Calculus Revisited #21: The Catch All

Welcome to Part 21 of our 21 part series on Calculus Series. Here are some interesting facts about calculus that were not discussed in the first 20 parts, but are worth paying attention to:

Implicit Differentiation

The key here is that when you are differentiating, you differentiate with respect to the specific variable. All other variables are treated as constants.

So for the function f(x,y):

d/dx f(x,y) means differentiate with respect to x, y is held constant

d/dy f(x,y) means differentiate with respect to y, x is held constant

Example: Let f(x,y) = ln(x + 2y) + y^2

d/dx f(x,y)
= d/dx ln(x + 2y) + d/dx y^2
= 1/(x + 2y) + 0
= 1/(x + 2y)

d/dx f(x,y)
= d/dy ln(x + 2y) + d/dy y^2
= 2/(x + 2y) + 2y

Arc Length

The arc length of a function f(x) is given by the integral:

b
∫ √( 1 + (f'(x))^2 ) dx
a

Usually, you will be finding the arc length by numerical methods and/or the calculator.

A numerical example: The arc length of f(x) = sin x from x = 0 to x = π/2 (to seven decimal places)

f(x) = sin x
f'(x) = cos x
(f'(x))^2 = cos^2 x

Then:

π/2
∫ (1 + cos^2 x) dx ≈ 1.9100989 (by calcualtor)

A Numerical Method of Integration: Simpson's Rule

For n partitions (n is even):

b
∫ f(x) dx
a

≈ (b - a)/3n * (f(a) + 4*f(x1) + 2*f(x2) + 4*f(x3) + 2*f(x4) + ... + 4*f(x_2k+1) + 2*f(x_2k+2) + .... + 4*f(x_n-1) + f(b))

where x_k = a + (b - a)/n * k

Error:

At maximum (b - a)^5/(180 * (2n)^4) * max|f''''(x)|

In general, the more complex the integral, the more terms needed

Example:
Use the Simpsons rule to calculate

2
∫ e^x dx with n = 4 (7 decimal places)
1

(b - a)/n = (2 - 1)/4 = 1/4

The integral is approximately:

≈ (2 - 1)/(3 * 4) * (f(1) + 4 * f(1.25) + 2 * f(1.5) + 4 * (1.75) + f(2))
≈ 4.6708749

(actual value is about 4.6707743)

Area Between Curves

Let f(x) and g(x) be two functions where f(x) ≥ g(x). The area between curves f(x) and g(x) is:

b
∫ f(x) - g(x) dx
a

Often a and b will be intersection points of f(x) and/or g(x).

Example:

Let f(x) = x^2 and g(x) = x. Find the area between the curves from x = 0 to x = 5.

When 0 ≤ x ≤ 5, f(x) ≥ g(x). Then the area between the curves is:

5
∫ x^2 - x dx = 175/6 ≈ 29.1666667
0

Volume of a Solid - Disk Method

The volume of a solid of revolution - method of discs revolving around the x-axis is:

b
∫ π * (f(x))^2 dx
a

Like arc length, you may be finding the volumes numerically and/or by calculator (like I will do by this example).

Example: Let f(x) = sin x from x = 0 to x = π

Then the volume is:

π
∫ π * sin^2 x dx ≈ 4.9348022 = π^2/2
0

If you are finding a volume with discs revolving around the y-axis then the volume would be:

b
∫ π * (g(y))^2 dy
a

where a and b are y-values.

Polar Equations and Polar Integral

To convert functions to their polar form and back, use the following:

x = r cos θ
y = r sin θ

r^2 = x^2 + y^2
θ = atan (y/x)

The Polar Integral is:

θ2
∫ 1/2 * (r(θ))^2 dθ
θ1

Example: Let r(θ) = 2 θ . Find the polar integral from θ = 0 to θ = 2π

Area:

2π
∫ 1/2 * (2 θ)^2 dθ = 16 π^3/ 3 ≈ 165.3668090
0


Parametric Equations

Parametric equations are in the form of ( x(t), y(t) ), where t is the independent variable and x and y are dependent variables. Each part can be separately differentiated and integrated.

Arc length of a parametric curve from t = a to t = b:

b
∫ √( (x'(t))^2 + (y'(t))^2 ) dt
a

Let:
x(t) = 2t
y(t) = t^3

Find (x, y) at t = 0 and t = 1. Find the arc length.

x(0) = 0, y(0) = 0
x(1) = 2, y(1) = 1

Arc length:

x'(t) = 2
y'(t) = 3t^2

(x'(t))^2 = 4
(y'(t))^2 = 9t^4

1
∫ √(4 + 9t^4) dt ≈ 2.3650656
0

Tips for Sketching a Curve y = f(x)

Find out where the extrema of f(x) are. Recall the critical points are found when:

f'(x) = 0

Let c be the critical points.

If f''(c) > 0, then x = c is strict minimum.

If f''(c) < 0, then x = c is a strict maximum.

If f'(c) > 0 when both f'(x) > 0 for x < c and x > c OR
f'(c) < 0 when both f'(x) < 0 for x < c and x > c,
then x = c is an inflection point.

When f'(x) > 0, f(x) is increasing.

When f'(x) < 0, f(x) is decreasing.

When f''(x) > 0, f(x) is concave up (holding water).

When f''(x) < 0, f(x) is concave down (spilling water).

If f(x) → +∞ or -∞ when x → a, then f(x) has a vertical asymptote at x = a.

If f(x) → b when x → +∞ or -∞, then f(x) has a horizontal asymptote at y = b.

Thank you as always. I hope you enjoyed the Calculus Revisited Series.

To all the students who are in calculus, good luck in finals!

Until next time,

Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #19: Tests of Convergence

Welcome to Part 19 of our 21 Part series: Calculus Revisited. We are in the middle of working with series. Today's session: Tests of Convergence.

In a calculus course, you are often asked if certain series converge - that is, does the series have a sum. You are not necessarily asked to find the sum itself, just to say if the series has one.

Hint: This is good to know, the p-series.

For a series

∞
∑ 1/n^p
n = 1

this series converges when p > 1 and diverges when p ≤ 1.

Now for the tests:

Comparison Test:

Give the infinite series ∑ a(n), find another comparable infinite series ∑ b(n). The series used for comparison dominates the series in question. Or:

b_n ≥ a_n for each n an integer

In addition, all terms both series, ∑ a(n) and ∑ b(n), are positive.

If ∑ b(n) converges, so does ∑ a(n).

However, if ∑ a(n) diverges, so does ∑ b(n).

There is a variant of the comparison test call the limit test. That is if

lim ( a_n / b_n) = L as n → ∞,

If ∑ b(n) converges and L < ∞, ∑ a(n) converges.
If ∑ b(n) diverges and L > 0, ∑ a(n) diverges.

Ratio Test:

For a given infinite series ∑ a(n), if

lim | a_n+1 / a_n | = L as n → ∞, and

L < 1,

then ∑ a(n) is absolutely convergent. If L = 1, the test is inconclusive. If L > 1, the series diverges.

Note: A series is absolutely convergent if

∑ |a(n)| = | a(1) | + | a(2) | + | a(3) | + ....

is convergent.

If ∑ a(n) converges but ∑ | a(n) | does not, then the series is said to be conditionally convergent.

Root Test:

For a given infinite series ∑ a(n), if

lim | a_n |^(1/n) = L as n → ∞, and

L < 1,

then ∑ a(n) is absolutely convergent. If L = 1, the test is inconclusive. If L > 1, the series diverges.

Alternating Series Test:

For the existing series:

∑ (-1)^n * a(n),

if the sequence {a_n} (ignoring the (-1)^n) is a strictly decreasing sequence of positive numbers (that is a_n+1 < a_n for all n), and

lim a_n = 0 as n → ∞,

the the alternating series is (at least) conditionally convergent.

Problems

1. Does this series converge?

∞
∑ n^2 / (2n - 1)!
n = 1

Use the ratio test. Then:

a_n = n^2 / (2n - 1)!

a_n+1 = (n + 1)^2 / (2(n + 1) - 1)! = (n + 1)^2 / (2n + 1)!

a_n+1 / a_n
= (n + 1)^2 / (2n + 1)! * (2n - 1)! / n^2
= (n + 1)^2 / (n^2 * (2n + 1)(2n))
= (n^2 + 2n + 1) / (4n^4 + 2n^3)

As n → ∞, a_n+1 / a_n → 0 < 1.

By the ratio test, this series converges.

Sometimes reasoning will be necessary.

2. Does this series converge?

∞
∑ 1 / (n * n^(1/n))
n = 1

Note that:
1 / (n * n^(1/n))
= 1 / (n^1 * n^(1/n))
= 1 / (n^(1 + 1/n))

Compare the series to the harmonic series:

∞
∑ 1/n
n = 1

where 1/n ≥ 1/(n^(1 + 1/n))

By Limit test:

lim ( 1/(n^(1 + 1/n)) / (1/n) ) as n → ∞
= lim ( n / n^(1+ 1/n) ) as n → ∞

As n grows large, 1 + 1/n → 1
Hence, n / n^(1 + 1/n) → n / n → 1

Since ∑ (1/n) diverges, so does ∑ 1/ (n * n^(1/n))

3. Does this series converge?

∞
∑ 1/(4n - 3)
n = 1

Use the limit test. Compare it to the harmonic series, which is known to diverge, and:

1/n ≥ 1/(4n - 3)

lim ( (1/(4n - 3)) / (1/n) ) as → ∞
= lim ( n / (4n - 3) ) as → ∞
= 1/4 > 0

Hence, the series in question diverges.

