Tag: derivative

How I Impressed My Wife: Part 6g

This series was inspired by a question that my wife asked me: calculate

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

Originally, I multiplied the top and bottom of the integrand by \tan^2 x and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.

green lineSince Q is independent of a, I can substitute any convenient value of a that I want without changing the value of Q. As shown in previous posts, substituting a =0 yields the following simplification:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}

= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}

= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}

= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}

= \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}

= 2\pi i \left[\displaystyle \frac{r_1}{r_1^2-1} + \displaystyle \frac{r_2}{r_2^2-1} \right],

where I’ve made the assumption that |b| < 1. In the above derivation, C_R is the contour in the complex plane shown below (graphic courtesy of Mathworld).

Also,

r_1 = \sqrt{1-b^2} + |b|i

and

r_2 = -\sqrt{1-b^2} + |b|i

are the two poles of the final integrand that lie within this contour.

It now remains to simplify the final algebraic expression. To begin, I note

\displaystyle \frac{r_1}{r_1^2-1} = \displaystyle \frac{\sqrt{1-b^2} + |b|i}{[\sqrt{1-b^2} + |b|i]^2 - 1}

= \displaystyle \frac{\sqrt{1-b^2} + |b|i}{1-b^2 + 2|b|i\sqrt{1-b^2} - |b|^2 - 1}

= \displaystyle \frac{\sqrt{1-b^2} + |b|i}{-2|b|^2 + 2|b|i\sqrt{1-b^2}}

= \displaystyle \frac{\sqrt{1-b^2} + |b|i}{2|b|i(|b|i +\sqrt{1-b^2})}

= \displaystyle \frac{1}{2|b|i}.

Similarly,

\displaystyle \frac{r_2}{r_2^2-1} = \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{[-\sqrt{1-b^2} + |b|i]^2 - 1}

= \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{1-b^2 - 2|b|i\sqrt{1-b^2} - |b|^2 - 1}

= \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{-2|b|^2 - 2|b|i\sqrt{1-b^2}}

= \displaystyle \frac{-\sqrt{1-b^2} + |b|i}{2|b|i(|b|i -\sqrt{1-b^2})}

= \displaystyle \frac{1}{2|b|i}.

Therefore,

Q = 2\pi i \left[\displaystyle \frac{r_1}{r_1^2-1} + \displaystyle \frac{r_2}{r_2^2-1} \right] = 2\pi i \left[ \displaystyle \frac{1}{2|b|i} + \frac{1}{2|b| i} \right] = 2\pi i \displaystyle \frac{2}{2|b|i} = \displaystyle \frac{2\pi}{|b|}.

green lineAnd so, at long last, I’ve completed a fifth different evaluation of Q.

How I Impressed My Wife: Part 6f

This series was inspired by a question that my wife asked me: calculate

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

Originally, I multiplied the top and bottom of the integrand by \tan^2 x and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.

green lineSince Q is independent of a, I can substitute any convenient value of a that I want without changing the value of Q. As shown in previous posts, substituting a =0 yields the following simplification:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}

= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}

= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}

= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}

= \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1},

where C_R is the contour in the complex plane shown below (graphic courtesy of Mathworld).

Amazingly, contour integrals can be simply computed by evaluating the residues at every pole located inside of the contour. (See Wikipedia and Mathworld for more details.) I have already handled the case of |b| = 1 and |b| > 1. Today, I begin the final case of |b| < 1.

Earlier in this series, I showed that

z^4 + (4b^2 - 2) z^2 + 1 = (z^2 + 2z \sqrt{1-b^2} + 1)(z^2 - 2z \sqrt{1-b^2} + 1)

if |b| < 1, and so the quadratic formula can be used to find the four poles of the integrand:

r_1 = \sqrt{1-b^2} + |b|i,

r_2 = -\sqrt{1-b^2} + |b|i,

r_3 = \sqrt{1-b^2} - |b|i,

r_4 = -\sqrt{1-b^2} - |b|i.

Of these, only two lie (r_1 and r_2) within the contour for sufficiently large R (actually, for R > 1 since all four poles lie on the unit circle in the complex plane).

