Apfelkuchen

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+104

0

Mathematisch kann 0 ⁰ nichtmal 0 sein.

Wir wissen: x ⁰ = x ^a-a = (x ^a)/(x ^a) = 1 ∀x≠0
Teilen durch 0 ist bekanntlich nicht erlaubt, daher wäre ein Ansatz wie (0 ^a)/(0 ^a) bereits nichtig.

Aber erinnern wir uns, ein Polynom nullten Grades ist per Def. eine Konstante. Wir bilden uns also eine Funktion f(x) = x ⁰ mit den oben genannten Eigenschaften.
Gesucht ist nun aber der Funktionswert an der Stelle x = 0.

Für solche Fälle ziehe ich gerne das Sandwich-Theorem vor. Notwendig hierfür ist nur eine ausreichende obere und untere Schranke, was in beiden Fällen mit x ⁰ bestens erfüllt ist und den Grenzwert für 0 ⁰ angibt.
Wir wissen:
1 ⁰ = 2 ⁰ = ... = x ⁰ = 1 ∀x>0
(-1) ⁰ = (-2) ⁰ = ... = x ⁰ = 1 ∀x<0

Nach dem Sandwich-Theorem gilt: f(x) ≤ g(x) ≤ h(x)
Dabei ist f(x) = x ⁰ ∀x<0 und g(x) = x ⁰ ∀x>0.

g(x) sei prinzipiell auch x ⁰, über den Limes nähern wir uns aber von links und von rechts an 0 an, daher wird der Ausdruck frech mit 0 ⁰ ersetzt.
Also x ⁰ ≤ 0 ⁰ ≤ x ⁰
Linke Seite:
lim x ⁰ = 1
x→-0

Rechte Seite:
lim x ⁰ = 1
x→+0

Wir nähern uns also von links (x → -0) und von rechts (x → +0) an die 0 an. Beide Grenzwerte ergeben den gleichen Wert, nämlich 1. Also muss
0 ⁰ ebenfalls 1 sein.

Andere herangehensweise:
0 ⁰ lässt sich auch ausdrücken mit dem Grenzwert der eulerschen Zahl.
Grenzwert e.png

Der Grenzwert von x → ∞ liefert für 1/x bekanntlich nahezu 0. cos(0) ist 1. 0 ⁰ ist also äquivalent zu (1-1/cos(0)) ⁰.

Und ja, ich versuche hier gerade mit den abstrusesten Möglichkeiten 0 ⁰ = 1 in der Analysis zu beweisen.

Apfelkuchen13.09.2013

+104

0

-10z ≤ 15 [/(-10)]

z ≥ -3/2

Apfelkuchen13.09.2013

+104

0

f(x) = -x ³ + bx ² + cx + d
f(x) = 0 with x ₀ = 0, x ₁ = 1, x ₂ = k

Because of x ₀ we know: d = 0

So we have the equation f(x) = -x ³ + bx ² + cx
f(1) = -(1 ³) + b(1 ²) + c1 = 0 b + c = 1 b = 1-c

Remainder of division is 8, so we conclude:
(-x ³ + bx ² + cx):(x-3) = -x ² + x(b-3) + (c + 3(b - 3)) + 3(c + 3(b - 3))/(x-3)

3(c + 3(b - 3)) = 8 = 3c + 9b - 27 3c + 9b = 35

Insert b = 1-c into the last equation and solve for c:
3c + 9(1-c) = 35 = 3c + 9 - 9c = 9 - 6c c = -26/6 = -13/3

Insert c = -13/3 into f(1):
b = 1-(-13/3) = 16/3

So we have the final equation: f(x) = -x ³ + (16x ²)/3 - 13x/3 = -x(x ² - 16x/3 + 13/3)

Now just solve x ² - 16x/3 + 13/3 = 0

[input]x^2 - 16x/3 + 13/3 = 0[/input]

Edit: Misread. Actual roots are x ₀ = 1, x ₁ = 2, x ₂ = k
So we have:
f(1) = -(1 ³) + b(1 ²) + c1 + d = 0 b + c + d = 1
f(2) = -(2 ³) + b(2 ²) + 2c + d = -8 + 4b + 2c + d = 0 4b + 2c + d = 8
f(k) = -(k ³) + b(k ²) + kc + d = 0

So the actual division would result into:
(-x ³ + bx ² + cx + d):(x-3) = -x ² + x(b-3) + (c + 3(b - 3)) + (3(c + 3(b - 3)) + d)/(x-3)

(3(c + 3(b - 3)) + d) = 8 9b + 3c + d = 35

Gauss-Jordan:
[1] 1 1 1 1
[2] 4 2 1 8
[3] 9 3 1 35

[2] - [1], [3] - [1]
[1] 1 1 1 1
[2] 3 1 0 7
[3] 8 2 0 34

[3] - 2x[2]
[1] 1 1 1 1
[2] 3 1 0 7
[3] 2 0 0 20 2b = 20 b = 10

Insert b = 10 into [2]
3*10 + c = 7 c = 7-30 = -23

Insert b = 10 and c = -23 into [1]
10 - 23 + d = 1 d = 14

Final equation f(x) = -x ³ + 10x ² - 23x + 14

Apfelkuchen13.09.2013

+104

0

Potenzsummen.png

Apfelkuchen10.09.2013

+104

0

So the square root of 2 would also be a rational number, because 2 is rational? Of course not.
6 is a rational number, thats correct.
But as we know, every square root of a number n is either an irrational number or an integer.