4. Does this series converge?

1 - 1/√3 + 1/√5 - 1/√7 + .... + (-1)^n/√(2n+1) + ....

Use the Alternating Series test.

Let a_n = 1/√(2n+1)

First a_n strictly decreases. Second, a_n → 0 as n → ∞.

By alternating series, the series conditionally convergent.

5. Does this series converge?

1/(2 ln 2) - 1/(3 ln 3) + 1/(4 ln 4) - 1/(5 ln 5) + ... + (-1)^n/(n ln n) + ....

Use the Alternating Series Test

Let a_n = 1/(n ln n),

a_n strictly decreases and a_n → 0 as n → ∞.

By alternating series, the series is conditionally convergent.

Next time we work with Taylor Series.

See you soon!

Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #18: Series

Welcome to Part 18 of our wonderful 21 part of our Calculus Revisited Series. I hope you are enjoying this series. An announcement: Part 21 will be a "catch all" section - covering some of the topics we don't get to in detail in this series.

The next three blog entries will be about series. Today, the basics.

Series:

A series is a sum of terms in a sequence. The series can be finite or infinite. An infinite series is convergent (has a value) if the following is true:

∞
∑ a_n = S
n=i

and S < ∞

Here are some famous series:

Arithmetic Series:

a + (a + d) + (a + 2d) + (a + 3d) + ...

Sum for a finite arithmetic series:

n-1
∑ a + k * d
k=0

= a * n + (d * n * (n -1))/2

Geometric Series:

a + a * r + a * r^2 + a * r^3 + ...

Sum for an infinite geometric series, provided that |r| < 1:

∞
∑ a * r^k = a / (1 - r)
k = 0

Harmonic Series:

∞
∑ 1/n
n=1

= 1 + 1/2 + 1/3 + 1/4 + 1/5 + ....

There is no closed formula for a finite harmonic series. In addition the infinite harmonic series diverges (has no sum).

Series Properties:

∑ c * a(n) = c * ∑ a(n) (c is a constant)

∑ a(n) + b(n) = ∑ a(n) + ∑ b(n)

n
∑ k = n * (n + 1)/2
k=1

n
∑ k^2 = n * (n + 1) * (2n + 1)/6
k = 1

n
∑ k^3 = n^2 *(n + 1)^2 / 4
k = 1


Subtracting from a whole:
If t > 1:

k
∑ a(n) =
n = t

∑( a(n) for n = 1 to k) - ∑( a(n) for n = 1 to t - 1)

Problems

1. Calculate

6
∑ 50 - 2k
k = 0

This is an arithmetic series with n = 6 + 1 = 7, a = 50, and d = -2.

Then the sum is:
(50)(7) + (-2 * 7 * 6)/2 = 308


2. Show why the (infinite) harmonic series is divergent.

Here is one way:

Let S be the sum of the harmonic series, that is:

S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...

Note that:
1 + 1/2 > 2/2
1/3 + 1/4 > 2/4
1/5 + 1/6 > 2/6
1/7 + 1/8 > 2/8
and so on...

Then we have 2/2 + 2/4 + 2/6 + 2/8 + ....

Which simplifies to 1 + 1/2 + 1/3 + 1/4 + .... = S

Which implies S > S, which is impossible.

The harmonic series is divergent.

3. Find the sum:

∞
∑ .5^n + .3^n
n=0

We can first break the sum up:

∑(.5^n for n=0 to ∞) + ∑(.3^n for n=0 to ∞)

Both terms are geometric series. r = .5 for the first term and r = .3 for the second term.

Then:

∑(.5^n for n=0 to ∞) + ∑(.3^n for n=0 to ∞)
= 1 / (1 - .5) + 1 / (1 - .3)
= 1 / .5 - 1 / .7
= 24/7 ≈ 3.42857

4. Find the sum:

∞
∑ 2 / (5^n)
n = 1

This looks like a geometric series with r = 1/5. If we can put the sum into it's proper form, perhaps by adding and subtracting 2 / 5^0:

∑( 2 / (5^n) for n = 1 to ∞)
= ∑ ( 2 / (5^n) for n = 0 to ∞) - 2 / 5^0
= ∑ ( 2 / (5^n) for n = 0 to ∞) - 2
= 2 / ( 1 - 1/5) - 2
= 2 / (4/5) - 2
= 5/2 - 2 = 1/2

5. Find the sum:

20
∑ k^2
k=10

Again, looks like a k^2 series, but use the subtraction from the whole technique and:

∑( k^2 for k = 10 to 20)
= ∑(k^2 for k = 1 to 20) - ∑(k^2 for k = 1 to 10 - 1)
= (20)(20 + 1)(2 * 20 + 1)/6 - (9)(9 + 1)(2 * 9 + 1)/6
= 17220 / 6 + 1710 / 6
= 15510 / 6 = 2585

Thank you once again for joining us. Next time we will work some series convergence tests.

Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #17: Sequences

This Part 17 of the Calculus Revisited series. Today we are dealing with sequences.

Sequence:

A sequence of numbers is exactly what you think of, a list of numbers. An infinite sequence is a list of numbers that goes on forever. In calculus, it is understood that unless specified, a sequence is an infinite one.

{a_n} = a_1, a_2, a_3, .... , a_n; n is an integer

Convergent Sequence:

A sequence {a_n} converges when

lim a_n = L
n → ∞

when L < ∞.

If the above condition is not met, then the sequence is said to diverge.

Bounded Sequence:

A sequence {a_n} is bounded if there exists a C such that

-C ≤ a_N ≤ C (alternatively |a_n| ≤ C)

If the above condition is not true, then the sequence is said to be unbounded.

Problems

1. Write the first five terms of the sequence {a_n} = 1 / n^3. Does this sequence converge?

a_1 = 1 / 1^3 = 1
a_2 = 1 / 2^3 = 1/8
a_3 = 1 / 3^3 = 1/27
a_4 = 1 / 4^3 = 1/64
a_5 = 1 / 5^3 = 1/125

lim 1 / n^3 = 0 as n → ∞. Conclude that the sequence converges.

Note that the sequence is also bounded. | 1 / n^3 | ≤ 1

2. Write the first five terms of {a_n} = (-1)^(n-1) * n^2. Does this sequence converge.

a_1 = (-1)^(1-1) * 1^2 = 1
a_2 = (-1)^(2-1) * 2^2 = -4
a_3 = (-1)^(3-1) * 3^2 = 9
a_4 = (-1)^(4-1) * 4^2 = -16
a_5 = (-1)^(5-1) * 5^2 = 25

lim (-1)^(n-2) * n^2 as n → ∞:

(-1)^(n-2) is bounded between -1 and 1.
However, n^2 grows to infinity.

Hence (-1)^(n-2) * n^2 → ∞ as n → ∞, the sequence diverges.

Also the sequence is unbounded.

3. Find the first six terms of the sequence {a_n} defined by the recursion formula:

a_0 = 1; a_n = 3 * a_n-1 + 2

a_0 = 1
a_1 = 3 (0) + 2 = 2
a_2 = 3 (2) + 2 = 8
a_3 = 3 (8) + 2 = 26
a_4 = 3 (26) + 2 = 80
a_5 = 3 (80) + 2 = 242
a_6 = 3 (242) + 2 = 728

Coming up: Series!

Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #16: Improper Integrals

Welcome to Part 16 of our 21 part Calculus Revisited series. Today we got down, dirty, and improper.

Improper Integrals


(I)

∞
∫ f(x) dx =
a

lim ( ∫(f(x) dx, a, t) as t → ∞

If f(r) is undefined:

(II)

a
∫ f(x) dx =
r

lim ( ∫f(x) dx, t, a) as t → r+ (t approaches r from the right side)

(III)

r
∫ f(x) dx =
a

lim ( ∫(f(x) dx, a, t) as t → r- (t approaches r from the left side)

Problems

1. Calculate:

4
∫ 2/(x - 2) dx
2

Note 2/(x - 2) is defined at x = 2. This is an improper integral.

∫( 2/(x - 2) dx, 2, 4)
= lim ∫( 2/(x - 2) dx, t, 4) as t → 2+
= lim (2 ln(4 - 2) - 2 ln(t - 2)) as t → 2+

Note: ln t has no limit as t → 0

Therefore the integral has no finite answer. Yes, that can happen.

2. Calculate:

3
∫ dx / √(3 -x)
0

Note that 1/√(3 - x) is undefined at x = 3. Another improper integral.

∫( 1 / √(3 - x) dx, 0, 3)
= lim ∫( 1 / √(3 - x) dx, 0, t) as t → 3-
= lim (- 2 * √(3 - t) + 2 * √(3 - 0) ) as t → 3-
= lim ( -2 * √(3 - t) + 2√3) as t → 3-
= 2√3 ≈ 3.46410

3. Calculate:

∞
∫ e^-x/2 dx
0

So:
∫ (e^(-x/2) dx, 0, ∞)
= lim (e^(-x/2) dx, 0, t) as t → ∞
= lim (-1/2* e^(-t/2) + 1/2 * e^0 ) as t → ∞
= 0 + 1/2
= 1/2

Next time we are going to work with sequences.

Until then, have a great day!

Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #15: Infinite Limits

We have arrived at the 15th blog entry of the 21 entry series of Calculus Revisited. It is all about infinity today, and some quick tips with deal with infinity and limits.

Infinite Limits

General Rules for Polynomials

Let p(x) be a polynomial with degree n. The lead term of p(x) is a_n*x^n.

