Calculus study realm

You're so mean. I curse you.

Yeah, well, that's how life is. Next question!

Do the whole assignment 10.

Ok, it's gonna take a while.

1) a) What is the difference between a sequence and a series? A sequence is a function of natural number to real number. A series is the sum of the terms in a sequence.
b) The sum of all terms in the sequence a_n is equal to 5. Thus, the series converges.
(what a wasting time questions)

2) a) Σ(n from 1 to ∞) 1/(4n-3)(4n+1)
consider 1/(4n-3)(4n+1) = A/(4n-3) + B/(4n+1) = (A(4n+1) + B(4n-3))/(4n-3)(4n+1)

n: 4A + 4B = 0; A = -B
1: A - 3B = 1; -B - 3B = -4B = 1; B = -1/4
A = 1/4

Σ(n from 1 to ∞) (1/4)/(4n-3) - (1/4)/(4n+1) = (1/4)/(1) - (1/4)/(5) + (1/4)/(5) - (1/4)/(9) + (1/4)/(9) - (1/4)/(13) + ... + (1/4)/(4n-3) - (1/4)/(4n+1) = 1/4 + (1/4)/(4n-3) = 1/4 + 0 = 1/4

b) Σ(n from 1 to ∞) π^n /3^(n+1) = π/9 (π/3)^(n-1)
This is a geometric series. Since -1 < r = π/3 > 1, the series diverge.

c) Σ(n from 1 to ∞) (3^n + 2^n)/6^n = Σ(n from 1 to ∞) 3/6 (3/6)^(n-1) + Σ(n from 1 to ∞) 2/6 (2/6)^(n-1)
Since both are geometric series, and -1 < r < 1,
Σ(n from 1 to ∞) (3^n + 2^n)/6^n = (3/6)/(1 - 3/6) + (2/6)/(1 - 2/6) = 3/2

3) a) Σ(n from 1 to ∞) 4^n x^n = Σ(n from 1 to ∞) 4x (4x)^(n-1)
This is a geometric series, thus -1 < r = 4x < 1; -1/4 < x < 1/4

b) Σ(n from 1 to ∞) (x - 4)^n = Σ(n from 1 to ∞) (x - 4)(x - 4)^(n-1)
This is a geometric series, thus -1 < r = x-4 < 1; 3 < x < 5

4) Test for convergence:
a) Σ(n from 1 to ∞) n/(n^2 + 4)
Integral test: let f(x) = x/(x^2 + 4). f(x) is continuous for all x (quotient of polynomials).
∫(1 to ∞) n/(n^2 + 4) = lim(c-->∞) ∫(1 to c) n/(n^2 + 4) = lim(c-->∞) 1/2 ln(n^2 + 4) |(1 to c) = lim(c-->∞) 1/2 (ln(c^2 + 4) - ln(5)) = ∞
Since the integral diverges, by the Integral test, the series diverges.

b) Σ(n from 1 to ∞) 1/n^(1/4)
This is a p-series with p = 1/4 < 1. Thus the series diverges.

c) Σ(n from 1 to ∞) 1/(n^3 +1)^(1/2)
Limit comparison test: let b_n = 1/n^(3/2)
lim(n-->∞) (1/(n^3 +1)^(1/2))/(1/n^(3/2)) = lim(n-->∞) n^(3/2)/(n^3 +1)^(1/2) = 1 > 0
Since Σ(n from 1 to ∞)b_n converges (p-series with p= 3/2 > 1), by the Limit comparison Test, Σ(n from 1 to ∞) 1/(n^3 +1)^(1/2) converges.

d) Σ(n from 1 to ∞) (2n^2 + 7n)/(3^n (n^2 + 4))
Ratio test: consider ((2(n+1)^2 + 7(n+1))/(3^(n+1) ((n+1)^2 + 4)))/((2n^2 + 7n)/(3^n (n^2 + 4)))
lim(n-->∞) r = 1/3 < 1
Thus, by the Ratio Test, the series converges.

