A CRUX MOVE is a useful math trick. I made a list of about twenty of them. This is number six.
As your mom probably told you, you are a special snowflake. A lie, of course, for if all of us were special snowflakes, then none of us would be. Sorry to break it to you kid, but you sprung from the loins of a devious Jezebel (LabKitty: putting the harsh in harsh mistress since 1979).
However, sometimes special snowflakes really are special snowflakes. Unlike you or me (okay, I admit I can't really say for sure about you. You could be Kari Byron, for example. Or Sarah Vowell. Although why Kari Byron or Sarah Vowell would be reading LabKitty is beyond my ken).
Stay with me: I assure you this is headed somewhere.
In more sober terms, if we're given an expression and told it holds for a bunch of x, then we are free to pick an x that makes life easy. Cuts to the chase, flushes out the result, gets us to the answer. Anytime you see the "for all" symbol (∀), it should tingle your KittySense. Say to yourself: this holds for all x, so it is true when x equals wawa. "Wawa" is your special snowflake (pronounced: "waa-waa"). Zero and one are often useful snowflakes, as we shall see.
E X A M P L E 1
Prove the trig identity: sin2(x) + cos2(x) = 1.
Let y(x) = sin2(x) + cos2(x). Take the derivative:
dy/dx = 2⋅sin(x)⋅cos(x) – 2⋅cos(x)⋅sin(x) = 0
if the derivative of a function is identically zero, we know the function must be equal to a constant. Call it c:
sin2(x) + cos2(x) = c (∀ x)
To find c, any x will do. Special snowflake x = 0 easily tells us c = 1. QED.
Footnote: Snowflakes are good for disproving a theory as well (although if it can be disproved, is it really a theory? A question beyond my pay grade). All it takes is one place wrong, one counterexample. Someone claims sin(x) + cos(x) = 1. It works for x = 0 and π/2, so maybe they're on to something. You could disprove it with 13π/5. But why not just use π?
E X A M P L E 2
Given: 1/x = Σk=0,∞ (–1)k (x–1)k (0 < x < 2), derive a series for ln(x).
It probably occurs to you that the derivative of the natural log is 1/x. Hence we can obtain the requested series by integrating both sides of the given series
∫ 1/x dx = ∫ Σk=0,∞ (–1)k (x–1)k dx
ln(x) + C = Σk=0,∞ ∫ (–1)k (x–1)k dx = Σk=0,∞ (–1)k (x–1)k+1 / (k+1)
There's an issue here of whether we're allowed to swap integration and summation. There's also an issue whether integration affects convergence. Long story short: yes and it can at the endpoints. However, the larger issue here is determining the constant of integration. We're free to pick any x in the interval of convergence. Using x = 1 works splendidly, as all the terms in the sum vanish. We're left with ln(1) + C = 0 from which we obtain C = 0.
E X A M P L E 3
Prove limx→∞ sin(x) / x = 0
We have sin(x) ≥ –1 and sin(x) ≤ 1. We have trapped sin(x) between two snowflakes:
–1 ≤ sin(x) ≤ 1
So, for x > 0 we have:
–1/x ≤ sin(x)/x ≤ 1/x<
It follows that:
limx→∞ –1/x ≤ limx→∞ sin(x)/x ≤ limx→∞ 1/x
But limx→∞ –1/x = limx→∞ 1/x = 0. So:
0 ≤ limx→∞ sin(x)/x ≤ 0
By the squeeze theorem, limx→∞ sin(x)/x = 0. QED.
E X A M P L E 4
The MacArthur–Wilson island metapopulation model can be written:
dp/dt = (a + bp)(1–p) – cp
where p = p(t) is the fraction of habitats occupied at time t (0 ≤ p(t) ≤ 1) and a, b, and c are (positive) model parameters. Show that the model exhibits a realistic steady state.
Solution: A stable state is a value of p for which dp/dt = 0 (often indicated p*). For this problem, a "realistic" (aka "biologically realistic" aka "feasible") steady state just means 0 ≤ p* ≤ 1 (because p is a proportion). Yes, we could set the LHS equal to zero and solve for p*. But snowflakes 0 and 1 make for less work. Let f(p) = (a + bp)(1–p) – cp. Note f(0) = a and f(1) = –c. But a > 0 and –c < 0 and f(p) is continuous (it's just a polynomial) so (by the intermediate value theorem) there must be a value of p between 0 and 1 for which f(p) = 0. There's your realistic steady state.
Footnote: We can squeeze more information out of the problem with just a little more effort. Technically, the intermediate value theorem only tells us there's an odd number of zero crossings between p = 0 and p = 1 (draw a picture). However, f(p) is quadratic so there can be at most two. Hence, there is one. Furthermore, everything to the left of the zero crossing is positive (because f(0) is positive) and everything to the right is negative (because f(1) is negative). If you know a little about qualitative differential equation analysis, you may recognize this derivative sign flip tells us p* is stable. We have proved the MacArthur–Wilson model exhibits a stable realistic steady state. And we didn't solve one equation!
