## Saturday, September 23, 2006

### Intermediate Value Property of Continuous Functions

Today's blog is another presentation of an elementary proof that I will use in my proof of the fundamental theorem of calculus.

The content in today's blog is taken from Edwards & Penney's Calculus and Analytic Geometry.

Theorem: Intermediate Value Property of Continuous Functions

If a function f is continuous on [a,b] and f(a) is less than K which is less than f(b), then K = f(c) for some number c in (a,b).

Proof:

(1) Let us start by defining three sequences {ai}, {bi}, and {Ii}

a1 = a
b1 = b

mi = midpoint of [ai,bi]

we can assume that f(mi) ≠ K since if it did, then c = mi and we are done with this proof.

if K is less than f(mi), then:

ai+1 = ai
bi+1 = mi

Otherwise K is greater than f(mi) and:

ai+1 = mi
bi+1 = bi

In this way, we can also define a sequence of nested intervals such that:
Ii = [ai,bi] and for all i, K lies between f(ai) and f(bi)

(2) Using the Nested Intervals Property of Real Numbers (see Lemma 1, here), we know that there exists a point c in I1 such that {c} = I1 ∩ I2 ∩ I3 ∩ ...

(3) We can see that both {ai} and {bi} have this point c as their limit. [See Definition 1, here for a definition of a limit]

(4) Since the function f(x) is continuous on (a,b), we know (see Definition 1, here for definition of continuous on a point) that:

for any positive real ε, there exists a positive real δ such that:

if abs(x-c) is less than δ, then abs(f(x) - f(c)) is less than ε

(5) By the definition of limits of a function (see Definition 1, here), it follows that the limit of {f(bn)} and the limit of {f(an)} is f(c).

(6) Since for all n, f(bn) is greater than K, it follows that f(c) ≥ K.

(7) Likewise for all n, f(an) is less than K so that f(c) ≤ K.

(8) But the only way that step #6 and step #7 can both be true is if f(c)=K.

QED

References

## Wednesday, September 20, 2006

### Properties of integrals

In today's blog, I use Reimann sums (see Definition 2, here) to establish some basic properties of integrals. I will later use these properties to present the fundamental theorem of calculus.

The content in today's blog is taken straight from Edwards and Penney's Calculus and Analytic Geometry.

Lemma 1: Integral of a Constant

∫ (b,a) c dx = c(b-a)

Proof:

In the case of a constant, one interval will do and the area is exactly c(b-a). [See here for a discussion of Riemann sums]

QED

Lemma 2: Constant Multiple Property

∫ (b,a) cf(x)dx = c ∫ (b,a) f(x)dx

Proof:

(1) Let R = the Riemann sum (see Definition 2, here) for this equation so that we have:

R = ∑ (j=1,n) c*f(cj)(xj - xj-1)

(2) Now, since c is constant, we have:

R = c*∑ (j=1,n) f(cj)(xj - xj-1)

(3) Now, using the definition of the integral as the limit of the Riemann sum R (see Definition 5, here), we get:

I = lim (n → ∞) R = lim (n → ∞) c*∑ (j=1,n) f(cj)(xj - xj-1) =

= c*lim(n → ∞) ∑ (j=1,n) f(cj)(xj - xj-1) = c*∫ (b,a) f(x)dx

QED

Lemma 3: Interval Union Property

If a is less than c is less than b, then:

∫ (a,b) f(x)dx = ∫ (c,a) f(x)dx + ∫ (b,c) f(x)dx

Proof:

(1) Let R1 = Riemann sum for f(x) on the interval [c,a] with Partition P1 and a set S1 of arbitrary points in each subinterval [See Definition 2, here for definition of Riemann sum]

(2) Let R2 = Riemann sum for f(x) on the interval [b,c] with Partition P2 and a set S2 of arbitrary points in each subinterval [See Definition 2, here for definition of Riemann sum]

(3) Let S = S1 ∪ S2

(4) Let P = P1 ∪ P2

(5) Let x1,i be the set of intervals that make up P1 [See Definition 1, here for definition of partition]

(6) Let x2,i be the set of intervals that make up P2 [See Definition 1, here for definition of partition]