As shown earlier in this series, the residue at each pole is given by

\displaystyle \frac{r^2 + 1}{2r^3 + (4b^2-2)r}

I’ll now simplify this considerably by using the fact that r^4 + (4b^2-2)r^2 + 1 = 0 at each pole:

\displaystyle \frac{r^2 + 1}{2r^3 + (4b^2-2)r} = \displaystyle \frac{r(r^2+1)}{2r^4+(4b^2)-r^2}

= \displaystyle \frac{r(r^2+1)}{r^4+r^4 + (4b^2)-r^2}

= \displaystyle \frac{r(r^2+1)}{r^4-1}

= \displaystyle \frac{r(r^2+1)}{(r^2+1)(r^2-1)}

= \displaystyle \frac{r}{r^2-1}.

Therefore, to evaluate the contour integral, I simply the sum of the residues within the contour and multiply the sum by 2\pi i:

Q = \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1} = 2\pi i \left[\displaystyle \frac{r_1}{r_1^2-1} + \displaystyle \frac{r_2}{r_2^2-1} \right].

green lineSo, to complete the evaluation of Q, I need to simplify the right-hand side. I’ll complete this in tomorrow’s post.

How I Impressed My Wife: Part 6d

This series was inspired by a question that my wife asked me: calculate

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

Originally, I multiplied the top and bottom of the integrand by \tan^2 x and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.

 nvenient value of a that I want without changing the value of Q. As shown in previous posts, substituting a =0 yields the following simplification:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}

= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}

= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}

= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}

= \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}

= 2\pi \displaystyle \left[ \frac{1-r_1^2}{-2r_1^3 + (4b^2-2) r_1} +\frac{1-r_2^2}{-2r_2^3 + (4b^2-2) r_2} \right]

where I’ve assumed |b| > 1, the contour C_R in the complex plane is shown below (graphic courtesy of Mathworld),

and the positive constants r_1 and r_2 are given by

r_1 = \sqrt{2b^2 - 1 + 2|b| \sqrt{b^2 - 1}},

r_2 = \sqrt{2b^2 - 1 - 2|b| \sqrt{b^2 - 1}}.

Now we have the small matter of simplifying our expression for Q. Actually, this isn’t a small matter because Mathematica 10.1 is not able to simplify this expression much at all:

ResidueCalculation

Fortunately, humans can still do some things that computers can’t. As observed yesterday, The numbers r_1 and r_2 are chosen so that \pm ir_1 and \pm ir_2 are the roots of the denominator z^4 + (4 b^2 - 2) z^2 + 1, so that

r_1^2 + r_2^2 = 4b^2 - 2,

r_1 r_2 = 1.

These relationships will be very handy for simplifying our expression for Q:

Q = 2\pi \displaystyle \left[ \frac{1-r_1^2}{-2r_1^3 + (4b^2-2) r_1} +\frac{1-r_2^2}{-2r_2^3 + (4b^2-2) r_2} \right]

= 2\pi \left[ \displaystyle \frac{1-r_1^2}{r_1 (-2r_1^2 + 4b^2-2)} + \frac{1-r_2^2}{r_2(-2r_2^2 + 4b^2-2)} \right]

= 2\pi \left[ \displaystyle \frac{1-r_1^2}{r_1 (-2r_1^2 +r_1^2 + r_2^2)} + \frac{1-r_2^2}{r_2(-2r_2^2 + r_1^2 + r_2^2)} \right]

= 2\pi \left[ \displaystyle \frac{1-r_1^2}{r_1 (-r_1^2 +r_2^2)} + \frac{1-r_2^2}{r_2(r_1^2 -r_2^2)} \right]

 = 2\pi \left[ \displaystyle \frac{r_1^2-1}{r_1 (r_1^2 -r_2^2)} + \frac{1-r_2^2}{r_2(r_1^2 -r_2^2)} \right]

= 2\pi \displaystyle \frac{(r_1^2-1)r_2 + r_1(1-r_2^2)}{r_1 r_2 (r_1^2 -r_2^2)}

 = 2\pi \displaystyle \frac{r_1^2 r_2- r_2 + r_1- r_1 r_2^2)}{r_1 r_2 (r_1^2 -r_2^2)}

= 2\pi \displaystyle \frac{r_1 - r_2 + r_1 r_2 (r_1 - r_2)}{r_1 r_2 (r_1-r_2)(r_1 + r_2)}

= 2\pi \displaystyle \frac{(r_1 - r_2)(1 + r_1 r_2)}{r_1 r_2 (r_1-r_2)(r_1 + r_2)}

= 2\pi \displaystyle \frac{1 + r_1 r_2}{r_1 r_2 (r_1 + r_2)}

= 2\pi \displaystyle \frac{1 + 1}{1 \cdot (r_1 + r_2)}

= \displaystyle \frac{4\pi}{r_1 + r_2}

To complete the calculation, I observe that

(r_1 + r_2)^2 = r_1^2 + 2r_1 r_2 + r_2^2 = 4b^2 -2 + 2 = 4b^2,

so that

r_1 + r_2 = 2|b|.