If n is even and a_n is positive:
* p(x) →+∞ as x → +∞
* p(x) → +∞ as x → -∞

If n is even and a_n is negative:
* p(x) → -∞ as x → +∞
* p(x) → -∞ as x → -∞

If n is odd and a_n is positive:
* p(x) → +∞ as x → +∞
* p(x) → -∞ as x → -∞

If n is odd and a_n is negative:
* p(x) → -∞ as x → +∞
* p(x) → +∞ as x → -∞

Generally: 1 / p(x) → 0 as x → +∞ or x → -∞

If p(x)/q(x) have the same degree, then

p(x)/q(x) → (lead coefficient of p(x)) / (lead coefficient of q(x))

as x → +∞ or x → -∞

General Rules for Other Functions

ln x → -∞ as x → -∞
ln x → +∞ as x → +∞

e^x → 0 as x → -∞
e^x → +∞ as x → +∞

e^-x → +∞ as x → -∞
e^-x → 0 as x → +∞

sin x has no limit as x → +∞ or x → -∞

cos x has no limit as x → +∞ or x → -∞

tan x has no limit as x → +∞ or x → -∞

Note that sin x and cos x are both bounded between -1 and 1.

L'Hospital's Rule

For indeterminate forms 0/0, +∞/+∞, -∞/-∞ you can use the rule:

lim f(x)/g(x) =
x → a

lim f'(x)/g'(x)
x → a

Let's go over some problems.

Problems

1. Find: lim (1/x * sin) as x → ∞

We know that:

lim 1/x = 0 as x → ∞ and sin x is bounded.

Then the overall limit is 0 * (a number between -1 and 1) = 0

2. Find lim (e^(-x^2)) as x → ∞

In general, e^-x → 0 as x → ∞

Even though x^2 grows, e^-x shrinks faster and is dominant.

Hence lim (e^(-x^2)) = 0 as x → ∞.

The next three problems uses the p(x)/q(x) rules.

3. lim (x^2 + 1)/(x - 2) as x → ∞

Degree of numerator = 2
Degree of denominator = 1

Numerator will grow faster than the denominator.

Then lim (x^2 + 1)/(x - 2) → ∞ as x → ∞

4. lim (x^2 + 1)/(x^2 - 1) as x → ∞

Degree of numerator = 2
Degree of denominator = 2

Then lim (x^2 + 1)/(x^2 - 1) = 1/1 = 1 as x → ∞

5. lim (x^2 + 1)/(x^3 - 1) as x → ∞

Degree of numerator = 2
Degree of denominator = 3

Denominator will faster than numerator.

Then lim (x^2 + 1)/(x^3 - 1) = 0 as x → ∞

The next two problems uses L'Hospital's Rule.

6. lim (x^2 / e^x) as x → ∞

lim (x^2 / e^x) as → ∞ = ∞/∞ (Indeterminate form)

Applying the rule:

lim (2x / e^x ) as x → ∞ = ∞/∞ (Indeterminate form)

One more time:

lim (2 / e^x) as x → ∞ = 0

Therefore, lim (x^2 / e^x) = 0 as x → ∞

7. lim (sin x / x) as x → 0

lim (sin x / x) as x → 0 = 0/0 (Indeterminate form)

Applying the rule:

lim (cos x/1) as x→ 0 = 1/1 = 1

Therefore, lim (sin x / x) = 1 as x → 0

This concludes today's session. Next time we work with improper integrals (yes, there are such things!).

Have a great day,

Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #14: Integration - Decomposition of Fractions

Welcome to Part 14 of our 21 part of our Calculus Revisited. Today's blog entry involves integration using decomposition of fractions.

Decomposition of Fractions

Here is the general strategy.

If the denominator has (x - c)^r then:

α / (x - c)^r = A1 / (x - c) + A2 / (x - c)^2 + ... + Ar / (x - c)^r

If the denominator has (x^2 + cx + d)^r then:

β / (x^2 + cx + d) = (A1x + B1) / (x^2 + cx + d) + (A2 x + B2) / (x^2 + cx + d)^2 + ... + (Ar x + Br) / (x^2 + cx + d)^r

Here are some examples.

Problems

Sometimes polynomial division is all you need.

1. ∫ (8x - 3) / (4x + 1) dx

First execute the division (8x - 3) / (4x + 1).

(8x - 3) / (4x + 1) = 2 - 5 / (4x + 1)

So

∫ (8x - 3) / (4x + 1) dx
= ∫ 2 - 5 / (4x + 1) dx
= 2x - 5/4 * ln(4x + 1)

Final: ∫ (8x - 3) / (4x + 1) dx = 2x - 5/4 * ln(4x+ 1) + C


Now here are two basic examples of decomposition of fractions.

2. ∫ dx / ((x + 1)(2x + 3))

Then:

A / (x + 1) + B / (2x + 3 ) = 1 / ((x + 1)(2x + 3))

A (2x + 3) + B (x + 1) = 1

2A(x) + 3A + B(x) + B = 1 + 0(x)

Look at the coefficients and match them.

Constant: 3A + B = 1
Coef. of x: 2A + B = 0

The next task is to solve for A and B.

3A + B = 1
2A + B = 0

Subtract the second equation from the first equation

A = 1

Find B

Using the first equation:
3(1) + B = 1
B = 1 -3
B = -2

Using the second equation:
2(1) + B = 0
2 + B = 0
B = -2

Conclude A = 1, B = -2

Back to our integral:

∫ dx / ((x + 1)(2x + 3))
= ∫ A / (x + 1) + B / (2x + 3) dx
= ∫ 1 / (x + 1) - 2 / (2x + 3) dx
= ln(x + 1) - ln(2x + 3)
= ln( (x + 1) / (2x + 3) )

Final: ∫ dx / ((x + 1)(2x + 3)) = = ln( (x + 1) / (2x + 3) ) + C

3. ∫ (2x + 1) / ( (x^2 + 1) (x - 1) ) dx

Decomposing:

(2x + 1) / ( (x^2 + 1) (x - 1) ) = (Ax + B) / (x^2 + 1) + C / (x - 1)

Then:

(Ax + B) (x - 1) + C (x^2 + 1) = 2x + 1
A(x^2) - A(x) + B(x) - B + C(x^2) + C = 2x + 1 + 0(x^2)
(A + C)x^2 + (-A + B)x + (-B + C) = (0)x^2 + (2)x + 1

Matching up terms of x^2, x, and the constant:
A + C = 0
-A + B = 2
-B + C = 1


From the first equation, C = -A (alternatively A = -C)

Using the second and third equations:
-A + B = 2
-B + C = 1

-A + B = 2
-B + (-A) = 1

-A + B = 2
-A - B = 1

Add the two equations together to get

-2A = 3
A = -3/2

Then: -A - B = 1
-(-3/2) - B = 1
- B = 1 - 3/2
B = 3/2 - 1
B = 1/2

So A = -3/2, B = 1/2, and C = 3/2

And:
∫ (2x + 1) / ( (x^2 + 1) (x - 1) ) dx
= ∫ (Ax + B) / (x^2 + 1) + C / (x - 1) dx
= ∫ ((-3/2)x + 1/2) / (x^2 + 1) + (3/2) / (x - 1) dx
= ∫ (-3/2)x / (x^2 + 1) + (1/2) / (x^2 + 1) + (3/2) / (x - 1) dx
= (-3/4) ln(x^2 + 1) + (1/2) atan x + (3/2) ln (x - 1)

Final:
∫ (2x + 1) / ( (x^2 + 1) (x - 1) ) dx = (-3/4) ln(x^2 + 1) + (1/2) atan x + (3/2) ln (x - 1) + C

That concludes our integration techniques section. Next time we move to infinite limits.

Until next time, take care,

Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #13: Integrals Involving Trigonometric Identities

We are at Part 13 of a 21 part series. This time we will work with integrals involving trigonometric identities.

Problems
All problems today will be indefinite integrals. Problems with definite integrals will be handled similarly.

Identities Involving
sin^2 x + cos^2 x = 1
cos (2x) = 1 - 2 sin^2 x
sin (2x) = 2 sin x cos x
sin^2 x = (1 - cos 2x)/2
cos^2 x = (1 + cos 2x)/2


1. ∫ cos^2 x dx

∫ cos^2 x dx
= ∫ 1 - sin^2 x dx
= ∫ 1 dx - ∫ sin^2 x dx
= x - ∫ 1/2 - 1/2 * cos 2x dx
= x - ( ∫ 1/2 dx - ∫ 1/2 * cos 2x dx )
= x - 1/2* x + 1/2 * ∫ cos 2x dx
= 1/2 * x + 1/2 * sin 2x * 1/2
= 1/2 * x + 1/4 * sin 2x

Final: ∫ cos^2 x dx = 1/2 * x + 1/4 * sin 2x + C

Note:
∫ cos (ax) dx = sin (ax)/a
∫ sin (ax) dx = -cos (ax)/a

2. ∫ sin^2 x cos^2 x dx

∫ sin^2 x cos^2 x dx
= ∫ (1 - cos 2x)/2 * (1 + cos 2x)/2 dx
= ∫ 1/4 * (1^2 - cos^2 (2x)) dx
= 1/4 * ( ∫ 1 - cos^2 (2x) dx )
= 1/4 * ( ∫ 1 - (1 + cos 4x)/2) dx )
= 1/4 * ( ∫ 1/2 - 1/2 * cos 4x dx)
= 1/4 * (1/2 * x - 1/8 * sin 4x)

Final: ∫ sin^2 x cos^2 x dx = 1/4 * (1/2 *x - 1/8 * sin 4x) + C

Trigonometric Substitutions
If the integrand has...