Claim: sqrt(6) is a rational number.
Therefore we could create an equation like sqrt(6) = n/k with n,k being integers > 1.
6 = (n/k)^2 = (n^2)/(k^2)
6*k^2 = n^2

We can see, that 6*k^2 is always even n^2 is also even n is even
So we rephrase the equation with 6*k^2 = n^2 = (2a)^2; n = 2a
(6*k^2)/2 = 2a^2 = 3*k^2

2a^2 is always even. Therefore 3*k^2 has to be even. This is only the case if k is even.
k = 2b
2a^2 = 3*k^2 = 3*(2b)^2 = 3*4b^2 = 12b^2
a^2 = 6b^2
etc.

sqrt(6) won't result in an integer. So there shouldn't be two even integers.

Apfelkuchen04.09.2013

+104

0

2^(999,999,999,999,999) is a logical left shift of 1 by 999,999,999,999,999.
Most common is an implementation of 32-bit blocks. So you'll need at least 999,999,999,999,999/32 ~ 31250000000000 blocks.

Apfelkuchen04.09.2013

+104

0

sqrt(6) = sqrt(2*3) = sqrt(2) * sqrt(3)
2 and 3 are prime numbers. You've got two prime numbers without an even exponent. The square root of every prime number is an irrational number sqrt(6) is an irrational number.

Apfelkuchen03.09.2013

+104

0

Per def. mean score: k = Σ(n_i, i = 1...k)/k

Σ(x_i, i = 1...6)/6 = (x_1 + x_2 + x_3 + x_4 + x_5 + x_6)/6 = 14.5 [*6]
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 14.5*6 = 87

7. game:
Σ(x_i, i = 1...7)/7 = (x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7)/7 = 16 [*7]
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 16*7 = 112
x_7 = 112 - (x_1 + x_2 + x_3 + x_4 + x_5 + x_6) = 112 - 87 = 25

Apfelkuchen03.09.2013

+104

0

Alan: a
Bob: b
Charlie: c

"Alan is twice the height of Bob" a = 2*b
"Charlie is 40cm taller than Bob" c = b+40cm

Total height: a+b+c = 410cm
2*b + b + b+40cm = 410cm = 4b + 40cm [-40cm]
4b = 410cm - 40cm = 370cm [/4]
b = 370cm/4 = 92.5cm

Insert b = 92.5cm into a = 2*b:
a = 2*92.5cm = 185cm

Insert b = 92.5cm into c = b+40cm:
c = 92.5cm + 40cm = 132.5cm

[input]solve(a+b+c=410[cm], a=2*b, c=b+40cm,a,b,c)[/input]

Apfelkuchen03.09.2013

+104

0

I suppose you mean F(x) = ∫f(x)dx

F(x) = ∫f(x)dx = ∫(x^4 - 2x^2)dx = (x^5)/5 - (2x^3)/3

Solve F(x) = g(x) for x:
F(x) = g(x) = (x^5)/5 - (2x^3)/3 = 2x^2 [-2x^2]
(x^5)/5 - (2x^3)/3 - 2x^2 = 2x^2 * (x^3/10 - x/3 - 1) = 0

So we get
x_0 = 0
x_1 = 0

For the other values you could use methods like completing the square or just the polynomial division.
[input]solve((x^3)/10 - x/3 - 1 = 0)[/input]

Apfelkuchen01.09.2013

+104

0

Logarithm to base 11:
[input]log(73,11)[/input]

Apfelkuchen01.09.2013

+104

0

x+y+z = f (fruits total)
x = oranges
y = apples
z = pears

"58% of the fruits at a stall are oranges" => x = f*0.58
"There are 240 more oranges than apples." => y = x - 240 = f*0.58 - 240
"The rest are apples and pears in the ratio of 3:4." => y*4/3 = (f*0.58 - 240)*4/3

x + y + z = (f*0.58) + (f*0.58-240) + (f*0.58-240)*4/3 = f
= -240-320 = f-f0.58-f0.58-f0.77333 = -0.9333333f
=> f = 560/0.93333333 = 600

Apfelkuchen01.09.2013