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These are lecture notes for a Math Subject GRE preparation course covering single variable calculus, multivariable calculus, linear algebra, differential equations, and complex analysis. The notes were written by Ian Coley for courses at UCLA in 2016 and 2018, and at Rutgers in 2021. The notes have been updated with feedback from mathematicians over the years and are as complete as possible. definitions, theorems, and problems with solutions.
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IAN COLEY
These are lecture notes I wrote for a Math Subject GRE preparation course I ran at UCLA in 2016 and 2018, then at Rutgers in 2021. They have been updated a number of times with feedback from many mathematicians over the years noticing small (or large) typos. As it stands, they are as complete as I would hope. The formatting is not exactly consistent throughout, but in here is definitely everything I want to say.
Last Updated: July 29, 2021. 1
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1.1. Basics.
Definition 1.1. Let f : R → R be a function. We say that lim x→a f (x) = L if for every
ε > 0, there exists δ > 0 such that whenever 0 < |x − a| < δ, |f (x) − L| < ε.
We will worry about limits of sequences on another day. To set some notation: a neighbourhood of x = a will be the set {x ∈ R : |x − a| < δ} for some δ > 0; if we specify the radius ε > 0, we will call it an ε-neighbourhood. A punctured neighbourhood of x = a will be the set {x ∈ R : 0 < |x − a| < δ} for some δ > 0. Thus we could rephrase the above definition in words: for every ε > 0, there exists a punctured neighbourhood of x = a that maps entirely to the ε-neighbourhood of L.
Definition 1.2. We say that f : R → R is continuous at x = a if lim x→a f (x) = f (a).
The condition |f (a) − f (a)| < ε is automatically satisfied for all ε > 0, so we can extend the demand of a punctured neighbourhood of x = a to a non-punctured one in the case that L = f (a).
Problem 1.3. At what points is the following function continuous?
f (x) =
x^2 x ∈ Q x/ 5 x ∈ R \ Q
This sort of problem (coupled with having taken real analysis) should remind you that the heuristic ‘continuous means you don’t pick up your pen’ doesn’t work in sophisticated situations. There is also a notion of right continuous and left continuous, where we use only one-sided limits. This is more easily drawn than defined (and the analogue in multi- variable calculus isn’t really helpful, so we will omit the full definition). If this were a blackboard, there would be a better example here. When are functions discontinuous? Jump discontinuities (almost always in piece- wise functions), infinite discontinuities, and removable discontinuities (holes). One could also consider a function discontinuous at the points where it ceases to exist, e.g. f (x) =
x is discontinuous at x ≤ 0, but this isn’t really necessary. For ex- amples of continuous functions, think of almost literally any function: polynomials, trigonometric functions, logarithms, exponential functions, etc.
MATH GRE BOOTCAMP: LECTURE NOTES 5
It is likely unimportant for the GRE, but it might be nice to recall the squeeze theorem just in case:
Theorem 1.4 (Squeeze Theorem). Suppose that f, g, h : R → R are three functions. If f (x) ≤ g(x) ≤ h(x) in a punctured neighbourhood of x = a, then
lim x→a f (x) ≤ lim x→a g(x) ≤ lim x→a h(x).
This is particularly useful when f (x) and h(x) are continuous at x = a (or one is even constant) but g(x) isn’t.
Problem 1.5. Compute lim x→ 0 x · sin
x
The second main definition we need for calculus is that of differentiable functions.
Definition 1.6. We say that f : R → R is differentiable at x = a if lim x→a
f (x) − f (a) x − a
exists. Alternatively, we can ask that lim h→ 0
f (a + h) − f (a) h
exists. We write f ′(a) for
this value.
Problem 1.7. Prove that these two definitions agree.
Problem 1.8. Prove that if f (x) is differentiable at x = a, it is continuous at x = a.
When are functions non-differentiable? By the previous exercise, when they’re discontinuous. It’s important to remember to always check continuity first, as in the following:
Problem 1.9. Describe the set of solutions (a, b, c) ∈ R^3 such that the following function is continuous and differentiable (everywhere):
f (x) =
ax^2 + bx + c x ≤ 1 x log x x > 1
Solution. The pair (a, b) determines the equality of derivatives at x = 1 and c determines the continuity. To check continuity, we have that
f (1) = a + b + c = 1 · log(1)
so that a + b + c = 0. Second, by the product rule we have (x log x)′^ = 1 + log x, so for differentiability we need f ′(1) = 2a + b = 1 + log(1) = 1, so that 2a + b = 1. We can solve b = 1 − 2 a, so a + b + c = a + (1 − 2 a) + c = 0 hence −a + c = −1 so c = a − 1 and b = 1 − 2 a. Thus the set of solutions can be written (a, 1 − 2 a, a − 1).