e) Σ(n from 2 to ∞) n/ln(n)
Test for divergence: let f(x) = x/lnx. f(x) is continuous for x ≥ 2 (quotient of continuous functions)
lim(x-->∞) x/lnx = (l'hopital) lim(x-->∞) 1/(1/x) = lim(x-->∞) x = ∞ ≠ 0
Thus, by the Divergence Test, the series diverges.

f) Σ(n from 1 to ∞) cos(nπ)/n^(3/4) = Σ(n from 1 to ∞) (-1)^n /n^(3/4)
This is an alternating series. It decreases and lim(n-->∞)(-1)^n /n^(3/4) = 0
Thus, by the Alternating Series Test, it converges.

g) Σ(n from 1 to ∞) n^3 /3^n
Ratio test: consider ((n+1)^3 /3^(n+1))/(n^3 /3^n)
lim(n-->∞) r = 1/3 < 1
Thus, by the Ratio Test, the series converges.

Give all the theoretical material that we need to memorize.

1) Definition of antiderivative.
A function such that its derivative is f(x).

2) Derive the integration by part rule with the product formula for differentiation.
When differentiating, we have
y = u·v; dy/dx = u·dv/dx + v·du/dx
dy = d(u·v) = u·dv + v·du
Thus ∫u·dv = ∫d(u·v) - ∫v·du = u·v - ∫v·du

3) Derive the reduction formula
i) ∫(sinx)^n dx = -cosx(sinx)^(n-1) + (n-1)∫(cosx)^2 (sinx)^(n-2) dx = -cosx(sinx)^(n-1) + (n-1)∫(1- (sinx)^2) (sinx)^(n-2) dx = -cosx(sinx)^(n-1) + (n-1)∫((sinx)^(n-2) - (sinx)^(n)) dx

n∫(sinx)^n dx = -cosx(sinx)^(n-1) + (n-1)∫(sinx)^(n-2) dx
∫(sinx)^n dx = -cosx(sinx)^(n-1) /n + (n-1)/n ∫(sinx)^(n-2) dx

f(x) = (sinx)^(n-1)
f(x) = (n-1)cosx (sinx)^(n-2)

g(x) = -cosx
g(x) = sinx

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ii) ∫(cosx)^n dx = sinx (cosx)^(n-1) + (n-1)∫(sinx)^2 (cosx)^(n-2) dx = sinx (cosx)^(n-1) + (n-1)∫(1 - (cosx)^2) (cosx)^(n-2) dx = sinx (cosx)^(n-1) + (n-1)∫((cosx)^(n-2) - (cosx)^n) dx

n∫(cosx)^n dx = sinx (cosx)^(n-1) + (n-1)∫(cosx)^(n-2) dx
∫(cosx)^n dx = sinx (cosx)^(n-1) /n + (n-1)/n ∫(cosx)^(n-2) dx

f(x) = (cosx)^(n-1)
f(x) = -(n-1)sinx (cosx)^(n-2)

g(x) = sinx
g(x) = cosx

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iii) ∫ln(x)^n dx = xln(x)^n - n∫ln(x)^(n-1)

f(x) = ln(x)^n
f(x) = nln(x)^(n-1) /x

g(x) = x
g(x) = 1

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iv) ∫(secx)^n dx = ∫((tanx)^2 + 1)(secx)^(n-2) dx = ∫(secx)^(n-2) (tanx)^2 dx + ∫(secx)^(n-2) dx = tanx(secx)^(n-2) /(n-2) - 1/(n-2) ∫(secx)^n dx + ∫(secx)^(n-2) dx

(1 + 1/(n-2)) ∫(secx)^n dx = (n-1)/(n-2) ∫(secx)^n dx = tanx(secx)^(n-2) /(n-2) + ∫(secx)^(n-2) dx
∫(secx)^n dx = tanx(secx)^(n-2) /(n-1) + (n-2)/(n-1) ∫(secx)^(n-2) dx

f(x) = tanx
f(x) = (secx)^2

g(x) = (secx)^(n-2) /(n-2)
g(x) = (secx)^(n-3) secx tanx

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v) ∫x^n e^x dx = x^n e^x - n∫x^(n-1) e^x dx

f(x) = x^n
f(x) = nx^(n-1)

g(x) = e^x
g(x) = e^x

4) Express a definite integral as a limit Riemann Sum and compute for a simple case.
E G: The function f(x) = x, between x =1 and x = 2.

