Extension One Revising Game (1 Viewer)

kaz1 · Jan 9, 2009

lim_h-->0 [(x+h)³-x³]/h
lim_h-->0 [x³+3x²h+3xh²+h³-x³]/h
lim_h-->0 [3x²h+3xh²+h³]/h
lim_h-->0 h(3x²+3xh+h²)/h
lim_h-->0 3x²+3xh+h²
3x²

gurmies · Jan 9, 2009

Moar question pl0x ^

shaon0 · Jan 9, 2009

Am I allowed to post up a question? If so, here is my question:

Consider f(x)=x^n.e^(-x) for x>0, given that f(x)>0 for all values of x in the given domain and n>0.
i) By considering the values of f(n), f(n-1) and f(n+1), prove that:
(1+[1/n])^n < e < (1-[1/n])^(-n)

The above part is easy. There's a harder part which comes later in the question.

bored of sc · Jan 17, 2009

If f(x) = 4x/(3-x⁷) find f''(x)

lisarh · Jan 19, 2009

bored of sc said:
If f(x) = 4x/(3-x⁷) find f''(x)

f(x) = 4x/(3-x7)

Now,
u = 4x
du/dx= 4

v = (3-x7)
dv/dx = -7x6

f'(x) = [ v.du/dx - u.dv/dx ] / v2
f'(x) = [(3-x7).4 - 4x.(-7x6 ] / (3-x7)2
f'(x) = (12 - 4x7+ 28x7) / (3-x7)2
f'(x) = (12 + 24x7)/ (3-x7)2

Now,

u = (12 + 24x7)
du/dx = 168x6

v = (3-x7)2
dv/dx = 2(3-x7).-7x6
dv/dx = -14x6(3-x7)

f''(x) = [ v.du/dx - u.dv/dx ] / v2
f''(x) = [ (3-x7)2.168x6 - (12 + 24x7).-14x6(3-x7) ] / (3-x7)4
f''(x) = [ 168x6(3-x7)2 - 12( 1 + 2x7).-14x6(3-x7) ] / (3-x7)4
f''(x) = [ 168x6(3-x7)2 + 168x6( 1 + 2x7)(3-x7) ] / (3-x7)4
f''(x) = 168x6(3-x7)[ (3-x7) + ( 1 + 2x7) ] / (3-x7)4
f''(x) = 168x6(3-x7)[ 3 - x7) + 1 + 2x7 ] / (3-x7)4
f''(x) = 168x6(3-x7)(4+x7) / (3-x7)4

lisarh · Jan 19, 2009

Next question ^^

Prove that

3ⁿ < n!

for all positive integers n ≥ 7

I don't know how to do this one so can someone please give the solutions to it. Ty

If not then,

Solve for x,

[ |x|/(x-1) ] > 2

cutemouse · Jan 19, 2009

Haha, I need to revise induction. Might be a good idea to do that soon

lisarh said:
If not then,

Solve for x,

[ |x|/(x-1) ] > 2

|x|/(x-1) ] > 2 b.s.*(x-1)²

|x|(x-1)>2(x-1)²
|x|(x-1)>2x²-4x+2

Case 1: x(x-1) > 2x²-4x+2
x² > 2x²-4x+2
x²-3x+2 <0
(x-2)(x-1)<0

By drawing a facilitating graph, 1 < x < 2

Case 2: x(x-1) < -2x²-4x+2
3x²+3x-2<0

Now, Δ<0, Therefore no real solution.

Overall solution:
1 < x < 2

Next question:

Solve for x: |5x-12| = -13

Trebla · Jan 19, 2009

lisarh said:
Prove that

3ⁿ < n!

for all positive integers n ≥ 7

I don't know how to do this one so can someone please give the solutions to it. Ty

For n = 7
LHS = 3⁷
= 2187
RHS = 7!
= 5040
LHS < RHS .: Statement is true for n = 7
Assume the statement is true for n = k
i.e. 3^k < k!
Required to prove it is true for n = k + 1
i.e. 3^{k + 1} < (k + 1)!
LHS = 3^{k + 1}
= 3.3^k
< 3k! by assumption
< (k + 1)! *
= RHS
*Since k > 7 => k + 1 > 8 > 3, hence 3 < k + 1 (then multiply both sides by k! giving 3k! < (k + 1)!)

The statement is true for n = k + 1, if it is true for n = k. Since it is true for n = 7, it follows by induction it is true for all other integers n > 7

Question:

One of the most famous equations used to describe quantum mechanics of particles is the Schrödinger equation which is given in the following one-dimensional simplified form:
(- ħ²/2m) (d²Ψ/dx²) = E.Ψ

where ћ and m are some constants and d²Ψ/dx² is the second derivative operated on Ψ with respect to x.

If Ψ = sin (πx/L) is a solution to this differential equation, find an expression for E.

PS: For those who do Physics m is the mass of the particle, E is the energy and ћ is actually Planck's constant divided by 2π (i.e. ћ = h/2π).

gurmies · Jan 19, 2009

jm01 said:
Next question:

Solve for x: |5x-12| = -13

Looks like no solution to me, as the graph of |5x-12| will always be above x-axis, and y = -13 is below...? Or you can interpet it to mean the absolute value of something yields a negative result which can't occur..

kaz1 · Jan 19, 2009

jm01 said:
Solve for x: |5x-12| = -13

How can an absolute number be negative?

shaon0 · Jan 19, 2009

Trebla said:
Question:

One of the most famous equations used to describe quantum mechanics of particles is the Schrödinger equation which is given in the following one-dimensional simplified form:
(- ħ²/2m) (d²Ψ/dx²) = E.Ψ

where ћ and m are some constants and d²Ψ/dx² is the second derivative operated on Ψ with respect to x.

