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Alright travis. I am going to split my answer into two parts. I think if you understand the logic behind Gauss sum, you can very easily apply the same logic to your original problem and arrive at the formulas you desire.
We will only regard even n in the following as the gauss sum for uneven n directly follows.
1) Gauss sum
The idea behind Gauss sum is to make pairs: You take the first and the last element (1 and n) , the second and the second to last element (2 , n-1) and so on. Notice that all these pairs have sum n+1. Now you have the n elements split up into n/2 pairs of sum n+1, whence the whole sum is (n+1)*n/2 , gauss sum.
2) partial sums
I advise you that if you understand 1), try to apply the same logic yourself before simply reading the solution.
+ Show Spoiler +
Now instead of summing 1+2+...+n you just havfe the sum 1+2+...n/2 . This means our pairs are (1 +n/2), (2+n/2 - 1) and so on. There are (n/2)/2 number of pairs as you have n/2 elements. this leaves us with (n/4)*(1+ n/2)=1/8 * n * (n+2)
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Also I wanted to share in this thread one of my favorite proofs. The problem is very simple and so is the proof, once you see it, though it takes a lot of creativity to solve it yourself. I dare you to try before reading the solution 
"For all prime numbers n with n greater or equal to 5 holds (n^2) - 1 is dividable by 24."
( In other words (n^2) - 1 mod 24 = 0 )
Proof+ Show Spoiler +
First we note that (n+1)*(n-1) = (n^2)-1. (*)
as n is prime we know that n is uneven. As every second even number (4,8,12,...) is dividable by 4, we know that either
n+1 or n-1 is divided by 4 and the other by 2. Hence we know that 8 divides (*). Now we regard n-1,n,n+1 as 3 neigbours. for three neighboured natural numbers, exactly one of them is divided by 3. And it is not n as n is prime, hence is is n-1 or n+1 . Therefore (*) is also divided by 3, therefore by 3*8=24.
I really like it because you can explain it to friends or family who dont know maths to explain what a proof is (or you can be that guy at a party).
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On June 28 2017 05:58 Kleinmuuhg wrote:Also I wanted to share in this thread one of my favorite proofs. The problem is very simple and so is the proof, once you see it, though it takes a lot of creativity to solve it yourself. I dare you to try before reading the solution "For all prime numbers n with n greater or equal to 5 holds (n^2) - 1 is dividable by 24." ( In other words (n^2) - 1 mod 24 = 0 ) Proof + Show Spoiler +
First we note that (n+1)*(n-1) = (n^2)-1. (*)
as n is prime we know that n is uneven. As every second even number (4,8,12,...) is dividable by 4, we know that either
n+1 or n-1 is divided by 4 and the other by 2. Hence we know that 8 divides (*). Now we regard n-1,n,n+1 as 3 neigbours. for three neighboured natural numbers, exactly one of them is divided by 3. And it is not n as n is prime, hence is is n-1 or n+1 . Therefore (*) is also divided by 3, therefore by 3*8=24.
I really like it because you can explain it to friends or family who dont know maths to explain what a proof is (or you can be that guy at a party).
By extension of your proof, its applicable to all odd numbers that are indivisible by 3 and not necessarily prime.
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sure, but that doesnt sound as nice ^^
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Hey, travis.
On June 28 2017 04:43 travis wrote:
Could someone help me with summation algebra
I keep getting problems where I am summing to some portion of n, say
the sum from i=1 to (n/2) of i
now, I know that the sum i=1 to n of i is gauss sum:
1/2n(n+1)
I know this because I have memorized it
but when we sum to n/2 it becomes 1/8n(n+2)
I know this because wolfram alpha tells me so
but how can I manually solve this sum? Is there some rule for what happens to the sum when I only sum to a portion of my "n" ?
It really isn't complicated. The formula for the finite sum as you know is this.
![[image loading]](http://i.imgur.com/JowFKPW.gif)
If the upper limit of the sum changes to n/2, then wherever we see an n in the formula, we replace it with n/2. Factor out a 1/2 to get Mathematica's answer.
![[image loading]](http://i.imgur.com/iGkK6LK.gif)
Sincerely,
Shalashaska_123
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How do you differentiate x^x with respect to x?
Do you - if so how - apply the chain rule?
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On July 28 2017 22:52 HKTPZ wrote:
How do you differentiate x^x with respect to x?
Do you - if so how - apply the chain rule?
You re-write your function as e^f(x) and you should be able to go on from there.
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I tried my luck googling and going through youtube videoes but I couldnt find anything so here goes:
When you are solving a seperable differential equation by seperating the variables, how is splitting up dy/dx justified (apart from the fact that the technique is without a doubt producing the right answer) and what is happening? Do you multiply each side of the equation by an infinitely small change of x?
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On August 03 2017 05:55 HKTPZ wrote:
I tried my luck googling and going through youtube videoes but I couldnt find anything so here goes:
When you are solving a seperable differential equation by seperating the variables, how is splitting up dy/dx justified (apart from the fact that the technique is without a doubt producing the right answer) and what is happening? Do you multiply each side of the equation by an infinitely small change of x?
Ah yes, this question touches on higher level calculus and my uni professor's answer to it in a calculus class was more or less "don't worry about it". If you want a fairly abstract and readable discussion on it, I'd recommend this stackexchange answer, or, if you prefer, this more technical blogpost.
