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 Calculus study realm

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 1:49

Dark, can you summarize all the way to test a sequence or a series's convergence and divergence. There are too many, and they are confusing me.
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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 3:17

No problem!

For sequence:

-Use the limit (i.e.: if lim(as n --> ∞) A_n = a finite real number, then the sequence converges to that number, otherwise it diverges)
-You can proof if it diverge by finding if it is bounded and monotonic (i.e.: every bounded monotonic sequence converges): if A_n < or > A_n+1, the for all n, the sequence is monotonic (proof algebraically, by calculating the ratio, or by finding the derivative)
-Oscillating sequence does not implies that it diverges (converges when lim = 0)
-Thus, if it is not bounded and monotonic, the sequence can also converge. Ultimately, you have to find the limit by using |A_n| (i.e.: if lim(as n--> ∞) |A_n| = 0, then lim(as n--> ∞) A_n = 0. Otherwise, the sequence diverges.)

For series:

-You can calculate the Riemann Sum as n goes from 1 to ∞, but if you can't:
-Create a sequence of partial sums and find the limit (i.e.: define Σ(n from 1 to ∞) A_n as S_n, such as S_1 = A_1, S_2 = A_1 + A_2, ..., S_n = A_1 + A_2 + ... + A_n, and if lim(as n --> ∞) S_n = a finite real number, then the series converges to that number, otherwise it diverges)
-For telescoping series, use partial fraction technique and calculate the Σ. Usually it converges, but there could be exceptions.
-All arithmetic series diverges (Σ(a+(n-1)d))
-For geometric series (Σar^(n-1)), if the common ratio -1 < r < 1, then the series converges to a/(1-r), otherwise, it diverges.
-For P-series (ΣA_n = Σ(1/n^p) such that p is a real number): p must be >1 for the series to converges. Otherwise, the series diverges.
-If ΣA_n converges, it must decrease somewhere. If it's increasing for all n, then it diverges.
-Test for divergence: If ΣA_n converges, limA_n = 0. Thus, if limA_n ≠ 0, then ΣA_n diverges
-Convergence test: integral test: If ΣA_n has positive terms and translate into y = f(x), which is continuous for x≥1, and there is a positive integer N such as f(x) is decreasing for all x≥N, then ΣA_n and ∫(1 to ∞) f(x) dx either both converges or both diverges. Thus, preform the improper integral. If the answer is a finite real number, then the series converges, otherwise it diverges.
-Comparison test: If the positive series ΣA_n ≤ ΣB_n and ΣB_n converges, ΣA_n also converges. If the positive series ΣA_n ≥ ΣB_n and ΣB_n diverges, ΣA_n also diverges. Thus, find a ΣB_n that is easy to proof its convergence or divergence and show that ΣA_n is bigger or smaller than ΣB_n according to the comparison test. So, you have to know the answer before. All you need to do is to proof your intuition.
-Limit comparison test: Suppose that ΣA_n and ΣB_n have positive terms. If lim(ΣA_n/ΣB_n) = c > 0, then both series converge or diverge. If c = 0, then if ΣB_n converges, ΣA_n converges. If c = ∞, then if ΣB_n diverges, ΣA_n diverges. Otherwise, the test fails, i.e.: use a another test.
-Ratio test; good for factorials; measure the rate of growth of ΣA_n by defining r_n = |(A_n+1)/A_n|. If r_n < 1, the series converges; if r_n > 1, the series diverges; if r_n = 1 the test fails.
-Root test; good for n powers; for ΣA_n, if lim(A_n)^(1/n) = L < 1, then the series converges. If L = ∞, the series diverges. If L = 1, the test fails.

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 3:19

Arg, 6 different tests for the series, how can you remember this? =.=
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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 3:20

I don't remember. Even if I memorize all this for the exam, it will only be short term memories.

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 15:29

I know, the sadness in our heart cannot be erased. But life still goes on. Let me substitute Spirit for the sake of our cal exam.
Good bless us all. Amen for the resting in peace one. 26

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 15:34

He's not dead, just hospitalize. But go ahead. Can you summarize the chapters on volume and area of revolution?
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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 15:35

What happened to Spirit? o_O

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 15:40

Sure, Sweety. That the application of integration. I'll post it shortly.

