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.

ApfelkuchenSep 13, 2013

+104

0

-10z ≤ 15 [/(-10)]

z ≥ -3/2

ApfelkuchenSep 13, 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

ApfelkuchenSep 13, 2013

+104

0

Potenzsummen.png

ApfelkuchenSep 10, 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.

ApfelkuchenSep 4, 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.

ApfelkuchenSep 4, 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.

ApfelkuchenSep 3, 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

ApfelkuchenSep 3, 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]

ApfelkuchenSep 3, 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]

ApfelkuchenSep 1, 2013

+104

0

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

ApfelkuchenSep 1, 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

ApfelkuchenSep 1, 2013