√(a^2 - x^2): use the substitution x = a sin u, dx = a cos u du

√(a^2 + x^2): use the substitution x = a tan u, dx = a sec^2 u du

√(x^2 - a^2): use the substitution x = a sec u, dx = a tan u sec u du

These can get messy!

Identity: 1 + tan^2 x = sec^2 x

3. ∫ 1 / √(4 - x^2) dx

a^2 = 4
a = 2
x = 2 sin u
dx = 2 cos u du

∫ 1/√(4 - x^2) dx
= ∫ (2 cos u) / √(4 - 4 sin^2 u) du
= ∫ (2 cos u) / (√4 * √(1 - sin^2 u) ) du
= ∫ (2 cos u) / (2 * √( cos^2 u ) ) du
= ∫ (2 cos u) / (2 * cos u) du
= ∫ 1 du
= u

x = 2 sin u implies that u = asin(x/2)

= asin(x/2)

Final: ∫ 1 / √(4 - x^2) dx = asin(x/2) + C

4. ∫ 1/ √(x^2 + 64) dx

a^2 = 64
a = 8
x = 8 tan u
dx = 8 sec^2 u du
u = atan(x/8)

∫ 1/ √(x^2 + 64) dx
= ∫ (8 sec^2 u)/(√(64 * tan^2 u + 64) du
= ∫ (8 sec^2 u)/(8 * √(sec^2 u)) du
= ∫ (8 sec^2 u)/(8 * sec u) du
= ∫ sec u du

∫ sec x dx = ln |sec x + tan x|

= ln |sec u + tan u|
= ln | sec(atan(x/8)) + tan(atan(x/8)) |

Refer to the right triangle below:

u = atan (x/8)

tan u = x/8

cos u = 8/√(x^2 + 64)

sec u = √(x^2 + 64)/8


Then:

ln | sec(atan(x/8)) + tan(atan(x/8)) |
= ln | √(x^2 + 64)/8 + x/8 |
= ln | √(x^2 + 64) + x | - ln 8

Final:
∫ 1 / √(x^2 + 64) dx
= ln | √(x^2 + 64) + x | - ln 8 + C

Note that ln 8 + C is a constant. Then.

∫ 1 / √(x^2 + 64) dx
= ln | √(x^2 + 64) + x | + C

5. ∫ 1 / (x^2 - 25)^(3/2) dx

a^2 = 25
a = 5
x = 5 sec u
u = asec(x/5) = acos(5/x)
dx = 5 sec u tan u du

1 + tan^2 x = sec^ x

Then:
∫ 1 / (x^2 - 25)^(3/2) dx
= ∫ (5 sec u tan u) / ( 25 sec^2 u - 25 )^(3/2) du
= ∫ (5 sec u tan u) / ( 25 tan^2 u )^(3/2) du
= ∫ (5 sec u tan u) / ( 25^(3/2) * tan^3 u ) du
= 5 / 25^(3/2) * ∫ (sec u tan u) / tan^3 u du
= 1/25 * ∫ cos u/sin^2 u du
= 1/25 * ∫ cot u csc u du

∫ cot x csc x dx = -csc x

5 / 25^(3/2) = 5 / ((5^2)^(3/2)) = 5 / (5^3) = 1 / 5^2 = 1/25

= 1/25 * -csc u
= -1/25 * csc(acos(5/x))

Refer to the triangle below:

x = 5 sec u

x/5 = sec u

5/x = cos u

√(x^2 - 25)/x = sin u

x/√(x^2 - 25) = csc u


Then:

-1/25 * csc(acos(5/x))
= -1/25 * x/√(x^2 - 25)

Final:
∫ 1 / (x^2 - 25)^(3/2) dx = -1/25 * x / √(x^2 - 25) + C

That concludes Part 13 of our series. Next time we will work with decomposition of fractions.

Have a good day,

Eddie


This blog is property of Edward Shore. © 2012

Calculus Revisited #12: Integration by Parts

Welcome to Part 12 of our 21-part Calculus Revisited series. Today we will cover Integration by Parts in our second of four integration techniques.

Integration by Parts

∫ u(x) * v'(x) dx = u(x) * v(x) - ∫ v(x) * u'(x) dx

Pick the parts u(x) and v'(x) dx. Find the derivative of u(x) and the integral of v'(x).

The trick is to pick the parts u(x) and v'(x) so that v(x) * u'(x) is easier to integrate than u(x) * v'(x).

In practice, you may find that somewhere down the line, u(x) * v'(x) = v(x) * u'(x). This comes up especially where you have sines and cosines.

You may have to repeat this procedure more than once. Let's get to some examples.

Problems

The key to this technique is to pick the correct parts. Again, practice, practice, practice.

If you can pick u(x) such that du/dx is a constant, or goes towards a constant, it will most likely help.

Indefinite Integrals

1. ∫ x sin x dx

u = x
dv = sin x dx

du = 1 dx (take the derivative of u(x))
v = - cos x (take the integral of v(x))

Putting it together:

∫ x sin x dx
= (x) * ( - cos x ) - ∫ (- cos x) (1) dx
= - x * cos x + ∫ cos x dx

The second term is easy to integrate.

= -x * cos x + sin x

For indefinite integrals, you can tack on the "C" at the end

Final Answer: ∫ x sin x dx = -x* cos x + sin x + C

2. ∫ x * 3^x dx

u = x
dv = 3^x dx

du = 1 dx
v = 3^x / ln 3

Then:

∫ x * 3^x dx
= ( x ) * ( 3^x / ln 3 ) - ∫ ( 3^x / ln 3) * 1 * dx
= ( x * 3^x / ln 3 ) - ( 1 / ln 3 ) * ∫ 3^x dx

Again, v(x) * u'(x) is easier to integrate than u(x) * v'(x), which is the idea.

= ( x * 3^x / ln 3 ) - ( 1 / ln 3) * ( 3^x / ln 3 )
= ( 3^x / ln 3) * (x - 1 / ln 3 )

Final Answer: ∫ x * 3^x dx = ( 3^x / ln 3) * (x - 1 / ln 3 ) + C

In the next two problems, integration by parts is used twice.

3. ∫ x^2 * e^x dx

Let's see if we can simplify that x^2 a bit..

u = x^2
dv = e^x dx

du = 2x dx
v = e^x

∫ x^2 * e^x dx
= x^2 * e^x - ∫ 2x* e^x dx

Looks like we have to use integration by parts again.

Concentrate on ∫ 2x * e^x dx:

u = 2x
dv = e^x dx

du = 2 dx
v = e^x

Then:
x^2 * e^x - ∫ 2x* e^x dx
= x^2 * e^x - ( 2x * e^x - ∫ 2 * e^x dx )
= x^2 * e^x - 2x * e^x + 2 ∫ e^x dx

The third term is straight-forward

= x^2 * e^x - 2x * e^x + 2 * e^x
= e^x * (x^2 - 2x + 2)

Final Answer: ∫ x^2 * e^x dx = e^x * (x^2 - 2x + 2) + C

4. ∫ e^x * sin x dx

u = e^x
dv = sin x dx

du = e^x dx
v = -cos x

∫ e^x * sin x
= -e^x * cos x - ∫-e^x * cos x dx
= -e^x * cos x + ∫e^x * cos dx

Let's do the technique again.

Concentrate on ∫ e^x * cos dx

u = e^x
dv = cos x dx

du = e^x dx
v = sin x

Then:
-e^x * cos x + ∫e^x * cos dx
= -e^x * cos x + (e^x * sin x - ∫e^x * sin x dx)

Note that:
∫ e^x * sin x dx = -e^x * cos x + e^x * sin x - ∫e^x * sin x dx

Combine the "∫e^x * sin x dx" terms, and solve for it.

2 * ( ∫ e^x * sin x dx ) = -e^x * cos x + e^x * sin x
∫ e^x * sin x dx = 1/2 * (-e^x * cos x + e^x * sin x)

Final Answer:
∫ e^x * sin x dx = e^x / 2 * (-cos x + sin x) + C


Definite Integrals

5. Calculate

3
∫ x^2 * ln x dx
1

First find the antiderivative F(x) - then we can worry about using the limits later. This is one way to do it.

F(x) = ∫ f(x) dx

We know

b
∫ f(x) dx = F(b) - F(a)
a

F(x) = ∫ x^2 * ln x dx

u = ln x
dv = x^2 dx

du = 1/x dx
v = x^3 / 3

Then:

F(x)
= x^3 / 3 * ln x - ∫ 1/x * x^3 /3 dx
= x^3 * ln x * 1/3 - ∫ x^2 / 3 dx
= 1/3 * x^3 *ln x - x^3 / 9

Now:

3
∫ x^2 ln x dx = F(3) - F(1)
1

= ( (1/3 * 3^3 * ln 3) - (3^3 / 9) ) - ( (1/1 * 1^3 * ln 1) - (1^3 / 9) )
= ( 9 * ln 3 - 3 ) - ( - 1/9 )
= 9 * ln 3 - 3 + 1/9
= 9 * ln 3 - 26/9 ≈ 6.9982

Next time we will be working with integrals involving trigonometric identities.

Until then, take care!

Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #11: Integration by Substitution

Welcome to Part 11 of our 21-part Calculus Revisited journey! Today we are continuing with integrals - and starting the first of four blog entries concerning techniques of integration. I will try my best to explain each one but I feel the best way to demonstrate these techniques is by example.