(7) Since a is less than c which is less than b, we have:

a=x1,0 less than x1,1 less than ... less than x1,n=c=x2,0 less than x2,1 less than ... less than x2,n = b

(8) From this, we can see that P is a partition on [a,b] (See Definition 1, here for definition of partition)

(9) Since S1 is a set of arbitary points in each subinterval on [a,c] and S2 is a set of arbitrary points in each subinterval on [b,c], it follows that S is set of arbitrary points on [a,c] [See Definition 2, here on Riemann sum]

(10) So, R is a Riemann sum since it is defined on [a,c] using a partition P and a set S of arbitrary points in each subinterval.

(11) Since f(x) is continuous, we know that R, R1, and R2 have limits (that is, their integrals exist. (See Theorem, here)

(12) Therefore, it follows that:

lim (n → ∞) R = lim (n → ∞)R1 + lim(n → ∞) R2

which using the definition of definite integrals in terms of Riemann sums (see Definition 5, here) gives us:

∫ (a,b) f(x)dx = ∫ (c,a)f(x)dx + ∫ (b,c) f(x)dx.

QED

Lemma 4: Comparison Property

If m ≤ f(x) ≤ M for all x in [a,b], then:

m(b - a) ≤ ∫ (b,a) f(x)dx ≤ M(b-a)

Proof:

(1) Let m be minimum of f(x) on [a,b] (See Theorem, here for proof of the existence of the minimum)

(2) Let M be ≥ maximum of f(x) on [a,b] (See Lemma 3, here for proof of the existence of the maximum)

(3) For any partition P on [a,b], m(b-a) ≤ L(P) [Since L(P) = min of f(x) on [a,b], see Definition 2, here]

(4) Likewise, M(b-a) ≥ U(P) [Since U(P) = max of f(x) on [a,b], see Definition 2, here]

(5) Since L(P) ≤ I ≤ U(P) [See Lemma 1, here], we have:

m(b-a) ≤ L(P) ≤ I ≤ U(P) ≤ M(b-a)

QED

References

### Fundamental Theorem of Calculus

In today's blog, I present the Fundamental Theorem of Calculus. This is the insight that the integral, the area under the curve, is the same as the antiderivative (that is, the inverse of the derivative) of the curve itself. This enables evaluation of the integral without relying on a Riemann sum.

The Fundamental Theorem is often presented as a single proof. In today's blog, I break it down into two theorems.

The content is taken from Edwards and Penney's Calculus and Analytic Geometry.

Definition 1: Average Value of a Function

If a function f is integrable on [a,b], then the average value y of y = f(x) on [a,b] is:

y = 1/(b-a) ∫ (b,a) f(x)dx

This definition becomes more intuitive, when we realize that:

∫ (b,a) f(x) dx = (b-a)y = (b-a)f(x)

This definition is used in the following theorem:

Theorem 1: Average Value Theorem

If a function f is continuous on [a,b], then:

f(x) = 1/(b-a) ∫ f(x)dx for some point x on [a,b]

Proof:

(1) Let m = f(c) be the minimum value of f on [a,b] (See Theorem, here for proof that it exists)

(2) Let M = f(d) be the maximum value of f on [a,b]. (See Lemma 3, here for proof that it exists)

(3) Using the Comparison Property (see Lemma 4, here),

m(b-a) ≤ ∫ (b,a) f(x)dx ≤ M(b-a)

Then, dividing all sides by (b-a) gives us:

m ≤ 1/(b-a) ∫ (b,a) f(x)dx ≤ M

(4) Because f is continuous and because y=[1/(b-a) ∫ (b,a) f(x)dx] lies between two values of f: m and M, we can use the Intermediate Value Property (see Theorem, here) to conclude that y must be a value of f.

(5) Thus, there must exists a point x where y = f(x) and where x lies between a and b.

QED

Definition 2: antiderivative

A function F is an antiderivative of a function f if and only if the derivative F'(x) = f(x).

The fundamental theorem of calculus (see Theorem 2 below) gives another notation for this same point:

F(x) = ∫ f(x)dx

Here is the proof.

