1.png

 

[whauk]

Could anyone give me an explanation why 0! is 1?

I might sound stupid...

you can also define factorials via the gammafunction: n!:=gamma(n+1)
and gamma(1)=1 therefore 0!=1
read more here: http://en.wikipedia.org/wiki/Gamma_function

if 0! would be 0, you would have some problems with things like 1/(n!) because you would always have to define the special case n=0 seperately. and as mathematicians are lazy...

 

[IanTheCuber]

Here's a wierd one:

Add parenthesees to make the expression 105: 9+8x7-6x5+4x3+2

I found it on the internet, they made it up-he knows there is a solution, but he forgot it a long time ago, so therefore I don't know the answer.

 

[tozies24]

Here's a wierd one:

Add parenthesees to make the expression 105: 9+8x7-6x5+4x3+2

I found it on the internet, they made it up-he knows there is a solution, but he forgot it a long time ago, so therefore I don't know the answer.

I got 103 with (9+8)*7 - 6x5+4x3+2 in about 2 minutes. I'll keep looking.

EDIT: FOUND IT: (9+8)*(7-6)*5+4*(3+2)

 

[ben1996123]

Here's a wierd one:

Add parenthesees to make the expression 105: 9+8x7-6x5+4x3+2

I found it on the internet, they made it up-he knows there is a solution, but he forgot it a long time ago, so therefore I don't know the answer.

((9+8)(7-6))5+4(3+2)

 

[samehsameh]

No one attempted my 'challenge'
\( \int_{-\infty}^{\infty} \! \frac{x\sin x}{x^2+a^2} \, \mathrm{d} x, a > 0 \)

p.s. i dont think it can be calculated by standard calculus, if that makes it more appealing to prove me wrong.

 

[ben1996123]

Here are a few random proofs for anyone who has thought about these things before.

\( \sqrt{i}=\frac{i+1}{\sqrt{2}} \)

Spoiler

Use Euler's formula: \( e^{ix}=cos(x)+isin(x) \) with \( x=\frac{\pi}{2} \)
\( e^{\frac{i\pi}{2}}=cos(\frac{\pi}{2})+isin(\frac{\pi}{2})=0+1i=i \)
Write \( \sqrt{i} \) as \( e^{\frac{i\pi}{2}\times\frac{1}{2}}=e^{\frac{i\pi}{4}} \)
\( e^{\frac{i\pi}{4}}=cos(\frac{\pi}{4})+isin(\frac{\pi}{4}) \)
\( \sqrt{i}=e^{\frac{i\pi}{4}}=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}=\frac{i+1}{\sqrt{2}} \)
Q.E.D

\( i^{i}=\frac{1}{\sqrt{e^{\pi}}} \)

Spoiler

\( i=e^{\frac{i\pi}{2}} \) (see previous proof)
\( i^{i}=e^{\frac{i\pi}{2}\times i}=e^{\frac{i^{2}\pi}{2}}=e^{\frac{-\pi}{2}}=\frac{1}{\sqrt{e^{\pi}}} \)
Q.E.D

\( \sum_{r=0}^{n}{^nC_r}=2^{n} \)

Spoiler

\( (a+b)^{n}={^nC_0}a^{n}+{^nC_1}a^{n-1}b+{^nC_2}a^{n-2}b^{2}+{^nC_3}a^{n-3}b^{3}+\cdots+{^nC_n}b^{n} \)
Let a = 1 and b = 1 so \( (a+b)^{n}=(1+1)^{n}=2^{n} \)
Therefore \( a^p=1 \) and \( b^q=1 \)
So \( 2^{n}={^nC_0}1^{n}+{^nC_1}1^{n-1}1+{^nC_2}1^{n-2}1^{2}+{^nC_3}1^{n-3}1^{3}+\cdots+{^nC_n}1^{n} \)\( ={^nC_0}+{^nC_1}+{^nC_2}+{^nC_3}+\cdots+{^nC_n}=\sum_{r=0}^{n}{^nC_r} \)
Q.E.D

 

[ben1996123]

I just found this: \( e^{x}\equiv sinh(x)+cosh(x) \)

 

[Christopher Mowla]

I just found this: \( e^{x}\equiv sinh(x)+cosh(x) \)

That's simply because of the definition of the hyperbolic functions:

\( \cosh \left( x \right)=\frac{e^{x}+e^{-x}}{2} \)

\( \sinh \left( x \right)=\frac{e^{x}-e^{-x}}{2} \)

 

[ben1996123]

That's simply because of the definition of the hyperbolic functions:

\( \cosh \left( x \right)=\frac{e^{x}+e^{-x}}{2} \)

\( \sinh \left( x \right)=\frac{e^{x}-e^{-x}}{2} \)

Oh yeah, didn't think of that, I did taylor series

 

[MalusDB]

Did A level, got a B. I was lazy. Did all sorts really. Pure maths becomes a bore eventually for me, always preferred physics. Knowledge without application is just wasteful to me. Not to say pure doesn't have applications, they just didn't present them at school. "Learn now, understand later" attitude. Sucked.

 

[JonWhite]

Here are a few random proofs for anyone who has thought about these things before.