Integration by Substitution

We have the integral ∫ f(x) dx.

Find a function u(x) such that

du = u'(x) dx and

∫ f(u) * du is easier to integrate than ∫ f(x) dx.

Then ∫ f(u) du = F(u) + C

For the definite integral

b
∫ f(x) dx
a

du = u'(x) dx

and the integral is re-written as

u(b)
∫ f(u) du = F(u(b)) - F(u(a))
u(a)

Let's get to the solved problems to demonstrate this technique.

Problems
Indefinite Integrals

1. ∫ √(4 + 5x) dx

Let u(x) = 4 + 5x
Then du = 5 dx

Note that √u = √(4 + 5x)
and dx = du/5

Rewriting the integral:

∫ √(4 + 5x) dx
= ∫ √u * 1/5 du
= 1/5 * ∫ √u du
= 1/5 * (u^(3/2))/(3/2) + C
= 1/5 * 2/3 * u^(3/2) + C
= 2/15 * u^(3/2) + C
When substitution is used, please don't not forget to state the answer in the original variable, in this case, x. u = 4 + 5x
= 2/15 * (4 + 5x)^(3/2) + C

2. ∫ sin x / √(1 + 2 cos x) dx

Let u = 1 + 2 cos x
Then du = - 2 sin x dx

Note - du/2 = sin x dx (which matches the numerator)

Rewriting the integral:
∫ sin x / √(1 + 2 cos x) dx
= ∫ 1/√u * -1/2 du
= -1/2 * ∫ u^(-1/2) du
= -1/2 * u^(1/2) * 2 + C
= -u^(1/2) + C
= -√u + C
Remember... the answer must contain the original variable!
= -√(1 + 2 cos x) + C

The right substitutions make integration much easier. The next two problems illustrate this.

3. ∫ sin^4 x * cos x dx

Let u = sin x
Then du = cos dx (note the cos x in the integral)

∫ sin^4 x * cos x dx
= ∫ u^4 du
= u^5/5 + C
= sin^5 x/5 + C

4. ∫ e^x / (e^(2x) + 1) dx

Let u = e^x
Then du = e^x dx (matches the numerator)

∫ e^x / (e^(2x) + 1) dx
= ∫ du/(u^2 + 1)
= atan u + C
= atan (e^(2x)) + C

Definite Integrals

If you use the substitution technique with indefinite integrals, do NOT forget to account for the limits! For definite integrals, you will recalculate the limits to account for the substitution. The next two problems illustrate this point.

5. Calculate

1
∫ cos (2x) dx
-1

Let u = 2x
Then du = 2 dx

Note du/2 = dx (try to get the term with the dx to match some part of the integral)

Accounting for the limits:
New Upper Limit = 2 * 1 = 2
New Lower Limit = 2 * -1 = -2

Putting it together:

1
∫ cos (2x) dx
-1

2
= ∫ cos u * 1/2 du
-2

= 1/2 * (sin(2) - sin(-2))

Recall sin(-x) = -sin x

= 1/2 * (sin(2) + sin(2)) = sin(2) ≈ 0.90930


Finding the right substitution may be tricky at times. Algebraic manipulation may be necessary. Here is a more difficult problem.

6. Calculate

9
∫ √t / (√t - 1) dt
4

Let u = √t - 1
Then du = 1/(2√t) dt

To help match the integral, (2√t) du = dt

Note that 2√t = 2(u + 1) = 2u + 2 (Look at the substitution and solve for √t).

So (2u + 2) du = dt

We can't forget about the limits, since we are working with a definite integral.
u = √t - 1

New Upper Limit: √9 - 1 = 4 - 1 = 3
New Lower Limit: √4 - 1 = 2 - 1 = 2

Putting this together:

9
∫ √t / (√t - 1) dt
4

2
= ∫ (u + 1)/u * (2u + 2) du
1

2
= ∫ (1 + 1/u) * (2u + 2) du
1

2
= ∫ 2u + 2 + 2 + 2/u du
1

2
= ∫ 2u + 4 + 2/u du
1

The antiderivative is u^2 + 4u + 2 ln u

= ( (2)^2 + 4(2) + 2 ln (2) ) - ( (1)^2 + 4(1) + 2 ln (1) )

= ( 4 + 8 + 2 ln 2 ) - ( 1 + 4 + 2 * 0 )

= 7 + 2 ln 2 ≈ 8.38629


Best thing to do if you are in a calculus course is to practice, practice, practice!

I hope these six example problems give you an understanding of the substitution technique. Mastery of this technique will come with time and practice - and will speed up integral evaluation.

Next time, we look at the famous (or maybe infamous) Integration by Parts technique.

See you soon,

Eddie


This blog is property of Edward Shore. © 2012

Links to the First Nine Parts of Calculus Revisited

#1: Functions
http://edspi31415.blogspot.com/2012/04/calculus-revisited-1-functions.html

#2: Trig Functions
http://edspi31415.blogspot.com/2012/04/calculus-revisited-2-working-with.html

#3: Logarithm and Exponential Functions
http://edspi31415.blogspot.com/2012/04/calculus-revisited-3-logarithmic-and.html

#4: Limits
http://edspi31415.blogspot.com/2012/04/calculus-revisited-4-limits.html

#5: Derivatives
http://edspi31415.blogspot.com/2012/04/calculus-revisited-5-derivatives.html

#6: Chain Rule
http://edspi31415.blogspot.com/2012/04/calculus-revisited-6-chain-rule.html

#7: Finding Extrema
http://edspi31415.blogspot.com/2012/04/calculus-revisited-7-finding-extrema.html

#8: Related Rates
http://edspi31415.blogspot.com/2012/04/calculus-revisited-8-related-rates.html

#9: Newton's Method
http://edspi31415.blogspot.com/2012/05/calculus-revisited-9-newton-method.html

Calculus Revisited #9: Newton's Method

Welcome to blog entry #9 of 21 of our Calculus Revisited. Today we will cover Newton's Method: an effective and powerful way of finding roots of functions.

Process: Newton's Method

Goal: Find roots of the function f(x). That is, solve for x: f(x) = 0

1. Start with an initial guess. Let this be x_n.

2. Calculate x_(n+1) = x_n - f(x_n)/f'(x_n) (function over its derivative)

3. If x_n+1 matches x_n, you are done, with the root being x_n+1. You determine how accurate the answer is by selecting how many decimal points to match. Calculator accuracy (what you see on the screen) varies from 10 to 12 decimal points.

Caution: The choice of initial value is important. Usually, selecting initial guesses close to the root works the best, but this does not always work.

Also, this method is not 100% in finding roots. If you get x_n and x_n+1 flopping between two distinct values, you are caught in a loop - a root will not be found.

I recommend that you use a calculator when working with Newton's Method. On scientific calculators, you may be able to take advantage of the last answer feature, setting x_n = ans.

Example: (TI -36X Pro, TI-30 Multiview, Casio fx-300, Casio fx-115ES, most graphing calculators)
2 ENTER (2 is stored in ans)
ans - f(ans)/f'(ans) (calculates x_1 with x_n = 2)
Keep pressing ENTER

Problems
Newton's Method will often be used to estimate nth roots (square root, cube root, etc). The first problem will be an example of such.

1. Estimate √6 to 5 decimal places.

Observe that √4 = 2 and √9 = 3, which means √6 lies in between 2 and 3. Let's make an initial guess 2.5.

Use the function f(x) = x^2 - 6. Why?

x^2 = 6
x^2 - 6 = 0

Finding the root to the above equation leads to √6.

Then f'(x) = 2x and

x_n+1 = x_n - (x^2 - 6)/(2x)

Having x_0 as the initial guess,
x_0 = 2.50000
x_1 = 2.45000
x_2 = 2.44949
x_3 = 2.44949

(x_1 = 2.5 - (2.5^2 - 6)/(2*2.5) = 2.45,
x_2 = 2.45 - (2.45^2 - 6)/(2*2.45) = 2.44949, and so on)

So to five decimal places, √6 ≈ 2.44949

2. Find a root of the equation e^x - x = 5 for any x > 0. Let x_0 = 2. (the initial guess)

Get the equation in form of f(x) = 0:
e^x - x = 5
e^x - x - 5 = 0

Then f(x) = e^x - x - 5, f'(x) = e^x - 1, and

x_n+1 = x_n - (e^x_n - x_n - 5)/(e^x_n - 1)

By calculation, we get:

x_0 = 2.00000
x_1 = 1.93911
x_2 = 1.93685
x_3 = 1.93685

Then a root of f(x) is x≈1.93685.

3. Solve x^4 - 4x - 4 = 0 with the condition -1 < x < 1. Find the root to five decimal places.

Hint: You may want to graph f(x) to estimate an appropriate initial guess.

Let x0 = -.5

Then f(x) = x^4 - 4x - 4, f'(x) = 4x^3 - 4, and

x_n+1 = x_n - (x_n^4 - 4x_n - 4)/(4x_n^3 - 4)

By calculation, we get:
x_0 = -0.50000
x_1 = -0.93056
x_2 = -0.86520
x_3 = -0.86198
x_4 = -0.86198

A root of f(x) where -1
Here is a situation where Newton's Method does not work for three picks. Not all f(x) will be successful.
4. Find a root of f(x) = √(x^2 - 4) Fortunately, we can find them using algebraic methods (x = ± 2).

f(x) = √(x^2 - 4)
f'(x) = 1/(2 * √(x^2 - 4) * 2x = x/√(x^2 - 4)
x_n+1 = x_n - (x^2 - 4)/x

If x_0 = 1, then subsequent calculations result in a loop between 1 and 4.
If x_0 = 1.5, then subsequent calculations result in a loop between 1.5 and 2.66667.
If x_0 = 1.75, then subsequent calculations result in a loop between 1.75 and 2.828571.