And Evilmashhi, don't worry, Spirit is in the arm of many female nurses. >0< He'll be happy in heaven.

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Dernière édition par Dark le Mar 17 Mai 2011 - 15:47, édité 1 fois
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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 15:41

Ok... XD 38

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 19:33

Area under a curve:
relative (or net) area = ∫(a to b) f(x) dx
total area = ∫(a to b) |f(x)| dx

Area between two curves:
If f(x) > g(x) between x=a and x=b, area = ∫(a to b) (f(x)-g(x)) dx

Volume of revolution: Base area x height
Using disks: volume = lim(as n-->∞) S_n = lim(as n-->∞) Σ(i from 1 to n) π(Radius)i^2 Δthickness = π∫(a to b) R^2 dt
Washers: volume = π∫R^2 th - π∫r^2 dt = π∫(R^2 - r^2) dt
Shells: volume = lim(as n-->∞) S_n = lim(as n-->∞) Σ(i from 1 to n) 2π(Radius)i (Height)i Δthickness = 2π∫(a to b) RH dt

Cross-section volume:
Volume = lim(as n-->∞) S_n = lim(as n-->∞) Σ(i from 1 to n) area(cross-section) Δthickness = ∫(a to b) A dt

Length of a curve:
Let a small section of the curve be Δs = (Δx^2 + Δy^2)^(1/2)
Length = L = lim(as n-->0) Σ(i from 1 to n) Δs = ∫(a to b) (1 + (dy/dx)^2)^(1/2)) dx or ∫(c to d) (1 + (dx/dy)^2)^(1/2)) dy
For parametric: ∫(t1 to t2) ((dx/dt)^2 + (dy/dt)^2)^(1/2)) dt

Surface of a volume of revolution:
Cut the entire object into bands. The area a a small section of the band is the same as a trapezoid, thus A = (B + b)h/2
Thus, total area of the band is A = 2π(R+r)h/2. Since R and r is almost indifferent, A = 2πRh
SA = lim(n-->∞) Σ2πR^2 h = 2π∫(a to b) R^2 ds

Average value of a function:
y_avg = (y1+ y2 + ... + yn)/n = lim(n-->∞) Σ(i from 1 to n) f(xi) 1/n
We know that, for the average between x=a and x=b, Δx=(b-a)/n. Thus 1/n= Δx/(b-a)
y_avg = lim(n-->∞) Σ(i from 1 to n) f(xi) Δx/(b-a) = 1/(b-a) ∫(a to b) f(x) dx (--> mean value theorem)

Center of mass:
x=(lim(n-->∞) Σ(i from 1 to n) (Mass)i xi)/(lim(n-->∞) Σ(i from 1 to n) (Mass)i) = (lim(n-->∞) Σ(i from 1 to n) (Density)i (Volume)i xi)/(lim(n-->∞) Σ(i from 1 to n) Density)i (Volume)i) = ∫(a to b) Dx dV / ∫(a to b) D dV
In R^3: dV = Area Δthickness
In R^2: dV = Height Δthickness
Same for y= ∫(a to b) Dy dV / ∫(a to b) D dV, and z= ∫(a to b) Dz dV / ∫(a to b) D dV

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 21:04

Even God is helping Spirit to find an excuse to escape from my hand. 028
Dark, you're the next in line. Can you tell me how to do #10 on the area exercise sheet?

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 21:23

#10
The key is to find the right formula for the second circle. From the look of picture, you know that it's a dy. So you have to isolate x..
x^2 + y^2 = 4x
x^2 - 4x + 4 - 4 + y^2 = 0
(x - 2)^2 + y^2 = 4
Thus, x = 2±(4 - y^2)^(1/2)

Since it's on the negative section of the graph, x = 2 - (4 - y^2)^(1/2)

For the other circle, it's on the positive section, so x = (4 - y^2)^(1/2)

Then you integrate:

∫(-√3 to √3) ((4 - y^2)^(1/2) - (2 - (4 - y^2)^(1/2))) dy = ∫(-√3 to √3) (2(4 - y^2)^(1/2) - 2) dy = 2∫(-√3 to √3) ((4 - y^2)^(1/2) - 1) dy

Notice that there is a symmetry about the x-axis.