Previous Move: Complete the Square
Next Move: Infinite Series Fu
As your mom probably told you, you are a special snowflake. A lie, of course, for if all of us were special snowflakes, then none of us would be. Sorry to break it to you kid, but you sprung from the loins of a devious Jezebel (LabKitty: putting the harsh in harsh mistress since 1979).
However, sometimes special snowflakes really are special snowflakes. Unlike you or me (okay, I admit I can't really say for sure about you. You could be Kari Byron, for example. Or Sarah Vowell. Although why Kari Byron or Sarah Vowell would be reading LabKitty is beyond my ken).
Stay with me: I assure you this is headed somewhere.
Crux Move #6
Use a Special Snowflake
Use a Special Snowflake
In more sober terms, if we're given an expression and told it holds for a bunch of x, then we are free to pick an x that makes life easy. Cuts to the chase, flushes out the result, gets us to the answer. Anytime you see the "for all" symbol (∀), it should tingle your KittySense. Say to yourself: this holds for all x, so it is true when x equals wawa. "Wawa" is your special snowflake (pronounced: "waa-waa"). Zero and one are often useful snowflakes, as we shall see.
E X A M P L E 1
Prove the trig identity: sin2(x) + cos2(x) = 1.
Let y(x) = sin2(x) + cos2(x). Take the derivative:
dy/dx = 2⋅sin(x)⋅cos(x) – 2⋅cos(x)⋅sin(x) = 0
if the derivative of a function is identically zero, we know the function must be equal to a constant. Call it c:
sin2(x) + cos2(x) = c (∀ x)
To find c, any x will do. Special snowflake x = 0 easily tells us c = 1. QED.
Footnote: Snowflakes are good for disproving a theory as well (although if it can be disproved, is it really a theory? A question beyond my pay grade). All it takes is one place wrong, one counterexample. Someone claims sin(x) + cos(x) = 1. It works for x = 0 and π/2, so maybe they're on to something. You could disprove it with 13π/5. But why not just use π?
E X A M P L E 2
Given: 1/x = Σk=0,∞ (–1)k (x–1)k (0 < x < 2), derive a series for ln(x).
It probably occurs to you that the derivative of the natural log is 1/x. Hence we can obtain the requested series by integrating both sides of the given series
∫ 1/x dx = ∫ Σk=0,∞ (–1)k (x–1)k dx
ln(x) + C = Σk=0,∞ ∫ (–1)k (x–1)k dx = Σk=0,∞ (–1)k (x–1)k+1 / (k+1)
There's an issue here of whether we're allowed to swap integration and summation. There's also an issue whether integration affects convergence. Long story short: yes and it can at the endpoints. However, the larger issue here is determining the constant of integration. We're free to pick any x in the interval of convergence. Using x = 1 works splendidly, as all the terms in the sum vanish. We're left with ln(1) + C = 0 from which we obtain C = 0.
E X A M P L E 3
Prove limx→∞ sin(x) / x = 0
We have sin(x) ≥ –1 and sin(x) ≤ 1. We have trapped sin(x) between two snowflakes:
–1 ≤ sin(x) ≤ 1
So, for x > 0 we have:
–1/x ≤ sin(x)/x ≤ 1/x<
It follows that:
limx→∞ –1/x ≤ limx→∞ sin(x)/x ≤ limx→∞ 1/x
But limx→∞ –1/x = limx→∞ 1/x = 0. So:
0 ≤ limx→∞ sin(x)/x ≤ 0
By the squeeze theorem, limx→∞ sin(x)/x = 0. QED.
E X A M P L E 4
The MacArthur–Wilson island metapopulation model can be written:
dp/dt = (a + bp)(1–p) – cp
where p = p(t) is the fraction of habitats occupied at time t (0 ≤ p(t) ≤ 1) and a, b, and c are (positive) model parameters. Show that the model exhibits a realistic steady state.
Solution: A stable state is a value of p for which dp/dt = 0 (often indicated p*). For this problem, a "realistic" (aka "biologically realistic" aka "feasible") steady state just means 0 ≤ p* ≤ 1 (because p is a proportion). Yes, we could set the LHS equal to zero and solve for p*. But snowflakes 0 and 1 make for less work. Let f(p) = (a + bp)(1–p) – cp. Note f(0) = a and f(1) = –c. But a > 0 and –c < 0 and f(p) is continuous (it's just a polynomial) so (by the intermediate value theorem) there must be a value of p between 0 and 1 for which f(p) = 0. There's your realistic steady state.
Footnote: We can squeeze more information out of the problem with just a little more effort. Technically, the intermediate value theorem only tells us there's an odd number of zero crossings between p = 0 and p = 1 (draw a picture). However, f(p) is quadratic so there can be at most two. Hence, there is one. Furthermore, everything to the left of the zero crossing is positive (because f(0) is positive) and everything to the right is negative (because f(1) is negative). If you know a little about qualitative differential equation analysis, you may recognize this derivative sign flip tells us p* is stable. We have proved the MacArthur–Wilson model exhibits a stable realistic steady state. And we didn't solve one equation!
Previous Move: Complete the Square
Next Move: Infinite Series Fu
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