You may try this with other x_0 and run into similar trouble.

Just be aware.

That wraps it up on Newton's Method. Next we head into integration! - Eddie


This blog is property of Edward Shore. © 2012

Calculus Revisited #7: Finding Extrema

We have arrived at the 7th in a 21 blog entry in our Calculus Revisited series.

Today - we will use the derivative to find local extrema. Extrema include the maximum and minimum values of a function f(x). Finding extrema determines where the function f(x) "turns".

First Derivative Test: A process for finding extreme of the function f(x):

1. Find the roots of f'(x). These roots are called critical points. Let c be a critical point.
2. Compute f'(x) where x is near c but not at x = c.

(A) If f'(x) < 0 when x < c and f'(x) > 0 when x > c, then f(x) has a minimum at x = c.

(B) If f'(x) > 0 when x < c and f'(x) < 0 when x > c, then f(x) has a maximum at x = c.

(C) If (A) or (B) do not apply, then f(x) has an inflection point at x = c.

Second Derivative Test: Another process, this one involving the second derivative:

1. Find the roots of f'(x). These roots are called critical points. Let c be a critical point.
2. Calculate f''(c) (second derivative).

(A) If f''(c) >0, then f(x) has a strict local maximum.

(B) If f''(c)<0, then f(x) has a strict local minimum.

(C) If (A) or (B) do not apply, the second derivative test is inconclusive.

Some observations:

1. If f'(x) > 0, f(x) is said to be increasing at x.
2. If f'(x) < 0, f(x) is said to be decreasing at x.
3. If f''(x) > 0, f(x) is said to be concave up at x. (think of a bowl holding water)
4. If f''(x) < 0, f(x) is said to be concave down at x. (think of a bowl turned upside down)

Problems
1. Find extrema of the function y(x) = -x^3 + 2x + 1

Find the critical points
y'(x) = 0
d/dx (-x^3 + 2x + 1) = 0
-3x^2 + 2 = 0
-3x^2 = -2
x^2 = 2/3
x = ±√(2/3)

The critical points are x = √(2/3) and x = -√(2/3).

By the second derivative test:
y''(x) = -6x

y''(√(2/3)) = -6 * √(2/3) < 0. f(x) has a maximum at x = √(2/3).

y''(-√(2/3)) = -6 * -√(2/3) = 6 * √(2/3) > 0. f(x) has a minimum at x = √(2/3).

√(2/3) ≈ 0.81650

2. An open box will be made out of a 12" x 10" sheet of cardboard. (see drawing below, not drawn to scale). A square of length x inches will be cut from each corner of the sheet. What should be the dimensions to construct a box with maximum volume?

x is going to be restricted. First of all x > 0. Second, x < 5. (because 10 - 2 * 5 = 0).

The volume of the box is:
V = (12 - (x + x)) * (10 - (x + x)) * x
= (12 - x^2) * (10 - 2x) * x
= 4x^3 - 44x^2 + 120x

To find the maximum volume, find the critical points. Use the derivative of V in terms of x.

dV/dx = 0
12x^2 - 88x + 120 = 0
By the quadratic formula:
x = (88 ± √(88^2 - 4*12*120))/(2*12)
where:
x ≈ 1.8107 and x ≈ 5.5226.

The latter measurement does not make sense since 10 - 2*5.5226 < 0, and negative dimensions do not make sense.

Test point x ≈ 1.8107.

The second derivative:
d^2V/dx^2 = 24x- 88

Then 24(1.8107) - 88 < 0. The test point is a local maximum.

The dimensions of this box are approximately 8.3786'' x 6.3786'' x 1.8107'' with a volume of about 96.7706 in^3.

3. Minimize the distance between the point (3,1) and the line y(x) = 2x + 5. (see diagram below).

The distance formula between two points (x1, y1) and (x2, y2).

D = (x2 - x1)^2 + (y2 - y1)^2

Find the distance between (3,1) and (x, 2x+5).

D = (x - 3)^2 + (2x + 5 - 1)^2
= (x - 3)^2 + (2x + 2)^2
= 5x^2 + 2x + 13

Taking the derivative D'(x)

D'(x) = 0
10x + 2 = 0
10x = -2
x = -1/5

Since D''(x) = 10 > 0, the point x is a strict local minimum.

The minimum distance is 5 * (-1/5)^2 + 2 * (-1/5) + 13 = 68/5 = 13.6

Next time we deal with problems involving related rates. Until next time, Eddie.


This blog is property of Edward Shore. © 2012

Calculus Revisited #6: The Chain Rule

Today we cover a very useful rule in Calculus: The Chain Rule. The rule deals with so many applications in higher mathematics. This is entry #6 in a series of 21.

The Chain Rule
d/dx f(g(x)) = f'(g(x)) * g'(x) = df/dg * dg/dx

If you learn nothing else in calculus - learn this rule! We will do some practice problems with chain rule today.

Problems
1. d/dx (3x^2 + 1)^2

Using the rule above, let f = g^2 and g = 3x^2 + 1.

Then d/dx f(g) = df/dg * dg/dx,
f = g^2
df/dg = 2g
g = 3x^2 + 1
dg/dx = 6x

Putting it all together:
d/dx (3x^2 + 1)^2 = 2g * 6x = 2(3x^2 + 1) * 6x = 36x^3 + 12x

2. d/dx cos(2x^3 - 1)

Let f = cos g and g = 2x^3 -1
Then:
df/dg = -sin g
dg/dx = 6x^2

Put it together:
d/dx cos(2x^3 - 1) = -sin g * 6x^2 = -sin(2x^3 -1) * 6x^2 = -6x^2 * sin(2x^3 - 1)

3. d/dx √(e^x + 1)

Note √x = x^(1/2)

Let f = √g and g = e^x + 1
Then:
df/dg = 1/2 * 1/√g
dg/dx = e^x

So:
d/dx √(e^x + 1) = 1/2 * 1/√g * e^x = e^x / (2 * √(e^x + 1))

Now that you get the though process, let's try another one.

4. d/dx cos^3 x = d/dx (cos x)^3

d/dx (cos x)^3 = 3 * (cos x)^2 * d/dx cos x = 3 * (cos x)^2 * -sin x = - 3 cos^2 x sin x

5. Evaluate the slope of the function y(x) = ln sin x at x = π/4

d/dx ln sin x
= 1 / sin x * d/dx (sin x)
= 1 / sin x * cos x
= cot x

To find the slope at x = π/4, find cot π/4 = 1.

6. The Chain Rule can be applied more than once. This problem illustrates it. Find d/dx ln (cos √x)

d/dx ln (cos √x)
= 1/(cos √x) * d/dx (cos √x)
= 1/(cos √x) * (-sin √x) * d/dx (√x)
= (-sin √x)/(cos √x) * 1/(2√x)
= (-tan √x)/(2√x)

Next time, we will find the extremes of functions.

This blog is property of Edward Shore. © 2012

Calculus Revisited #5: Derivatives

Welcome to entry #5 of 21 in our Calculus Revisited series. Today, we tackle derivatives!

First the formal definition.

Derivative:

df/dx = f'(x) =

lim ( f(x + δx) - f(x) ) / δx
δx → 0

Note that both symbols for derivatives, the quotient-like symbol and the use of the prime symbol are used interchangeably.

Use the derivative to:
1. Find the instantaneous rate of change at point x0 using the function f(x).
2. Find the slope at given point.
3. When given an equation dealing with the position of an object, f(t), you can find the velocity of an object by calculating f'(t).

But we are going to jump into doing the derivatives. Here is a basic of derivatives:

d/dx ( f(x) + g(x) ) = f'(x) + g'(x)
d/dx ( f(x) * g(x) ) = f'(x) * g(x) + f(x) * g'(x)
d/dx ( f(x) / g(x) ) = (g(x) * f'(x) - g'(x) * f(x))/(g(x)^2)
d/dx a = 0 (a is a constant)
d/dx x = 1
d/dx x^n = n * x^(n-1)
d/dx sin x = cos x
d/dx cos x = -sin x
d/dx tan x = sec^2 x
d/dx e^x = e^x
d/dx a^x = a^x ln a (a is a constant)
d/dx ln x = 1/x
d/dx asin x = 1/√ (1 - x^2)
d/dx acos x = -1/√ (1 - x^2)
d/dx atan x = 1/(x^2 + 1)

TIP: For d/dx ( f(x) / g(x) ), let N = f(x) (numerator) and D = g(x) (denominator). Then d/dx N / D = ( D * N' - D' * N)/(D^2)

Higher Order Derivative: Repeat the derivative operation on f(x).

f''(x) = d^2/dx^2 is the second derivative: take the derivative of f(x), twice.
f'''(x) = d^3/dx^3 is the third derivative: take the derivative of f(x), thrice (three times).
f^(n)(x) = d^n/dx^n is the nth derivative: take the derivative of f(x) n times.

Problems
1. Find the derivative of f(x) = x^2 + 2x + 3.

We can use the addition property to help us.

d/dx (x^2 + 2x + 3)
= d/dx (x^2) + d/dx (2x) + d/dx (3)
= 2x + 2 + 0
= 2x + 2

Remember, the derivative of a constant is 0.