Theorem 2: The integral is the antiderivative of a continuous function.

d/dx (∫ (x,a) f(t)dt) = f(x) where f(x) is a continuous function

Proof:

(1) Let F(x) be a continuous function defined on [a,b] by:

F(x) = ∫ (x,a) f(t)dt

(2) By the definition of the derivative (see Definition 1, here):

F'(x) = lim (Δx → 0) [F(x + Δx) - F(x)]/Δx

= lim (Δx → 0) (1/Δx)(∫ (x+Δx,a) f(t)dt - ∫ (x,a) f(t)dt)

(3) Using the interval union property (see Lemma 3, here), we know that:

∫ (x+Δx,a) f(t)dt = ∫ (x,a) f(t)dt + ∫(x+Δx, x) f(t)dt

(4) Thus:

F'(x) = lim (Δx → 0) (1/Δx) ∫ (x+Δx,x) f(t)dt

(5) The average value theorem (see Thereom 1, above) gives us:

1/Δx ∫ (x+Δx,x) f(t)dt = f(t) for some number t in [x, x + Δx].

(6) It is clear that as Δx → 0, t → x.

(7) Thus, because f is continuous, we see that:

F'(x) = lim(Δx → 0) (1/Δx) ∫ (x+Δx,x) f(t)dt = lim (Δx → 0) f(t) = lim (t → x) f(t) = f(x).

(8) Hence the function F defined above is indeed the antiderivative of f.

QED

Theorem 3: Evaluation of Definite Integrals

The definite integral of a continuous function G'(x) is equal to the difference of the antiderivatives:

∫ (b,a) G'(x)dt = G(b) - G(a)

Proof:

(1) Let F'(t)=f(t) be a continous function on [a,b].

NOTE: For purposes of clarity, I use f(t) and F'(t) when refering to the continuous function itself and F(x),G(x) when refering to the antiderivative. The reason for this is to make clear the definite integral:

F(x) = ∫ (x,a) f(t)dt.

(2) F(x) is then its antiderivative with:

F(x) = ∫ (x,a) f(t)dx

(3) Let G be any antiderivative of F'(t) such that:

G(x) = F(x) + C on [a,b] where C is a constant.

Since F'(t) is the continuous function, F(x) is the antiderivative and the set of all antiderivatives is characterized by F(x) + C since d/dx(F(x) + C) = d/dx(F(x)) + d/dx(C) = F'(x) + 0 = F'(x) [See Lemma 1, here for details on why F(C)=0]

(4) Since F(a) = ∫ (a,a) f(t)td = 0, we know that:

G(a) = F(a) + C = 0 + C = C

(5) Combining this with step #1 gives us:

G(x) = F(x) + G(a) for all x in [a,b].

which is the same as:

F(x) = G(x) - G(a)

(6) If x=b, this gives us:

G(b) - G(a) = F(b) = ∫ (b,a) f(x)dx.

QED

References

## Tuesday, September 19, 2006

### Summation Formula

It is not very difficult to verify the simplest summation formulas but it raises the question how these formulas were derived.

In today's blog, I will first use induction to prove the summation formulas for ∑ x, ∑ x2, and
∑ x3. Then, I will show how it is possible to derive each of these formulas.

I will use the summation formula ∑ x3 in my example of using a Riemann sum to calculate the area under a simple curve.

Lemma 1:

(a) (n-1)2 = n2 - 2n + 1

(b) (n-1)3 = n3 - 3n2 + 3n - 1

(c) (n-1)4 = n4 -4n3 + 6n2 -4n + 1

Proof:

(a) This follows straight from multiplying (n-1)(n-1)

(b) (n-1)3 = (n-1)(n2 - 2n + 1) = n3 - 2n2 + n -n2 + 2n - 1 = n3 - 3n2 + 3n - 1

(c) (n-1)4 = (n-1)(n3 - 3n2 + 3n - 1) = n4 -3n3 + 3n2 - n -n3 +3n2 -3n + 1 = n4 -4n3 + 6n2 -4n + 1

QED

Lemma 2: ∑(i=1,n) i = (1/2)n2 + (1/2)n

Proof:

(1) ∑ (i=1,1) i = 1 = (1/2)(1) + (1/2)(1) = 1

(2) Since this is true for n=1, let's assume it's true up to n-1.