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On August 03 2017 06:58 Warfie wrote:Show nested quote +On August 03 2017 05:55 HKTPZ wrote:
I tried my luck googling and going through youtube videoes but I couldnt find anything so here goes:
When you are solving a seperable differential equation by seperating the variables, how is splitting up dy/dx justified (apart from the fact that the technique is without a doubt producing the right answer) and what is happening? Do you multiply each side of the equation by an infinitely small change of x? Ah yes, this question touches on higher level calculus and my uni professor's answer to it in a calculus class was more or less "don't worry about it". If you want a fairly abstract and readable discussion on it, I'd recommend this stackexchange answer, or, if you prefer, this more technical blogpost.
It could also be fun to note that in algebraic geometry you introduce [Kähler] differentials on curves and surfaces basically by saying "let's consider the vector space of things of the form dx" and just treating it formally, and it ends up behaving quite like the differential geometrical notions you're discussing
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There is a lot of difference between the answers when studying maths and when studying physics. If you are studying physics, the answer is: Because it works.
If you are studying maths, you have to get into what total differentials mean.
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I'd guess your first step is to learn about limits from a formal perspective, and then to learn about different kinds of spaces and representations of them and how you end up doing limits in them
Maybe a course on analysis is sufficient?
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Actually, mathematically this is not permitted, as it is not formally correct. Often times you would rather use substitutions that are permitted. Or you can (and this is common practice) informally calculate a solution using that trick and then afterwards show that it solves the diff eq. This however is no better than saying:oh look this function magically appeared and now I can prove it solves my issue.
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On August 05 2017 01:52 Kleinmuuhg wrote:
Actually, mathematically this is not permitted, as it is not formally correct. Often times you would rather use substitutions that are permitted. Or you can (and this is common practice) informally calculate a solution using that trick and then afterwards show that it solves the diff eq. This however is no better than saying:oh look this function magically appeared and now I can prove it solves my issue.
Which for differential equations is usually how it works anyways. You have theorems that prove that if a solution with the necessary amount of degrees of freedom exists, it is the only one. So if you have one, no matter where you got it from, that is all the information you need.
And i think if you interpret the "dx" and "dy" as total differentials, you can actually do make that approach work.
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On August 05 2017 05:06 Simberto wrote:Show nested quote +On August 05 2017 01:52 Kleinmuuhg wrote:
Actually, mathematically this is not permitted, as it is not formally correct. Often times you would rather use substitutions that are permitted. Or you can (and this is common practice) informally calculate a solution using that trick and then afterwards show that it solves the diff eq. This however is no better than saying:oh look this function magically appeared and now I can prove it solves my issue. Which for differential equations is usually how it works anyways. You have theorems that prove that if a solution with the necessary amount of degrees of freedom exists, it is the only one. So if you have one, no matter where you got it from, that is all the information you need. And i think if you interpret the "dx" and "dy" as total differentials, you can actually do make that approach work.
Even if interpreted as total differentials there is no general concept of simplifying differentials in general, I think. Im not quite sure on that though, since "total differential" is something that I dont really meet in the kind of functional analysis im usually concerned with..
I seem to remember an exercise from a couple of years ago that gave an example in which simplifying the application of dx/dy * dy/dz * dz/dx to an equation to 1 would yield a wrong solution. Ill see if I can dig it up.
From my first couple of classes on ordinary differential equations I seem to remember that the proofs of the proper mathematical way of solving these problems involved integration of an inverse function and applying it to the original problem, which relies heavily on the involved functions being smooth homeomorphisms (at least locally) and ranges/domains getting along. These two things are often times not checked at all by engineers/physicists that Ive been working with.
Edit: Found it!
By an application of the implicit function theorem, you can show that for the function F with open domain Omega and for positive constants a,b,R
F: \R^3\supseteq \Omega \to \R, (p,v,T)\mapsto [p+a/(v^2)](v-b)-RT
the identity
\diff v/\diff T * \diff T/\diff p * \diff p/\diff v = -1
holds on D=\set{ (p,v,T)\in\R^3 : v>b, \diff F/\diff v(p,v,T) \neq 0}\subseteq \Omega.
(Here it is assumed that the function has a zero in D, i think).
Interpreting p as pressure, v as molar volume and T as absolute temperature, the equation F(p,v,T)=0 (or p=RT/(v-b) - a/(v^2) ) should be the Van-der-Vals-equation, which approximates the behaviour of real gases, iirc.
So yeah, in this case, the differentials cannot be simplified, as that would yield 1 instead of -1 in the above equation.
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TL are my bros so yall get this secret:
The best math resource I have EVER found is this blog:
https://betterexplained.com/cheatsheet/
(I have a Masters in Chemistry with Math and moonlight as a high school Math teacher, and this shit is so legit it helps my students and helps me like crazy)
If we want science to go really fast, and get to the future sooner, we need to massively speed up our math education (especially in the West). We have the Education knowledge of 'how to teach well', we just need to apply that to our math classes.
This BetterExplained blog does that brilliantly, and I strongly suggest everyone check it out.
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Back in school, doing a linear algebra review assignment.
It has me plot a couple vectors in R2 and then the vectors under a transformation.
The transformation matrix is [0 1; 1 0] --->(the ; denotes the next line of the matrix)
So I can see that what is happening to my vectors is that the x and y coordinates are swapped.
The assignment wants me to "Describe geometrically what T does to each vector x in R2". Is my answer that the coordinates are swapped an acceptable answer? Is there some other way to describe what this transformation matrix is doing? Thank you.
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Swapping coordinates doesn't sound very geometrically.
I assume the answer that is expected is "mirrored on the straight line that goes from (0/0) to (1/1) (and further)" (Don't know what it is called in english, it has a special name in german. Could be something like "first Median")
Obviously any other straight line that goes in a 45° angle between the coordinate axis would also be fine if you are only talking about vectors. If you are talking about points, you need exactly that one.
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EPT