There are two basic theorems about continuous and differentiable functions we will state now, and a third later.
MATH GRE BOOTCAMP: LECTURE NOTES 7
d dx
cos(x) = − sin(x)
d dx
sinh(x) = cosh(x)
d dx
cosh(x) = sinh(x)
d dx
ex^ = ex
d dx
log(x) =
x Other trigonometric functions can be computed using the quotient rule. Perhaps it is also worth remembering the less-used but GRE-noteworthy formula for the second derivative of a function:
Problem 1.14. Prove that, if f ′(x) is differentiable, then
f ′′(x) = lim h→ 0
f (x + h) + f (x − h) − 2 f (x) h^2
Solution. We can rewrite this expression as follows:
f (x + h) + f (x − h) − 2 f (x) h^2
f (x+h)−f (x) h −^
f (x)−f (x−h) h h
If we take lim h→ 0 of the top, we obtain f ′(x) − f ′(x − h), and we can certainly agree
that
f ′′(x) = lim h→ 0
f ′(x) − f ′(x − h) h
It is a fun fact that the righthand side (the symmetric second derivative) of the above can exist even when f ′′(x) itself does not exist – the example given on Wikipedia is the sign function
σ(x) =
− 1 x < 0 0 x = 0 1 x > 0
What other techniques do we use for computing derivatives? Computing the derivatives of inverse functions can be difficult, specifically when we don’t have a closed formula for the inverse. What circumstances are those?
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Definition 1.15. Let f : R → R be a function, and suppose that X ⊂ R is a set on which f is one-to-one. Then we say that f is invertible on X, and write f −^1 (y) for the inverse, which is defined by f −^1 (y) = x if and only if f (x) = y.
Problem 1.16. Suppose that f : R → R is a function and that for all x ∈ X ⊂ R, f ′(x) > 0. Then f is invertible on X.
Problem 1.17. Suppose that f : R → R is a differentiable invertible function. Then f −^1 : R → R is also differentiable, except at those y = f (x) ∈ R such that when f ′(x) = 0.
How do we compute the derivative of the inverse? Write f −^1 = g for simplicity, Then f (g(y)) = y, so taking derivatives and using the chain rule we have
(f ◦ g)′(y) = f ′(g(y)) · g′(y) = 1 =⇒ g′(y) =
f ′(g(y))
So as long as we can figure out g(y) and f ′(x), we can figure out g′(y).
Problem 1.18. Compute the derivative of tan−^1 (y).
Solution. By the formula, we have that the inverse should be the derivative of tan(x) evaluated on tan−^1 (y). The derivative of tan(x) is sec^2 (x), so we need to figure out sec^2 (tan−^1 (y)). This is done with a technique I personally call ‘draw the triangle’. We know that tan−^1 (y) = x for some x, so we need to draw the triangle in which x is one of the angles. We know only that tan(x) = y, so we may draw a right triangle:
1
√ 1+y^2
y
x
Therefore sec^2 (x) we can compute as the hypotenuse squared over the adjacent side squared, that is sec^2 (tan−^1 (y)) = 1 + y^2. Therefore the derivative of tan−^1 (y) is 1 1 + y^2
This method can be used to compute the derivatives of the other inverse trigono- metric and hyperbolic trigonometric functions on the fly, so you don’t need to nec- essarily memorise all of them. That said, you should definitely memorise the above example. Logarithmic differentiation is a useful technique, and it also recalls implicit differ- entiation.
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1.3. Applications of Derivatives. Related rates problems, which involve para- metric functions as discussed above, show up quite often in the single-variable cur- riculum. Classic examples include balloons filling up or deflating and basins filling up or emptying of water. Ladders sliding down a wall or shadows lengthening are also common.
Problem 1.23. Suppose we have a right conical coffee filter which is 8cm tall with a radius of 4cm. The water drips through at a constant rate of 2 cubic centimetres per second. When there is one eighth of the original water remaining, how fast is the water level dropping?