(3) Then ∑ (i=1,n) i = ∑(i=1,n-1)i + n = (1/2)(n-1)2 + (1/2)(n-1) + n = (1/2)(n2 - 2n + 1) + (1/2)(n) - (1/2) + n = (1/2)n2 + (1/2)n

(4) Which proves its true by mathematical induction.

QED

Corollary 2.1: ∑(i=1,n) i = (1/2)n(n+1)

Proof:

(1) From Lemma 2, we have:

∑(i=1,n) i = (1/2)n2 + (1/2)n

(2) (1/2)n2 + (1/2)n = (1/2)[n2 + n] = (1/2)[n(n + 1)]

QED

Lemma 3: ∑ (i=1,n) i2 = (1/3)n3 + (1/2)n2 + (1/6)n

Proof:

(1) ∑ (i=1,1) i 2 = 1 = (1/3)(1) + (1/2)(1) + (1/6)(1) = 1

(2) Since this is true for n=1, let's assume it's true up to n-1.

(3) Then ∑(i=1,n) i2 = ∑(i=1,n-1)i2 + n2 = (1/3)(n-1)3 + (1/2)(n-1)2 + (1/6)(n-1) + n2

(4) Applying lemma 1 above gives us:

(1/3)(n-1)3 + (1/2)(n-1)2 + (1/6)(n-1) = (1/3)(n3 - 3n2 + 3n - 1) + (1/2)(n2 - 2n + 1) + (1/6)n - (1/6) + n2 = (1/3)n3 -n2 +n - (1/3) + (1/2)n2 -n + (1/2) + (1/6)n - (1/6) + n2 = (1/3)n3 + (1/2)n2 + (1/6)n

(5) Which proves its true by mathematical induction.

QED

Lemma 4: ∑ (i=1,n) i3 = (1/4)n4 + (1/2)n3 + (1/4)n2

Proof:

(1) ∑ (i=1,1) i3 = 1 = (1/4)(1) + (1/2)(1) + (1/4)(1) = 1

(2) Since this is true for n=1, let's assume it's true up to n-1.

(3) Then ∑ (i=1,n) i3 = ∑(i=1,n-1)i3 + n3 = (1/4)(n-1)4 + (1/2)(n-1)3 + (1/4)(n-1)2 + n3

(4) Applying lemma 1 above gives us:

(1/4)(n-1)4 + (1/2)(n-1)3 + (1/4)(n-1)2 + n3 = (1/4)(n4 -4n3 + 6n2 -4n + 1) + (1/2)( n3 - 3n2 + 3n - 1) + (1/4)(n2 - 2n + 1) + n3 =

= (1/4)n4 + (1/2)n3 + (1/4)n2

(5) Which proves its true by mathematical induction.

QED

I now present a method for deriving each of these formulas. The content in this section is taken from Boris Spokoinyi's web site (see references below)

Example 1: ∑ i = (1/2)n2 + (1/2)n

This formula comes from Carl Friedrich Gauss. The story goes that he came up with it when he was a young schoolboy. He was told to add up all the numbers from 1 to 100. Gauss came up with the formula and calculated the sum in less than five minutes.

(1) The sum = 1 + 2 + 3 + 4 + 5 + ...

(2) If we line this up with n going the other way we have:

1 + 2 + 3 + 4 + 5 + ....
n + (n-1) + (n-2) + ....

So that we have (1 + n) + (2 + n - 1) + (3 + n - 2) + ...

(3) This means that addition of the two sums ∑ (i=1,n) (n)*2 = (n)(n+1) and dividing both sides by 2 gives us:

∑ (i=1,n) (n) = (1/2)(n)(n+1) = (1/2)(n2 n) = (1/2)n2 + (1/2)n.