2. Find the slope of f(x) = -x^5 + 2x - 1 at x = 1

First find the derivative
d/dx (-x^5 + 2x - 1)
= d/dx (-x^5) + d/dx(2x) - d/dx(1)
= -5x^4 + 2 - 0
= -5x^4 + 2
The slope is determined by calculating f'(1).
f'(1) = -5(1)^4 + 2 = -3

The slope of f(x) at x = 1 is -3.

3. Find the derivative of f(x) = x^2 * ln x.

Use the product rule: x^2 multiplied by ln x.

d/dx (x^2 * ln x)
= d/dx (x^2) * ln x + x^2 * d/dx (ln x)
= 2x * ln x + x^2 * 1/x
= 2x * ln x + x

4. Find the derivative of f(x) = (x^2 -1)/(x + 2)

Use the division rule: numerator is x^2 -1 and denominator is x + 2.

d/dx ((x^2 - 1)/(x + 2))
= ((x + 2) * d/dx(x^2 - 1) - (x^2 - 1) * d/dx(x + 2))/(x + 2)^2
= ((x + 2) * 2x - (x^2 - 1) * 1)/(x + 2)^2
= (2x^2 + 4x - x^2 + 1)/(x + 2)^2
= (x^2 + 4x + 1)/(x + 2)^2

5. Higher order derivatives: Find the first, second, and third derivative of f(x) = 3x^6 + 7x and g(x) = sin x respectively.


f(x) = 3x^6 + 7x
First Derivative
f'(x) = 18x^5 + 7
Second Derivative
f''(x) = 90x^4
Third Derivative
f'''(x) = 360x^3

g(x) = sin x
g'(x) = cos x
g''(x) = -sin x
g'''(x) = -cos x

6. An interesting property: d/dx( asin x + acos x)

d/dx(asin x + acos x)
= d/dx asin x + d/dx acos x
= 1/√(1-x^2) + (-1)/√(1-x^2)
= 0

This implies that the function asin x + acos x for all x. Bonus question: what is that constant?

Thanks for joining us. Next time we will work with an important rule in calculus: The Chain Rule.

This blog is property of Edward Shore. © 2012

By the way, the answer to the bonus question: asin x + acos x = π/2

Calculus Revisited #4: Limits

Welcome to Calculus Revisited! This is blog entry #4 of a 21 blog entry series. Today, we will cover the basics of limits.

Limit: The limit of a function f(x) is a value of which a function approaches as x creeps closer and closer to it's target x=a. This is not the same as f(a).

In notation:

lim f(x) = L
x→a

The limit may or may not exist.

Left Side Limit: The limit of a function f(x) as x approaches a from the left side. Hence, x < a.

In notation:

lim f(x) = L
x→ a-

Right Side Limit: The limit of a function f(x) as x approaches a from the right side. Hence, x > a.

In notation:

lim f(x) = L
x→ a+

If the left side limit is equal to the right side limit, then the general limit exists at L.

Continuous Function: A function f(x) is continuous if a limit exists for each x in the domain (or specified interval), and that limit is the same as the function's value. It has been often said that you can graph continuous functions without lifting a pencil.

Common continuous functions include:
f(x) = p(x) (polynomials a_n * x^n + ... + a_0)
f(x) = e^x
f(x) = ln x (for x>0 only)
f(x) = sin x
f(x) = cos x

Properties of Limits

lim (f(x) + g(x)) = lim f(x) + lim g(x)

lim c *f(x) = c * lim f(x), c is a constant


Common Ways to Attack Limits

1. If f(x) is continuous, then

lim f(x) = L
x →a

for all x.

2. "Calculator Method": plug in various x_i as x approaches closer and closer to x = a (but not at x = a), observe the results and make a educated conclusion.

3. Graph the function. If graphing calculators are allowed, this is the time to use them.

Problems
1. Find

lim x^2 + 1
x→2

Using the "calculator method":

Left Side Limit: (x approaches 2 with x < 2)
f(2 - .01) = 4.9601
f(2 - .001) = 4.99601
f(2 - .0001) = 4.99960001
f(2 - 10^-9) = 4.999999996
The value is getting close to 5.

Right Side Limit: (x approaches 2 with x > 2)
f(2 + .01) = 5.0401
f(2 + .001) = 5.004001
f(2 + .0001) = 5.00040001
f(2 + 10^-9) = 5.000000004
The value is getting close to 5.

We can reasonably conclude that

lim x^2 + 1 = 5 as x → 2

We could have also observed that x^2 + 1 is continuous everywhere and figured the limit out by plugging in 2 for x.

Now let's go to a case where f(x) is not continuous everywhere.

2. Find

lim 1/(x-2)
x→2

f(2) = 1/0. So plugging in x=2 does not work here.

Left Side Limit: (x approaches 2 with x < 2)
f(2 - .001) = -1,000
f(2 - .0001) = -10,000
f(2 - .00001) = -100,000
f(2 - 10^-9) = -1,000,000,000
Note that f(x) is getting to be a very large negative number, towards -∞

Right Side Limit: (x approaches 2 with x > 2)
f(2 + .001) = 1,000
f(2 + .0001) = 10,000
f(2 + .00001) = 100,000
f(2 + 10^-9) = 1,000,000,000
Note that f(x) is getting to be a very large positive number, towards ∞

But -∞ ≠ ∞

3. Find

lim (sin x)/x
x→ 0

Again, f(0) = 0/0, the plugging it won't work.

Left Side Limit:
f(-.001) = .9999998333
f(-.0001) = .9999999983
f(-.00001) = 1 (calculator returns 1)

Right Side Limit:
f(.001) = .9999998333
f(.0001) = .9999999983
f(.00001) = 1 (calculator returns 1)

From the "Calculator Method":

lim (sin x)/x = 1
x → 0

The next time we will working with derivatives. See you next time, Eddie.

This blog is property of Edward Shore. © 2012

Calculus Revisited #3: Logarithmic and Exponential Functions

Welcome to Part 3 of the 21 part series: Calculus Revisited. Today will we talk about exponential and logarithmic functions.

Exponential and Logarithmic Functions

The function y(x) = e^x is a function where the constant e is taken to the power of x. To 25 decimal places, e = 2.7182818284590452353602874. e is known as the Euler's Number or Napier's Constant. (Wikipedia Source)

The function y(x) = ln x is known as the natural logarithmic function (base e), and is the inverse function of the exponential function. In computer mathematical software and Microsoft Excel ln x is referred to as log x.

Note: On calculators, the function log x refers to a common logarithmic function, not natural. log x uses base 10.

Graphs of both e^x and ln x follow.

Common Properties of the Exponential and Logarithmic Functions

e^x * e^y = e^(x + y)
e^x / e^y = e^(x- y)
(e^x)^y = e^(x * y)

ln(x * y) = ln x + ln y
ln(x / y) = ln x - ln y
ln(x^y) = y * ln x

e^(ln x) = x
ln (e^x) = x

log_n θ = ln θ / ln n (logarithm to base n)

Problems
1. Solve 2^x = 68

2^x = 68
Take the logarithm of both sides
ln (2^x) = ln 68
x ln 2 = ln 68
x = ln 68/ln 2 ≈ 6.08746

2. Solve e^(2x) = 2 * e^(3x)

e^(2x) = 2 * e^(3x)
e^x ≠ 0 for all x, so we can divide.
Divide by e^(2x)
1 = 2 * e^(3x) / e^(2x)
1 = 2 * e^(3x - 2x)
1 = 2 * e^x
1/2 = e*^x
x = ln (1/2) ≈ -.69315

3. Solve e^(2x + 1) = 2^(2x - 3)

e^(2x + 1) = 2^(2x - 3)
ln e^(2x + 1) = ln 2^(2x - 3)
ln e = 1
2x + 1 = (2x - 3) * ln 2
2x + 1 = (2 ln 2)x - (3 ln 2)
(2 - 2 ln 2)x = -3 ln 2 - 1
(2 - 2 ln 2)x = -(3 ln 2 + 1)
x = -(3 ln 2 + 1)/(2 - 2 ln 2) ≈ -5.01778

Next time, we tackle limits. Until next time, take care! Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #2: Working with Trigonometric Functions

Welcome to Part 2 of a series of the 21 Part Series Calculus Revisited, where we cover some of the major topics. Today's and tomorrow's blog entries will be cover the various transcendental functions: today trigonometric functions, tomorrow exponential and logarithmic functions.

Trigonometric Functions

Trigonometric Functions include sine, cosine, and tangent. (abbreviated sin, cos, and tan, respectively) They have corresponding reciprocal functions, cosecant, secant, and cotangent (abbreviated csc, sec, and cot respectively).

The relationship between the trigonometric and their reciprocal functions:

csc x = 1 / sin x
sec x = 1 / cos x
cot x = 1 / tan x

The inverse functions of sine, cosine, and tangent are the arcsine, arccosine, and arctangent functions, respectively. (abbreviated asin or sin^-1, acos or cos^-1, atan or tan^-1 respectively)

sin (asin x) = x
asin (sin x) = x
cos (acos x) = x
acos (cos x) = x
tan (atan x) = x
atan (tan x) = x


The graphs of sin x, cos x, and tan x follow.

Remember: sin^2 x = (sin x)^2. sin^-1 x = asin x. This applies for all trigonometric functions.