Base of rectangle = (2-1)/n = 1/n

lim(n-->∞) Σ(i from 1 to n) i/n 1/n = lim(n-->∞) 1/n^2 Σ(i from 1 to n) i = lim(n-->∞) 1/n^2 n(n+1)/2 = lim(n-->∞) (n+1)/2n = 1/2

5) (Do I need to explain this? -_-lll)

6) Mean Value theorem for Integral Cal
If y = f(x) is continuous on the closed interval [a,b], then there is some number c in the interval [a,b], such that ∫(a to b) f(x) dx = f(c)·(b-a)
Proof: Since y = f(x) is continuous on [a,b], by the Extreme Value Theorem, the function attains a maximum and a minimum value M and m respectively in the interval [a,b]. Thus, there exist two point x1 and x2 such that f(x1) = M and f(x2) = m. Thus for all values of x in [a,b], we have m ≤ f(x) ≤ M.
From the properties of definite integral, we have
m(b-a) ≤ ∫(a to b) f(x) dx ≤ M(b-a)
m ≤ 1/(b-a) ∫(a to b) f(x) dx ≤ M
f(x2) ≤ 1/(b-a) ∫(a to b) f(x) dx ≤ f(x1)
Assume that x1>x2 (the similar argument can be made if the opposite is true). The above shows that 1/(b-a) ∫(a to b) f(x) dx is a number between f(x2) = m and f(x1) = M. Since y = f(x) is continuous, by the Intermediate Value Theorem, there exist a number c in ]x2, x1[ such that f(c) = 1/(b-a) ∫(a to b) f(x) dx. Therefore, there exist a value c such that ∫(a to b) f(x) dx = f(c)·(b-a).

7) Fundamental Theorem of Calculus (Part I)
If y = f(x) is continuous on [a,b], then the function g(x) = ∫(a to x) f(t) dt is continuous on [a,b] and differentiable on ]a,b[, and g'(x) = f(x).
Proof: If x and x + Δx are in ]a,b[, then
g(x + Δx) - g(x) = ∫(a to x + Δx) f(t) dt - ∫(a to x) f(t) dt = ∫(x to x + Δx) f(t) dt
Hence, for Δx ≠ 0, (g(x + Δx) - g(x))/Δx = 1/Δx ∫(x to x + Δx) f(t) dt
Since y = f(x) is continuous on [a,b], the Mean Value Theorem of Integral Calculus guarantees a number c in [a,b] such that
∫(x to x + Δx) f(t) dt = f(c) (x + Δx - x) = f(c) Δx
Hence, g'(x) = lim(Δx-->0) (g(x + Δx) - g(x))/Δx = lim(Δx-->0) f(c) = f(x)
This implies that y = g(x) is differentiable for any x on ]a,b[, and thus it must be continuous on [a,b]

Fundamental Theorem of Calculus (Part II)
If y = f(x) is continuous on [a,b], then ∫(a to b) f(x) dx = F(b) - F(a), where y = F(x) is any antiderivative of y = f(x), that is a function such as F'(x) = f(x).
Proof: Let g(x) = ∫(a to x) f(t) dt, we know from part I that g'(x) = f(x); that is g'(x) is an antiderivative of y = f(x). If y = F(x) is another antiderivative, then we know from a far away theorem that y = F(x) and y = g(x) differ by a constant:
F(x) = g(x) + C, for a < x < b
But both y = F(x) and y = g(x) are continuous. If we take the limit on both side of the equation (as x --> a^+ and x --> b^-) we see that it is also hold when x = a and x = b.
F(x) = g(x) + C, for a ≤ x ≤ b
If we put x = a in the formula for y = g(x), we get ∫(a to a) f(t) dt = 0
Using the 3 equation, we get
F(b) - F(a) = [g(b) + C] - [g(a) + C] = ∫(a to b) f(t) dt

QED

8) State de relationship among a sequence being bounded, monotonic, and convergent.
Bounded and monotonic sequence converges.