Example 2: ∑ (i=1,n) i2

(1) ∑ (i=1,n) (i+1)3 = ∑ (i=1,n)(i3 + 3i2 + 3i + 1)

(2) ∑ (i=1,n)(i3 + 3i2 + 3i + 1) = ∑ (i=1,n) i3 + 3∑(i=1,n) i2 + 3∑(i=1,n) i + ∑ (i=1,n) 1

(3) From Lemma 2 (see Example 1 for deriving it), we have:

∑ (i=1,n) i = (n)(n+1)/2

(4) ∑ (i=1,n) 1 = 1 + 1 + ... + 1 = n

(5) Applying step #3 and step #4 to step #2 gives us:

∑ (i=1,n) i3 + 3∑(i=1,n) i2 + 3∑(i=1,n) i + ∑ (i=1,n) 1 = ∑ (i=1,n) i3 + 3∑(i=1,n) i2 + 3(n)(n+1)/2 + n

(6) ∑(i=1,n) (i+1)3 = ∑ (i=2,n+1) i3 = ∑ (i=1,n) i3 + (n+1)3 - 13

(7) So that combining step #5 and step #6 gives us:

∑ (i=1,n) i3 + 3∑(i=1,n) i2 + 3(n)(n+1)/2 + n = ∑ (i=2,n+1) i3 = ∑ (i=1,n) i3 + (n+1)3 - 1

After subtracting ∑ (i=1,n) i3 from both sides, we get:

3∑(i=1,n) i2 + 3(n)(n+1)/2 + n = (n+1)3 - 1

(8) Subtracting 3(n)(n+1)/2 + n from both sides gives us:

3∑(i=1,n) i2 = (n+1)3 - 1 - 3(n)(n+1)/2 -n

So that we have:

∑(i=1,n) i2 = (1/3)[(n+1)3 - 1 - 3(n)(n+1)/2 -n]

(9) Since (n+1)3 = n3 + 3n2 + 3n + 1, we have:

(1/3)[(n+1)3 - 1 - 3(n)(n+1)/2 -n] = (1/3)[n3 + 3n2 + 3n + 1 - 1 - (1/2)3n2 - (1/2)(3n) - n] =

= n3/3 + n2/2 + n/6

Example 3: ∑ (i=1,n) i3

(1) Since (i+1)4 = (i4 + 4i3 + 6i2 + 4i + 1), we have:

∑ (i=1,n) (i+1)4 = ∑ (i=1,n) (i4 + 4i3 + 6i2 + 4i + 1) =

= ∑ (i=1,n) i4 + 4∑(i=1,n)i3 + 6∑(i=1,n)i2 + 4∑(i=1,n)i + ∑(i=1,n)1

(2) Likewise, we have:

∑ (i=1,n) (i+1)4 = ∑ (i=2,n+1) i4 = ∑ (i=1,n) i4 + (n + 1)4 - 14

(3) So that we get:

∑ (i=1,n) i4 + 4∑(i=1,n)i3 + 6∑(i=1,n)i2 + 4∑(i=1,n)i + ∑(i=1,n)1 = ∑ (i=1,n) i4 + (n + 1)4 - 1

And subtracting ∑ (i=1,n) i4 from both sides gives us:

4∑(i=1,n)i3 + 6∑(i=1,n)i2 + 4∑(i=1,n)i + ∑(i=1,n)1 = (n + 1)4 - 1

(4) Which means that:

∑ (i=1,n)i3 = (1/4)[(n + 1)4 - 1 - 6∑(i=1,n)i2 - 4∑(i=1,n)i - ∑(i=1,n)1]

(5) Now, we know that:

∑ (i=1,n) = n

∑ (i=1,n)i = (1/2)n2 + (1/2)n [See Lemma 2 above and Example 1 above]

∑ (i=1,n) i2 = (1/3)n3 + (1/2)n2 + (1/6)n [See Lemma 3 above and Example 2 above]

(4) So that we have:

∑ (i=1,n)i3 = (1/4)[(n + 1)4 - 1 - 6(1/3)n3 - 6(1/2)n2 - (6)(1/6)n - 4(1/2)n2 - 4(1/2)n - n] = (1/4)[n4 + 4n3 + 6n2 + 4n + 1 - 1 -2n3 -3n2 -n -2n2 -2n -n] =

= (1/4)[n4 + 2n3 + n2] = (1/4)n4 + (1/2)n3 + (1/4)n2

References