Therefore,

Q = \displaystyle \frac{4\pi}{r_1 + r_2} = \displaystyle \frac{4\pi}{2|b|} = \displaystyle \frac{2\pi}{|b|}.

green line

In tomorrow’s post, I’ll present another way to simplify this nasty algebraic expression.

How I Impressed My Wife: Part 6c

This series was inspired by a question that my wife asked me: calculate

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

Originally, I multiplied the top and bottom of the integrand by \tan^2 x and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.

green lineSince Q is independent of a, I can substitute any convenient value of a that I want without changing the value of Q. As shown in previous posts, substituting a =0 yields the following simplification:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}

= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}

= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}

= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}

= \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1},

where C_R is the contour in the complex plane shown below (graphic courtesy of Mathworld).

Amazingly, contour integrals can be simply computed by evaluating the residues at every pole located inside of the contour. (See Wikipedia and Mathworld for more details.) I handled the case of |b| = 1 in yesterday’s post. Today, I’ll begin the case of |b| > 1.

To find the poles of the integrand, I use the quadratic formula to set the denominator equal to zero:

z^4 + (4 b^2 - 2) z^2 + 1 = 0

z^2 = \displaystyle \frac{2-4b^2 \pm \sqrt{(4b^2-2)^2 - 4}}{2}

z^2 = \displaystyle \frac{2-4b^2 \pm \sqrt{16b^4 - 16b^2 + 4 - 4}}{2}

z^2 = \displaystyle \frac{2-4b^2 \pm \sqrt{16b^4 - 16b^2}}{2}

z^2 = \displaystyle \frac{2-4b^2 \pm 4|b| \sqrt{b^2 - 1}}{2}

z^2 = 1-2b^2 \pm 2|b| \sqrt{b^2 - 1}

As shown earlier in this series, the right-hand side is negative if |b| > 1. So, for the sake of simplicity, I’ll define

r_1 = \sqrt{2b^2 - 1 + 2|b| \sqrt{b^2 - 1}},

r_2 = \sqrt{2b^2 - 1 - 2|b| \sqrt{b^2 - 1}},

so that the four poles of the integrand are ir_1, ir_2, -ir_1, and -ir_2. Of these, only two (ir_1 and ir_2) lie within the contour for sufficiently large R, and so I’ll need to compute the residues for these two poles.

Before starting that task, I notice that

z^4 + (4 b^2 - 2) z^2 + 1 = (z - ir_1)(z + ir_1)(z - ir_2)(z + ir_2),

or

z^4 + (4b^2 - 2)z^2 + 1 = (z^2 + r_1^2)(z^2 + r_2^2),

or

z^4 + (4b^2 -2)z^2 + 1 = z^4 + (r_1^2 + r_2^2) z^2 + r_1^2 r_2^2.

Matching coefficients, I see that

r_1^2 + r_2^2 = 4b^2 - 2,

r_1^2 r_2^2 = 1.

These will become very handy later in the calculation.

The integrand has the form \displaystyle g(z)/h(z), and each pole has order one. As shown earlier in this series, the residue at such pole is equal to

\displaystyle \frac{g(r)}{h'(r)}.

In this case, g(z) = 2(1+z^2) and h(z) = z^4 + (4b^2-2)z^2 + 1 so that h'(z) = 4z^3 + 2(4b^2-2)z, and so the residue at r_1 and r_2 are given by

\displaystyle \frac{2(1+[ir_1]^2)}{4 [ir_1]^3 + 2(4b^2-2) [ir_1]} = \displaystyle \frac{-i(1-r_1^2)}{-2r_1^3 + (4b^2-2) r_1}

and

\displaystyle \frac{2(1+[ir_2]^2)}{4 [ir_2]^3 + 2(4b^2-2) [ir_2]} = \displaystyle \frac{-i(1-r_2^2)}{-2r_2^3 + (4b^2-2) r_2}

Finally, to evaluate the contour integral, I simply the sum of the residues within the contour and then multiply the sum by 2\pi i:

Q = \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1} = 2\pi i \left[\frac{-i(1-r_1^2)}{-2r_1^3 + (4b^2-2) r_1} +\frac{-i(1-r_2^2)}{-2r_2^3 + (4b^2-2) r_2} \right]

= 2\pi \displaystyle \left[ \frac{1-r_1^2}{-2r_1^3 + (4b^2-2) r_1} +\frac{1-r_2^2}{-2r_2^3 + (4b^2-2) r_2} \right]

 

green lineSo, to complete the evaluation of Q, I’m left with the small matter of simplifying the right-hand side. I’ll tackle this in tomorrow’s post.