Solution. We begin by noting that V (t) =
π 3
· h(t) · r(t)^2. But because our cone
is conical, we also know that the height and radius of the cone are in a fixed ratio: h(t) = 2 · r(t). Because we are solving for h′(t), we will substitute in r(t) = h(t)/2. Hence
V (t) =
π 3
· h(t) ·
h(t)^2 4
π 12
· h(t)^3.
Taking the time derivative,
V ′(t) =
π 4
h(t)^2 · h′(t)
Letting t = t 0 be the time at which we would like to find the change in height, we know that V ′(t 0 ) = −2 no matter what. Therefore we need to find h(t 0 ) to complete
the problem. We know that V (t 0 ) = V (0)/8, and that V (0) =
π 3
128 π 3
. So
as V (t 0 ) = 12 π · h(t 0 )^3 = 163 π , it’s pretty clear that h(t 0 ) = 4. Therefore
−2 = V ′(t) =
π 4
· 42 · h′(t 0 ) =⇒ −
2 π
= h′(t 0 ).
We can now turn to optimisation. This is certainly a favourite in the undergraduate curriculum and appears sometimes in the GRE. Why does it work?
Theorem 1.24 (Extreme Value Theorem). Let f : [a, b] → R be a continuous func- tion. Then the set f ([a, b]) has both a maximum value and a minimum value.
Note that it’s necessary that [a, b] be a closed interval. The image of an open interval under a continuous function may have neither a maximum nor a minimum, e.g. f (x) = 1/x and the interval (0, ∞). We can even say more:
Definition 1.25. Let f : R → R be a differentiable function. We say that a ∈ R is a critical point if f ′(a) = 0.
MATH GRE BOOTCAMP: LECTURE NOTES 11
We might also include the case that f : R → R is a function which is differentiable except at finitely many points and call those critical points as well. Their utility is the following theorem, credited to Fermat by various sources.
Theorem 1.26 (Fermat’s Boring Theorem). Let f : [a, b] → R be a differentiable function (except at the endpoints). Then the maxima and minima of f (x) occur at critical points and end points.
There are many different kinds of optimisation problems (as there are related rates problems), but all have the same basic process: we are given both a function to optimise and a constraint. The function to optimise is going to be some F (x, y) in two variables, and the constraint is going to be some equation g(x, y) = c. After substituting, we’ll have a one-variable problem. Sometimes we will have endpoints, and sometimes it will be implicit that as your one variables gets too big or small that there is no extremum to be found. We then find the critical points and exhaustively check which is the largest and which is the smallest.
Problem 1.27. Suppose we are constructing a window comprised of a semicircle sitting atop a rectangle. Given that the perimeter of the window must be 4 meters, what is the maximum area?
Solution. Let us call x the width and y the height of the rectangle. We know that
the area is given by the sum xy +
π 2
x 2
, the first for the rectangle and the second
for the semicircle. Our constraint is 2y + x +
πx 2
= 4, the bottom three sides of the
rectangle and the arc of the semicircle. It looks like making x our sole variable will
be the best path to success, so we will substitute y = 2 −
2 + π 4
· x. Our function is
thus
A(x) = x ·
2 + π 4
· x
π 2
x 2
= 2x −
4 + π 8
· x^2.
Note that this is the equation of a downward-facing parabola, so if x is too big or too small we have A(x) < 0. This is obviously a nonsense answer to the question, so what we’re looking for is the critical point giving the vertex of the parabola – its maximum.
We now need to solve A′(x) = 2 −
4 + π 4
· x = 0, so x =
4 + π
. We can now
compute the maximum area:
A
4 + π
4 + π
(not a typo).
Problem 1.28. What is the minimum distance between the curve y = 1/x and the origin?
MATH GRE BOOTCAMP: LECTURE NOTES 13
1.5. L’Hˆopital’s Rule. The last topic worth remembering L’Hˆopital’s rule, which comes surprisingly in the second quarter of calculus at UCLA but we can recall now.
Theorem 1.32. Let f, g : R → R be two functions and a ∈ R ∪ {±∞}. If lim x→a f (x) = lim x→a g(x) = c where c ∈ { 0 , ±∞}, then
lim x→a
f (x) g(x)
= lim x→a
f ′(x) g′(x)
Expressions of the form
and ±
are called indeterminate forms. L’Hˆopital’s
rule can also be used to solve problems which are not immediately in an indeterminate form. The easiest case is expressions of the form f (x) · g(x) which give rise to the indeterminate form 0 · ∞. By rearranging to
f (x) 1 /g(x)
or
g(x) 1 /f (x)
we will obtain an actual indeterminate form. Which one to choose depends on whether 1/f (x) or 1/g(x) is easier to differentiate.
