#gauß

[official-german-puns]

Gauß ist der Smooth Operator.

[honig-prinzessin]

Finishing my exercises in maths - finally my motivation is back 😊💪

[katslgr]

Got a chance to visit Göttingen for a talk. Observatory and workplace of Carl Friedrich Gauß.

[clyde420-69-666]

die Vermessung der Wel

[mzimmer275]

Das Ergebnis habe ich schon, jetzt brauche ich nur noch den Weg, der zu ihm führt.

Carl Friedrich Gauß (1777 - 1855), deutscher Mathematiker, Astronom, Geodät, Physiker

[cashgeeh]

1 is 'im fine' and im like ⌊0,9⌋

[hippasos]

Carl Friedrich Gauß was a great German mathematician of the 18th/19th century who founded the non-Euclidian Geometry and explored many theorems of Number Theory, normal distribution and integral calculus. We can see him on the 10 Deutsche Mark banknote. Many people know the story of the little Gauß. When Gauß was 9 years old his teatcher Büttner gave his pupils the task to calculate the sum of the numbers from 1 to 100. Gauß surprised his teacher by solving the problem in a very short time. He formed 50 pairs of the sum 101 (1 + 100, 2 + 99, 3 + 98,...) and got 5050 as a solution. With the words "Ligget se" (Low German: "Here it is") he put the paper on the teacher's desk.

This pattern of adding numbers is the same as in the Gaussian sum-formula

Normally you get this formula first by intuition, i.e. you add a few numbers and look at it carefully to get the general idea of the formula. Then you prove it by induction. We will see, how to solve such a formula by induction.

(Experts might skip the following proof. ;-))

Proof: First we have the beginning of induction. We show that the formula is true for one n in the natural numbers. When we are done, we will show that the formula is true for any n by proving that the formula is true for n+1. In this case (n+1), we assume that the formula is true for one n in the natural numbers. But if we have shown that the formula is true for one n in the naturals and for any n+1, then we can see the following. If it is true for n = 1, then it's also true for n = 1 + 1 = 2. But then it is true for n = 2 + 1 = 3, but then also for n = 3 + 1 = 4 and so on...

So we will have shown that the formula is true in general. That's the idea behind it. This is enough explanation, let's start the proof. ^^

Begining of Induction for n = 1: We just fill in n = 1:

(1*(1 + 1))/2 = (1*2)/2 = 2/2 = 1.

The formula is correct. :)

Induction step: We assume the formula is correct for one n in the natural numbers and prove the formula for n + 1, i.e.

((n+1) * (n + 1 + 1))/2 = ((n + 1)*(n + 2))/2 = (n² + 3n + 2)/2 = (n² + n)/2 + (2n + 2)/2 = (n² + n)/2 + (n + 1)

Now we can use our assumption: 1 + ... + n = (n² + n)/2.

(n² + n)/2 + (n + 1) = 1 + ... + n + (n + 1).

So we have: 1 + ... + n + (n + 1) = ((n+1) * ((n + 1) + 1))/2 and this is the formula for n + 1.    Q.E.D.

I think some people, who had mathematics in school, know this proof by induction. But next time I will describe an other way of proving this with a machinery of discrete differential and integral calculus.

We will construct an analogon to d/dx, the derivative of a function, and one to the integral.

I will use the following notation:

sum_{i = 1}^{n} i := 1 + 2 + ... + n for the sum from i = 1 to n and int_{x}^{y} for the integral in the interval [x,y].

It would be mean to stop here without giving you an intuition how the analoga look like. ;-)

Maybe you have already seen this, but we know that

sum_{i = 1}^{n} i = (n² + n)/2 = n²/2 + n/2.

If we look at the first summand n²/2 it reminds us of the usual antiderivate of f(x) = x, which is F(x) = x²/2 (Remark: because F'(x) = f(x)).

Let's look at sum_{i = 1}^{n} i² = 1 + 4 + 9 + 16 + ... + n². The formula is given by

(square pyramidal number)

It reminds us of the antiderivate of f(x) = x², which is F(x) = x³/3.

This shows that the sum has something to do with the integral and we will make this "has something to do with" more precise next time.

But if the sum is somthing like the integral, what's the analogon to the derivative of a function? O.o

That's a great question, but we will see, how we construct a discrete analogon to the derivative

Why should we construct these tools?

With these tools we will, find a formula for any sum which look like the two above (cf. e.g. Faulhaber's formula). Well, we can proof the formula for sums with induction (this could be an exercise for the second formula), but we always must have an idea of the general formula. With the tool we construct, we won't have to be creative. ;-)

This is a great advantage! See you then!

tl;dr

The Gaussian sum formula is given by

and can be used for the short addition of the natural numbers from 1 to 100. The proof is just induction. We can find discrete analogons to the derivate and the integral of a function in differential and integral calculus. The analogon of the integral is the sum. Next time we will construct an analogon to the derivate. These analogons will help us to find very easily “closed” formulas for diverse sums.

[fragile-splendors]

So does anybody want to explain the Gaussian elimination to me