More Common Trig Identities

tan x = sin x/cos x
cot x = cos x/tan x

sin^2 x + cos^2 x = (sin x)^2 + (cos x)^2 = 1
tan^2 x + 1 = sec^2 x
1 + cot^2 x = csc^2 x

sin(π - x) = sin x
cos(π - x) = -cos x

sin 2x = 2 sin x cos x
cos 2x = 1 - 2 sin^2 x

sin(a + b) = cos a sin b + sin a cos b
cos(a + b) = cos a cos b - sin a sin b

Problems

1. Let y(x) = a sin (bx), where a and b are any real numbers. What is the significance of a and b?

a is the amplitude of the sine function, where a is the maximum value of y(x) and -a is the minimum value of y(x).

b involved in the period of y(x). The period is π/b.

Note: The variables of a and b have the same significance in the function y(x) = a cos (bx).

2. Is sin x an even or odd function? What about cos x and tan x?

The function f(x) is an even function if for every x, f(-x) = f(x). The function f(x) is an odd function if for every x, f(-x) = -f(x).

Using the graphs above as a reference, sin(-x) = -sin x and cos(-x) = cos x. Note that tan x = sin x/cos x. Then tan(-x) = sin(-x)/cos(-x) = - sin x/cos x = - tan x.

Therefore, sin x is an odd function, cos x is an even function, and tax x is an odd function.

3. Solve cos^2 x - cos x = sin^2 x.

Recall sin^2 x + cos^2 x = 1
Then sin^2 x = 1 - cos^2 x

cos^2 x - cos x = sin^2 x
cos^2 x - cos x = 1 - cos^2 x
2 cos^2 x - cos x - 1 = 0
2 (cos x)^2 - (cos x) - 1 = 0
This is quadratic equation in terms of cos x. This implies that
(cos x - 1) * (2 cos x + 1) = 0
Hence:
cos x - 1 = 0
cos x = 1
x = acos 1 = 0
and
2 cos x + 1 = 0
2 cos x = -1
cos x = -1/2
x = acos (-1/2) = 2π/3

The solutions are x = 0, 2π/3

4. Solve 2 sin x = sin 2x

Recall sin 2x = 2 sin x cos x

2 sin x = sin 2x
2 sin x = 2 sin x cos x
Since sin x can be zero, don't divide by sin x. Instead
2 sin x - 2 sin x cos x = 0
Factor out 2 sin x...
(2 sin x)(1 - cos x) = 0
Then
2 sin x = 0
sin x = 0
x = asin 0 = 0
and
1 - cos x = 0
1 = cos x
x = acos 1 = 0

The solution is x = 0.

Next time we will work exponential and logarithmic functions. Until then, take care! Eddie


This blog is property of Edward Shore. © 2012

Calculus Revisited #1: Functions

Welcome to Calculus Revisited

Greetings. Over the next 21 blog entries (Monday - Friday), Eddie's Math and Calculator Blog will revisit and review some of the major topics in one-variable calculus. I may not get to everything, but I am going to hit a lot of the major topics. Let's start off with some pre-calculus and then head into derivatives and integrals. This blog is going to serve as a reference: whether you are a student or a professional. Enjoy, and let's get started!

Sources Used for this Series

Ross, Debra Anne. "Master Math: Calculus" Career Press: Franklin Lakes, NJ 1998

Silverman, Richard A. Ph.D "Calculus with Analytic Geometry" Prentice-Hall Inc., Englewood Cliffs, NJ 1985


With the preliminairies out of the way...

Function Basics

Function: Is an expression (formula or equation) that associates each element of the domain to a corresponding element of the range. A true function is an expression where every element in the domain corresponds to 1 and only 1 element in a range. So no one element in domain can correspond to two (or more) elements in the range.

Domain: The initial set to which a function is applied. The variable that describes the domain is the independent variable.

Range: The set of results. The variable that describes the range is the dependent variable.

So...

domain → Function → range
x → y = f(x) → y

Note: For the purpose of this series, I will use x, t, θ (the Greek letter theta) for independent variables and y, f, and g for dependent variables. Obviously any letter or description can be used for variables.

The set of real numbers: that is all numbers including integers, decimals, fractions, and irrational numbers. Real numbers just don't include any complex or imaginary number. So, if √-1 (labeled i or j) is involved, the number is not a real number.

This series will focus exclusively on the set of real numbers.

---------------


1. Find the domain and range of the function y(x) = 2/(x-4).

Note that y(x) is not defined at x=4. Why? y(4) = 2/(4-4) = 2/0. We all know that you can't divide by zero. You can calculate y(x) at various points to verify, however y(x) returns answers that cover the set of real numbers.

Domain: The set of real numbers where x ≠ 4
Range: The set of real numbers

2. Find the domain and range of the function y(x) = √ (16 - x^2).

At first glance, y(x) is defined for all x that make 16 - x^2 nonnegative (positive or zero). To calculate the appropriate domain:

Set:
16 - x^2 ≥ 0
16 ≥ x^2
Take the square root of both sides.
x ≥ -4 and x ≤ 4
The proper domain is
-4 ≤ x ≤ 4

Observe that y(x) hits its maximum point when x = 0
y(0) = √ (16 - 0^2) = √ 16 = 4
Minimum points:
y(4) = √ (16 - 4^2) = 0 and
y(-4) = √ (16 - (-4)^2) = 0

Domain: -4 ≤ x ≤ 4
Range: 0 ≤ y ≤ 4

Composite Functions: Basically, a function of a function. Often noted with a raised circle.

Let f(x) and g(x) be two functions. Then f(g(x)) and g(f(x)) are two composite function. In f(g(x)), the g(x) is treated as the independent variable, and vice versa.

3. Let f(x) = x^2 and g(x) = sin x. Find the composite functions f(g(x)) and g(f(x)).

f(g(x)) = f(sin x) = (sin x)^2 = sin^2 x

Caution: Calculators and most mathematical (if not all) software will not allow the user to enter expressions such as sin^2 x. Use the expression (sin x)^2 instead.

g(f(x)) = g(x^2) = sin (x^2)

Inverse Functions: A function that describes a rule from an element of the range to the element of the domain. If y = f(x), then x = f^-1(y).

4. Let f(x) = 7x + 1. Find the inverse function.

Set y as y = 7x + 1. Solve for x.
y = 7x + 1
y - 1 = 7x
(y - 1)/7 = x

Hence the inverse function is f^-1(y) = (y - 1)/7

Tomorrow: Working with Trigonometric Functions

This blog is property of Edward Shore. © 2012

Calculus Revisited # 20: Taylor Series and Maclaurin Series

Welcome to Part 20 of 21: Taylor and Maclaurin Series.

The Taylor and Maclaurin series are representation of the function f(x) by using an infinite series. If we use a finite number of terms, the series can (I stress can), but a good approximation f(x).

A Taylor series of f(x) is centered on a focus point x = a. Generally, approximations are best when x is around a, and gets worse the further x gets from a.

Taylor Series:

About the point x = a:

f(x) = f(a) + f'(a) * (x - 1) + f''(a) * (x - a)^2 / 2! + f'''(a) * (x - a)^3 / 3! + ....

If the series is cut off at n terms, the final term of the Taylor series is:

f^(n+1)(t) / (n+1)! * (x - a)^(n+1). This is known as the error term.

Maclaurin Series:

The Taylor series with a = 0.

f(x) = f(0) + f'(0) * x + f''(0) * x^2/2! + f'''(0) *x^3/3! + ...

With error term f^(n+1)(0) / (n+1)! * x^(n+1)

Some famous series:

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + .... + x^n/n! + ....

sin x = x - x^3/3! + x^5/5! - x^7/7! + .... + ( (-1)^n * x^(2n+1) ) / (2n+1)! + ....

cos x = 1 - x^2/2! + x^4/4! - x^6/6! + ... + ( (-1)^n * x^(2n) ) / (2n)! + ....

Problems

1. Find a Macluarin series for

f(x) = ln(x + 1) to four terms

f(x) = ln(x + 1)
f(0) = ln(0 + 1) = 0

f'(x) = 1/(x+1)
f'(0) = 1/1 = 1

f''(x) = -1/(x+1)^2
f''(0) = -1/(1^2) = -1

f'''(x) = 2/(x+1)^3
f'''(0) = 2/(1^3) = 2

Then:

ln(x + 1) = 0 + 1 * x + (-1) * x^2/2! + 2 * x^3/3! + ....
= x - x^2/2! + 2x^3/3! + ....

2. Find the Macluarin series for

f(x) = e^(-x^2) to three nonzero terms. Approximate its integral.

f(x) = e^(-x^2)
f(0) = 1

f'(x) = e^(-x^2) * (-2x)
f'(0) = 0

f''(x) = e^(-x^2) * (4x^2 - 2)
f''(0) = -2

f'''(x) = e^(-x^2) * (-8x^3 + 12x)
f'''(0) = 0

f''''(x) = e^(-x^2) * (16x^4 - 48x^2 + 12)
f''''(0) = 12

Then:

e^(-x^2) = 1 - 2 * x^2/2! + 12 * x^4/4! + ....
= 1 - x^2 + x^4/2 + ...

∫ e^(-x^2) dx = x -x^3/3 + x^5/10 + .... + C

3. Find the Taylor series for cos x at a = π/4 to three nonzero terms.

f(x) = cos x
f(π/4) = √2/2

f'(x) = - sin x
f'(π/4) = -√2/2

f''(x) = -cos x
f''(π/4) = -√2/2

Then:

cos x = √2/2 - √2/2 * (x - π/4) + √2/2 * (x - π/4)^2/2! + ....

That concludes our section on Taylor and Maclaurin series. The next and last entry will be our "catch all" entry.

Thank you as always,

Eddie

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