Problem 1.33. Compute the following limit:
lim x→ 2
x − 2
2 x x^2 − 4 Solution. If you plug in x = 2, we don’t obtain an indeterminate form, but we obtain something that looks like ∞ − ∞. This is our clue to combine the fractions into an indeterminate form. With a common denominator,
x + 2 x^2 − 4
2 x x^2 − 4
−x + 2 x^2 − 4
−(x − 2) (x − 2)(x + 2)
x + 2
In this case we don’t even need to use L’Hˆopital’s rule to finish the problem since we had some nice cancellation. In other cases we might not be so lucky.
Problem 1.34. Compute the following limit:
lim x→∞
x^1 /x
Solution. These problems are also related to L’Hˆopital’s rule as well. Plugging in, we obtain ∞^0. We notice that if we took the log of this expression, log
x^1 /x
= (^1) x · log x
yields the form 0 · ∞. We can now rearrange it to
log x x
and finish the problem:
lim x→∞ log
x^1 /x
= lim x→∞
log x x
L’H = lim x→∞
x
But of course this solves the wrong question. If we say L = limx→∞ x^1 /x, then log L = 0 (as log is a continuous function so commutes with limits). Thus L = 1.
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This same type of solution works if we have the form 1∞, as log(1∞) = ∞ · log 1 yields ∞ · 0. This concludes the differential side of single-variable calculus.
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We will almost never use the right and left Riemann sums, as these limits are not calculable in practice, but it’s important to know a few things. If a function is increasing, then the left Riemann sum will always underestimate the actual value of the integral, and the right Riemann sum will always overestimate it. This definitely comes up on the GRE.
Problem 2.3. Let f (x) = 1/x^2. For the interval [1, 2], order the following terms:
L 5 f ,
1
f (x) dx, R 5 f.
Solution. We know that 1/x^2 is decreasing on the interval [1, 2], so the left Riemann sum will always overestimate it and the right Riemann sum will always underestimate it. The actual integral will land in the middle. So while the hasty student is still putting numbers into their calculator (which isn’t allowed on the test anyway) to try
to crunch these Riemann sums, we can finish the problem: L 5 f >
1
f (x) dx > R 5 f.
The Riemann integral is not guaranteed to exist for an arbitrary function, but it must exist for our favourite functions.
Proposition 2.4. If f : [a, b] → R is bounded and continuous at all but finitely
many points, then
Z (^) b
a
f (x) dx exists.
We could assign this as an exercise, but it’s a bit difficult and not necessary at all for the GRE. This isn’t a necessary and sufficient condition, but it is certainly good enough for almost all purposes. The integral is linear, just as the derivative was. In particular, this means that Z (^) b
a
c · f (x) dx = c ·
Z (^) b
a
f (x) dx
and (^) Z b
a
f (x) ± g(x) dx =
Z (^) b
a
f (x) dx ±
Z (^) b
a
g(x) dx
Moreover, recalling the definition via Riemann sums, we can always split an inte- gral into intermediate chunks. For any c ∈ [a, b], we have Z (^) b
a
f (x) dx =
Z (^) c
a
f (x) dx +
Z (^) b
c
f (x) dx
By definition we will say
Z (^) a
a
f (x) dx = 0, so using the additivity above Z (^) b
a
f (x) dx = −
Z (^) a
b
f (x) dx
MATH GRE BOOTCAMP: LECTURE NOTES 17
This way we can make sense of integrals where our interval [a, b] happens to be oriented the wrong way (i.e. a > b). There’s a useful computational trick that is best explained using Riemann sums, so we include it here.
Definition 2.5. A function f : R → R is called odd if f (x) = −f (−x) for all x ∈ R. A function g : R → R is called even if g(x) = g(−x) for all x ∈ R.
Odd functions include polynomials with only odd-degree terms, sin(x), and tan(x). Even functions include polynomials with only even-degree terms (including the con- stant term) and cos(x). Compositions of even and odd functions work like multi- plication – odd composed with odd is odd, even composed with odd is even, etc. Products of even and odd functions work like addition: odd times odd is even, even times odd is odd, etc. We mention all this because of the following trick:
Proposition 2.6. Let f : R → R be an odd, integrable function. Then Z (^) R
−R
f (x) dx = 0
for any R > 0.