How I Impressed My Wife: Part 5b

Amazingly, the integral below has a simple solution:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.

Even more amazingly, the integral Q ultimately does not depend on the parameter a. For several hours, I tried to figure out a way to demonstrate that Q is independent of a, but I couldn’t figure out a way to do this without substantially simplifying the integral, but I’ve been unable to do so (at least so far).

So here’s what I have been able to develop to prove that Q is independent of a without directly computing the integral Q.

green lineEarlier in this series, I showed that

Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}

= 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}

= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }

= \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{(a^2 + b^2) v^2 + b^2 }

Yesterday, I showed used the substitution w = (a^2 + b^2) v to show that Q was independent of a. Today, I’ll use a different method to establish the same result. Let

Q(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+b^2) dv}{(a^2+b^2)^2 v^2 + b^2 }.

Notice that I’ve written this integral as a function of the parameter a. I will demonstrate that Q'(a) = 0, so that Q(c) is a constant with respect to a. In other words, Q(a) does not depend on a.

To do this, I differentiate under the integral sign with respect to a (as opposed to x) using the Quotient Rule:

Q'(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{ 2a \left[ (a^2+b^2)^2 v^2 + b^2\right] - 2 (a^2+b^2) \cdot (a^2+b^2) v^2 \cdot 2a }{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv

Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{(a^2+b^2)^2 v^2 + b^2- 2 (a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv

Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{b^2-(a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv

I now apply the trigonometric substitution v = \displaystyle \frac{b}{a^2+b^2} \tan \theta, so that

(a^2+b^2)^2 v^2 = (a^2+b^2)^2 \displaystyle \left[ \frac{b}{a^2+b^2} \tan \theta \right]^2 = b^2 \tan^2 \theta

and

dv = \displaystyle \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta

The endpoints of integration change from -\infty < v < \infty to -\pi/2 < \theta < \pi/2, and so

Q'(a) = \displaystyle 4a \int_{-\pi/2}^{\pi/2} \frac{b^2- b^2 \tan^2 \theta}{\left[ b^2 \tan^2 \theta + b^2 \right]^2} \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta] \sec^2 \theta}{\left[ \tan^2 \theta +1 \right]^2} d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\left[ \sec^2 \theta \right]^2} d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\sec^4 \theta} d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta]}{\sec^2 \theta} d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [1- \tan^2 \theta] \cos^2 \theta \, d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [\cos^2 \theta -\sin^2 \theta] d\theta

= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \cos 2\theta \, d\theta

= \displaystyle \left[ \frac{2ab^3}{a^2+b^2} \sin 2\theta \right]^{\pi/2}_{-\pi/2}

= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ \sin \pi - \sin (-\pi) \right]

= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ 0- 0 \right]

= 0.

green line

I’m not completely thrilled with this demonstration that Q is independent of a, mostly because I had to do so much simplification of the integral Q to get this result. As I mentioned in yesterday’s post, I’d love to figure out a way to directly start with

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

and demonstrate that Q is independent of a, perhaps by differentiating Q with respect to a and demonstrating that the resulting integral must be equal to 0. However, despite several hours of trying, I’ve not been able to establish this result without simplifying Q first.

Proving theorems and special cases (Part 15): The Mean Value Theorem

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first 10^{316} cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.

However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.

6. Theorem (Mean Value Theorem). If f is a continuous function on the interval [a,b] which is differentiable on the interior (a,b), then there is a point c \in (a,b) so that

f'(c) = \displaystyle \frac{f(b)-f(a)}{b-a}

In other words, there is a point c in (a,b) so that the slope of the tangent line at c is the same as the slope of the line segment connecting the endpoints.

This is a consequence of the following lemma.

Lemma (Rolle’s Theorem). If f is a continuous function on the interval [a,b] which is differentiable on the interior (a,b) so that f(a) = 0 and f(b) = 0, then there is a point c \in (a,b) so that f'(c) = 0.

Notice that Rolle’s Theorem is really a special case of the Mean Value Theorem: if f(a) = 0 and f(b) = 0, then the right-hand side of the conclusion of the Mean Value Theorem becomes

\displaystyle \frac{f(b)-f(a)}{b-a} = \displaystyle \frac{0-0}{b-a} = 0,

thus matching the conclusion of Rolle’s Theorem.