If Ψ = sin (πx/L) is a solution to this differential equation, find an expression for E.

PS: For those who do Physics m is the mass of the particle, E is the energy and ћ is actually Planck's constant divided by 2π (i.e. ћ = h/2π).

S -ћ^2/2m d^2Ψ/dx^2 = EΨ
=> - ћ^2/2m (dΨ/dx) = EΨ
=> (- ћ^2/2m)*Ψ = EΨx
=> ћ^2/2m = -Ex
Don't know if this is right though

Trebla · Jan 19, 2009

shaon0 said:
S -ћ^2/2m d^2Ψ/dx^2 = EΨ
=> - ћ^2/2m (dΨ/dx) = EΨ
=> (- ћ^2/2m)*Ψ = EΨx
=> ћ^2/2m = -Ex
Don't know if this is right though

Integration won't work because you have no information to evaluate the constants (which you excluded)

independantz · Jan 19, 2009

Trebla said:
Question:

One of the most famous equations used to describe quantum mechanics of particles is the Schrödinger equation which is given in the following one-dimensional simplified form:
(- ħ²/2m) (d²Ψ/dx²) = E.Ψ

where ћ and m are some constants and d²Ψ/dx² is the second derivative operated on Ψ with respect to x.

If Ψ = sin (πx/L) is a solution to this differential equation, find an expression for E.

PS: For those who do Physics m is the mass of the particle, E is the energy and ћ is actually Planck's constant divided by 2π (i.e. ћ = h/2π).

Ψ=sin(nx/L)
dΨ/dx=(n/L)cos(nx/L), assuming n and L are constants- if not then this is prob wrong lol
d²Ψ/dx²=-(n²x)/L²)sin(nx/L)

therefore, E.sin(nx/L)=(- ħ²/2m) .-(n²x)/L²)sin(nx/L)
E=(- ħ²/2m) .-(n²x)/L²)
=ħ²n²x/2mL²

shaon0 · Jan 19, 2009

Trebla said:
Integration won't work because you have no information to evaluate the constants (which you excluded)

Yeah, i had a feeling it was wrong.

Trebla · Jan 20, 2009

independantz said:
Ψ=sin(nx/L)
dΨ/dx=(n/L)cos(nx/L), assuming n and L are constants- if not then this is prob wrong lol
d²Ψ/dx²=-(n²x)/L²)sin(nx/L)

therefore, E.sin(nx/L)=(- ħ²/2m) .-(n²x)/L²)sin(nx/L)
E=(- ħ²/2m) .-(n²x)/L²)
=ħ²n²x/2mL²

There shouldn't be an x in the equation and btw it's π (pi) not n. Though the method is correct...

shaon0 · Jan 20, 2009

Trebla said:
There shouldn't be an x in the equation and btw it's π (pi) not n. Though the method is correct...

Is x=(L/pi)arcsin(Ψ)?

untouchablecuz · Jan 20, 2009

Trebla said:
For n = 7
LHS = 3⁷
= 2187
RHS = 7!
= 5040
LHS < RHS .: Statement is true for n = 7
Assume the statement is true for n = k
i.e. 3^k < k!
Required to prove it is true for n = k + 1
i.e. 3^{k + 1} < (k + 1)!
LHS = 3^{k + 1}
= 3.3^k
< 3k! by assumption
< (k + 1)! *
= RHS
*Since k > 7 => k + 1 > 8 > 3, hence 3 < k + 1 (then multiply both sides by k! giving 3k! < (k + 1)!)

The statement is true for n = k + 1, if it is true for n = k. Since it is true for n = 7, it follows by induction it is true for all other integers n > 7

Question:

One of the most famous equations used to describe quantum mechanics of particles is the Schrödinger equation which is given in the following one-dimensional simplified form:
(- ħ²/2m) (d²Ψ/dx²) = E.Ψ

where ћ and m are some constants and d²Ψ/dx² is the second derivative operated on Ψ with respect to x.

If Ψ = sin (πx/L) is a solution to this differential equation, find an expression for E.

PS: For those who do Physics m is the mass of the particle, E is the energy and ћ is actually Planck's constant divided by 2π (i.e. ћ = h/2π).

[(h^2)(pi^2)]/(2mL) = E ?

Do you just sub Ψ = sin (πx/L) into the equation, then find the second derivative of sin (πx/L) with respect to x, then just simplify?

Trebla · Jan 20, 2009

untouchablecuz said:
[(h^2)(pi^2)]/(2mL) = E ?

Do you just sub Ψ = sin (πx/L) into the equation, then find the second derivative of sin (πx/L) with respect to x, then just simplify?

Like independantz, correct method but slightly incorrect answer lol check your working...

jet · Jan 20, 2009

Ψ = sin (πx/L)
Therefore
dΨ/dx = (π/L)cos(πx/L)
d^2Ψ/dx^2 = -(π^2/L^2)sin(πx/L)

Hence, Esin(πx/L) = [(h^2)(π^2)/2mL^2]sin(πx/L)
Hence, E = [(h^2)(π^2)/2mL^2]

untouchablecuz · Jan 21, 2009

ooooo yer, L^2

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