\( \sqrt{i}=\frac{i+1}{\sqrt{2}} \)

Spoiler

Use Euler's formula: \( e^{ix}=cos(x)+isin(x) \) with \( x=\frac{\pi}{2} \)
\( e^{\frac{i\pi}{2}}=cos(\frac{\pi}{2})+isin(\frac{\pi}{2})=0+1i=i \)
Write \( \sqrt{i} \) as \( e^{\frac{i\pi}{2}\times\frac{1}{2}}=e^{\frac{i\pi}{4}} \)
\( e^{\frac{i\pi}{4}}=cos(\frac{\pi}{4})+isin(\frac{\pi}{4}) \)
\( \sqrt{i}=e^{\frac{i\pi}{4}}=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}=\frac{i+1}{\sqrt{2}} \)
Q.E.D

i'm sorry to say that you have foolishly taken one of the longest paths to show this. A sixth grader could easily cross multiply and square. Unless you're in sixth grade, then your knowledge (not its application though) is impressive.

I'm having trouble with optimization questions where I have to inscribe an object inside another. Hopefully, if someone could solve this for me, I can do the other ones.
Determine the area of the largest rectangle that can be inscribed inside a semicircle with a radius of 10 units. Place the length of the rectangle along the diameter.

Spoiler: Answer

"100 square units when 5√2"

Set the center of the semicircle at (0, 0). Let the upper right corner of the rectangle be (x, y). Note that this falls on the circle; thus, \( x^2 + y^2 = 100 \). The height of the rectangle is y, the length is 2x, and the area is 2xy. Now apply AM-GM:

\( xy = \sqrt{x^2y^2} \le \frac{x^2+y^2}{2} = 50 \)
\( 2xy \le 100. \)

The equality condition occurs if both terms are equal, so x = y. From there we have \( x = y = 5\sqrt{2} \).

 

[aronpm]

always preferred physics. Knowledge without application is just wasteful to me.

You realise how much of physics has no relevant applications for you right now, right?

And it's certainly not wasteful. Science is about the process of discovery and learning and understanding (just like mathematics is). Application is completely irrelevant and the demand that science needs to have an application is actually pretty insulting in my opinion.

I'm studying physics at university because I want to learn more and have a better understanding of how the universe works, not because I want a qualification to get a job. If I wanted to study something that would be "useful" in applying to the real world, I would have picked engineering.

i'm sorry to say that you have foolishly taken one of the longest paths to show this. A sixth grader could easily cross multiply and square. Unless you're in sixth grade, then your knowledge (not its application though) is impressive.

I disagree. Taking an expression like sqrt(i) and then changing it to the form (i+1)/sqrt2 is a much better (read: more satisfying) proof than assuming sqrt(i)=(i+1)/sqrt(2) and showing that it's true. It's the difference between testing a statement and deriving a statement. I think what you said sounded really rude.

re: the optimisation question, it would probably be better to show him a calculus solution because he can apply the ideas to more situations, as opposed to using the AM-GM inequality.

Using calculus I would do:

Spoiler

set (0,0) as the origin, (x,y) is the point of the the corner on the rectangle that lies on the semi-circle

\( A = 2xy \)
From trig:

\( x = 10sin(\theta) \)
\( y = 10cos(\theta) \)
\( A = 100*2sin(\theta)cos(\theta) = 100sin(2\theta) \)
Where \( \theta \) is the angle between the positive x-axis and the line from (0,0) to (x,y).

If we plot a graph of \( y=100sin(2x) \) (not the same x/y in the question!) we can see that as x varies in [0, pi/2], y varies from [0, unknown maxima], being 0 at both ends, so if we find where the derivative is zero, that will be the maxima. If you're writing for a test or exam you'll want to explain that better, probably using the second derivative test for maxima/minima.

\( \frac{dA}{d\theta} = \frac{d}{d\theta}(100sin(2\theta)) = 200cos(2\theta) \)

By setting that to zero we get

\( 0 = 200cos(2\theta) \)
\( 0 = cos(2\theta) \)

which has a solution for theta (between 0 and pi/2) where
\( 2\theta = \frac{\pi}{2} \)

Substituting this back into the formula for the area
\( A = 100sin(2\theta) = 100sin(\frac{\pi}{2}) = 100 \)

If you're also asked for the dimensions of the rectangle, just substitute theta back into x and y.

Alternative solution would be to say that

Spoiler

\( A^2 = 4x^2y^2 = 4x^2(100-x^2) \) (from \( x^2 + y^2 = 100 \))
then realising that because A is an area, it only varies between 0 and the maximum, so A squared will vary between 0 and maximum^2. If you differentiate \( A^2 \) and solve for x, you get \( x=5\sqrt{2} \) which can be plugged back into \( A^2 = 4x^2(100-x^2) \) and you can solve A.

I think the alternative is shorter but I would prefer to do the first one.

 

[Hyrtsi]

Anyone into prime numbers and big numbers in general? Looking for primes (6^n + 1) I found out that 6^100352 + 1 may be a prime. Primality tests for a number like this take a lot of time. Any ideas to reduce the calculating time? Rather than calculating 6^100352 + 1, I thought of changing it into this: (((((6^2)^2)^2)...)^2) + 1.