4∫(0 to √3) ((4 - y^2)^(1/2) - 1) dy = 4∫(0 to √3) (4 - y^2)^(1/2) dx - 4y|(0 to √3)
Do a trig sub
Let y=2sin$
dy=2cos$ d$
And change the limits
When y=√3, $=π/3
When y=0, $=0

4∫(0 to π/3) (4 - 4(sin$)^2)^(1/2) 2cos$ d$ - 4√3 = 16∫(0 to π/3) (cos$)^2 d$ - 4√3 = 16∫(0 to π/3) (1/2 + 1/2 cos2$) d$ - 4√3 = 16(1/2 $ 1/4 sin2$ |(0 to π/3) - 4√3 = 8π/3 - 2√3 sq. unit

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 21:36

Thanks.
Can you do # 11 and 12, too?

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 21:48

No problem

#11 It's an ellipse with center at the origin. Thus, there is symmetry about the x-axis and the y-axis.

You can chose to dx or dy.

For dx: 4∫(0 to 1) 4 sinΦ dx
We know that x = cosΦ, thus dx/dΦ = -sinΦ, and dx = -sinΦdΦ
Change your limits; when x=1, Φ=0; when x=0, Φ=π/2

So, it becomes -16∫(π/2 to 0) (sinΦ)^2 dΦ = 16∫(0 to π/2) (1/2 - 1/2 cos2Φ) dΦ = 16(1/2 Φ - 1/4 sin2Φ)|(0 to π/2) = 4π sq. units

For dy: 4∫(0 to 4) cosΦ dy
Again, change dy to dΦ and change your limits.
dy = 4cosΦdΦ
when y=0, Φ=0; when y=4, Φ=π/2

It becomes 16∫(0 to π/2) (cosΦ)^2 dΦ = 16∫(0 to π/2) (1/2 + 1/2 cos2Φ) dΦ = 16(1/2 Φ + 1/4 sin2Φ)|(0 to π/2) = 4π sq. units

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 22:00

#12 This is easier if you dx, cuz dy is more complicated and you probably don't want to know that.

Again, do the same thing as in #11.
dx = 1-cosΦ dΦ
When x=0, Φ=0; when x=2π, Φ=2π

∫(0 to 2π) (1-cosΦ)^2 dΦ = ∫(0 to 2π) (1 - 2cosΦ + (cosΦ)^2) dΦ = (Φ - 2sinΦ)|(0 to 2π) + ∫(0 to 2π)(1/2 + 1/2 cos2Φ) dΦ = 2π + (Φ/2 + 1/4 sin2Φ)|(0 to 2π) = 3π sq. units

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 22:16

Can you dy?

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 22:27

XD why do you even want to know that?

The problem with dy is that it give you the area of the piece that is enclose by the curve with the y-axis. So you have to shift the graph or find the area of the rectangle and minus that area.

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MessageSujet: Re: Calculus study realm    Mar 17 Mai 2011 - 22:28

-_-lll Ok, you can't do it.

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MessageSujet: Re: Calculus study realm    Mer 18 Mai 2011 - 2:06

Looks like Spirit is not going to survive. 37
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MessageSujet: Re: Calculus study realm    Mer 18 Mai 2011 - 2:08

What do you mean? You got the result? What did your dad say?

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MessageSujet: Re: Calculus study realm    Mer 18 Mai 2011 - 2:13

There is only 30% of chance of recovery. But even he can recover, his lower body may become paralyzed.
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MessageSujet: Re: Calculus study realm    Mer 18 Mai 2011 - 2:15

Noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo wayyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Poor Spirit! He was so innocent!!!!!!!!!! 19


ARE YOU SERIOUS! 2 2 2

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MessageSujet: Re: Calculus study realm    Mer 18 Mai 2011 - 2:15

DO I LOOK LIKE JOKING?!
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MessageSujet: Re: Calculus study realm    Mer 18 Mai 2011 - 2:16

YOU ARE SO CALM! I DON'T BELIEVE YOU! I GOING TO THE HOSPITAL TO SEE IT FOR MYSELF!

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MessageSujet: Re: Calculus study realm    Mer 18 Mai 2011 - 2:20

Is this true???? And why are we discussing this on the forum? It seems so not serious.
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