I won’t type out the proofs of Rolle’s Theorem and the Mean Value Theorem here, since Wikipedia has already done that very well. Suffice it to say that Rolle’s Theorem logically comes first, and then the Mean Value Theorem can be proven using Rolle’s Theorem. The main idea is to assume that the function f satisfies the hypotheses of the Mean Value Theorem and then define

g(x) = f(x) - f(a) - \displaystyle \frac{f(b)-f(a)}{b-a} (x-a)

It’s straightforward to show that g satisfies the hypotheses of Rolle’s Theorem and conclude that there must be a point so that g'(c) = 0, from which we obtain the conclusion of the Mean Value Theorem.

Proving theorems and special cases (Part 14): The Power Law of differentiation

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first 10^{316} cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.

However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.

5. Theorem. For any rational number r, we have \displaystyle \frac{d}{dx} x^r = r x^{r-1}.

This theorem is typically proven using the Chain Rule (in the guise of implicit differentiation) and the following lemma:

Lemma. For any integer n, we have \displaystyle \frac{d}{dx} x^n = n x^{n-1}.

Clearly, the lemma is a special case of the main theorem. However, the lemma can be proven without using the main theorem:

Proof of Lemma (Case 1). If n is a positive integer, then

\displaystyle \frac{d}{dx} x^n = \displaystyle \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}

= \displaystyle \lim_{h \to 0} \frac{x^n + n x^{n-1} h + \frac{1}{2} n(n-1) x^{n-2} + \dots + h^n - x^n}{h}

= \displaystyle \lim_{h \to 0} \left[ n x^{n-1} + \frac{1}{2} n(n-1) x^{n-2} h + \dots + h^{n-1} \right]

= n x^{n-1} + 0 + \dots + 0

= n x^{n-1}

Case 1 can also be proven using the Product Rule and mathematical induction.

Proof of Lemma (Case 2). If n = 0, then the theorem is trivially true since x^0 = 1, and the derivative of a constant is zero.

Proof of Lemma (Case 3). If n is a negative integer, then write n = -m, where m is a positive integer. Then, using the Quotient Rule,

\displaystyle \frac{d}{dx} x^n = \displaystyle \frac{d}{dx} \left( x^{-m} \right)

= \displaystyle \frac{d}{dx} \left( \frac{1}{x^m} \right)

= \displaystyle \frac{0 \cdot x^m - 1 \cdot m x^{m-1}}{x^{2m}}

= -m x^{-m - 1}

= n x^{n-1}

QED

Now that the lemma has been proven, the main theorem can be proven using the lemma.

Proof of Theorem. Suppose that r = p/q, where p and q are integers. Suppose that y = x^r = x^{p/q}. Then:

y = x^{p/q}

y^q = \displaystyle \left[ x^{p/q} \right]^q

y^q = x^p

Let’s now differentiate with respect to x:

q y^{q-1} \displaystyle \frac{dy}{dx} = p x^{p-1}

\displaystyle \frac{dy}{dx} = \displaystyle \frac{p x^{p-1}}{q y^{q-1}}

\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} \frac{x^{p-1}}{[x^{p/q}]^{q-1}}

\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} \frac{x^{p-1}}{x^{p - p/q}}

\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} x^{p-1 - (p-p/q)}

\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} x^{p - 1 - p + p/q}

\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} x^{p/q - 1}

\displaystyle \frac{dy}{dx} = r x^{r-1}

 

QED

Day One of My Calculus Class: Index

I’m using the Twelve Days of Christmas (and perhaps a few extra days besides) to do something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on what I teach my students on the first day of calculus in order to start the transition from Precalculus and to get them engaged for what we’ll be doing throughout the semester.

Part 1: The two themes of calculus: Approximating curved things by straight things and passing to limits.

Part 2: Using the distance-rate-time formula to estimate how fast an accelerating object lands when dropped from a tall building.

Part 3: Passing to limits to precisely calculate the above velocity.

Part 4: Using rectangles to estimate the area under a parabola.

Part 5: Passing to limits to precisely calculate the area under a parabola.

Part 6: Final comments: these two questions apparently have nothing to do with each other, but are in fact highly interrelated. The connection between these two topics, the Fundamental Theorem of Calculus, is one of greatest discoveries in the history of mankind, which my students are now privileged to understand at the ripe old age of 18 or 19 years old.