Proof. Using the midpoint Riemann sum, we can give a computation: Z (^) R
−R
f (x) dx = lim n→∞
n
X^ n−^1
i=
f
2 n
· (2i + 1)
The list of values −R + R(2 ni+1) is completely symmetric about the y-axis and if n is odd we get 0 in the middle. Since f (x) is odd, we have f (xi) = −f (−xi), so these terms of the Riemann sum will cancel each other out. Thus
R − (−R) n
nX− 1
i=
f
2 n
· (2i + 1)
so the integral will be zero as well. □
Problem 2.7. Compute the integral Z (^1)
− 1
sin(x^3 ) + sin^3 (x) dx
Solution. There are both odd functions (since they are compositions of the odd functions sin(x) and x^3 ), so they both integrate to zero.
Besides the above tricks, we will always compute the integral using the Funda- mental Theorem of Calculus.
MATH GRE BOOTCAMP: LECTURE NOTES 19
substitution as well, but it’s much more annoying (and the GRE is all about saving time where you can).
In the same way, we can compute the area between curves with integrals. We need only to compute the area under the upper curve and subtract the area under the lower curve. The main question in situations like this is which curve is on top and which is on bottom.
Problem 2.12. Compute the area between the curves y =
x and y = x^2 in the first quadrant.
Solution. Thinking on the graphs, we can recall that
x is above x^2 in the region [0, 1] where these curves intersect. Therefore the integral we need to compute is Z (^1)
0
x − x^2 dx.
Luckily, we can get antiderivatives of these functions very easily: Z (^1)
0
x − x^2 dx =
2 x^3 /^2 3
x^3 3
1 0
The other application is to surfaces and volumes of revolution. Supposing that we rotate a curve y = f (x) around the x-axis, the height of the curve becomes the radius of a disc, which we then need to integrate along the interval [a, b] in question, whatever it is:
V = π
Z (^) b
a
f (x)^2 dx
Problem 2.13. What is the volume of the region created by rotating y = log x around the x-axis between x = 1 and x = e^2?
Note that you’ll need to use integration by parts for this one (see below). It’s harder to compute the volume when we rotate y = f (x) around the y-axis. Rather than use the method of discs, we use the method of cylindrical shells. In this case, we are computing the area of a cylinder of radius r and height h, which is given by 2π · r · h. In our case, the radius is x and the height is f (x), hence
V = 2π
Z (^) b
a
x · f (x) dx
Problem 2.14. Compute the volume of the region created by rotating y = 1 − 2 x + 3 x^2 − 2 x^3 from [0, 1] around the y-axis.
We can also repeat the above problems when asked to revolve the area between two curves around either the x- or the y-axis. These follow the general formulas, for
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f (x) ≥ g(x):
V = π
Z (^) b
a
f (x)^2 − g(x)^2 dx
and for the method of cylindrical shells:
V = 2π
Z (^) b
a
x(f (x) − g(x)) dx
Arc length of a curve is another application. If we are looking at the infinitesimal change in the length of a curve, it travels dx in the x direction and dy in the y direction. Therefore its total length is ds =
p (dx)^2 + (dy)^2. In the case that y = f (x), we have
s =
ds =
Z s (dx)^2 +
dy dx
Z (^) b
a
p 1 + f ′(x)^2 dx
Problem 2.15. Compute the arc length of y = cosh x over the interval [0, 2].
Sometimes we also want to calculate the surface areas of regions of revolution, not just the volumes. In this case, we again have a formula: the infinitesimal amount of surface area is given by the arc length along the surface times the circumference of the disc, which in the case of rotation around the x-axis is 2πf (x). Thus
S = 2π
f (x)
p 1 + f ′(x)^2 dx
Problem 2.16. Compute the surface area of a sphere of radius R, using the curve y =
R^2 − x^2.
2.3. Integration techniques. Now besides guessing at antiderivatives, what are our other integration techniques? The first is u-substitution, which is is our answer to the chain rule.
Theorem 2.17 (u-substitution). Suppose that h(x) is a continuous function and we can write h(x) = f ′(g(x))g′(x). Then Z (^) b
a
h(x) dx =
Z (^) g(b)
g(a)
f (u) du.
Problem 2.18. Evaluate the integral of f (x) = 2x · sin(x^2 ) on [0,
p π/2].
Solution. We see that g(x) = x^2 and f (u) = sin u is a good choice. Hence Z √π/ 2
0
2 x · sin(x^2 ) dx =
Z (^) π/ 2
0
sin u du = − cos u
π/ 2 0