9) If -1 < r < 1, then lim(n-->∞) r^n = 0
If r < -1 or r > 1, then lim(n-->∞) r^n = ∞
If r = 1, lim(n-->∞) r^n = 1
If r = -1, lim(n-->∞) r^n = Ø

10) i) lim(n-->∞) (1 + x/n)^n = e^x
Let L = lim(n-->∞) (1 + x/n)^n
lnL = ln(lim(n-->∞) (1 + x/n)^n) = lim(n-->∞) ln(1 + x/n)^n = lim(n-->∞) n ln(1 + x/n) = lim(n-->∞) ln(1 + x/n) / (1/n) = (l'hopital rule) lim(n-->∞) [(-x/n^2)/(1 + x/n)]/[-1/n^2] = lim(n-->∞) -n^2(-x/n^2)/(1 + x/n) = lim(n-->∞) x/(1 + x/n) = x
L = e^x

ii) lim(n-->∞) r^n /n! = 0
If r < -1 or r > 1, then lim(n-->∞) r^n /n! = 0/∞ = 0
If r = 1, lim(n-->∞) r^n /n! = 1/∞ = 0
If r = -1, lim(n-->∞) r^n /n! = ±1/∞ = 0
If r < -1 or r > 1, then 1< |r| < ∞, thus lim(n-->∞) r^n /n! = (r...rr...)/(1·...·r·(r+1)...) = (r...r)/(1·...·r) r^∞/(r·(r+1)·(r+2)...) = (r...r)/(1·...·r) ·0 = 0

11) State the definition of convergence and divergence of a series in terms of its sequence of partial sums.
Given a sequence A_n = {a1, a2, a3, a4, ..., an}, we define a series attaches as Σ(n from 1 to ∞) A_n = a1 + a2 + a3 + a4 + ... + an
If the sum gives a finite number, the series is said to converge. If not, it is said to diverge.

12) See Dark's post.

Still asking questions.

Orz, I saw the calender on my computer and it say may 19th. I was like "damn! Did I missthought the date?"

XD you thought "tomorrow" is Friday?

Yep. But it is. Anyway. You know what I mean.

BUT GOD! I'M DOOMED FOR THE EXAM. SO NOT READY!!!

Me too!
So, stop wasting time now. Continue studying.
Although, I think my score won't even pass the average.

I studied all day. I didn't have life all week.
Well, except that I was watching MV at the same time.

Watching MV? You still have the mood for that?

I can't help it. I feel too lonely studying only with the sound of the computer buffering.
But what do you mean still have the mood. Are you that much stressed?

Aren't you? Isn't it natural to be stressed before the final exam? How can you be so relaxed?

I'm stressed at the end of the exam. When its almost time to hand in the exam. Or after the exam, while waiting for the result. In other time, I'm cool.

I envy your calming ability. I'm stressed from the beginning to the end. After the exam, I'm totally depressed.
Ok, let's get back to math now.

Dude, we should get some shut eye. It's TOMORROW the exam!

We still have at least 24 hours.
But, you're right. We have to wake up early. So, this morning, at 8AM, ultimate final study realm: everyone be here to see if anyone missed anything.

Yo, I wake up at noon! I'll only get 7 hours of sleep if I have to wake up at 8.

What about Friday, you need to wake up at 5.
So get prepared now.
8AM!

So, who's here?

I am.

That's it? -__-lll