Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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If \( {a}^{m} \equiv 1\left( {\;\operatorname{mod}\;N}\right) \) and \( {a}^{n} \equiv 1\left( {\;\operatorname{mod}\;N}\right) \) then\n\n\[
{a}^{\gcd \left( {m, n}\right) } \equiv 1\;\left( {\;\operatorname{mod}\;N}\right)
\] | This is just the famous fact that \( \gcd \left( {{a}^{m} - 1,{a}^{n} - 1}\right) = {a}^{\gcd \left( {m, n}\right) } - 1 \) phrased using modular arithmetic. | Yes |
Show that there are infinitely many primes which are congruent to 1 modulo \( n \) for any positive integer \( n \) . | Solution. Suppose there were only finitely many such primes \( {p}_{1},{p}_{2},\ldots ,{p}_{N} \) . Look at the number\n\n\[ M = {\Phi }_{n}\left( {n{p}_{1}{p}_{2}\ldots {p}_{N}}\right) \]\n\nAs a polynomial, \( {\Phi }_{n}\left( X\right) \) has roots which are all roots of unity (meaning they have norm 1), so its cons... | No |
Problem 6.17. Prove that modulo any prime \( p \) there exists a primitive root! | From [4]: Consider the cyclotomic polynomial \( {\Phi }_{p - 1}\left( X\right) \mid {X}^{p - 1} - 1 \) . Show that it factors completely modulo \( p \), and pick any root. | No |
Find all functions \( f : \mathbb{R} \rightarrow \mathbb{R} \) such that\n\n\[ f\left( {f{\left( x\right) }^{2} + f\left( y\right) }\right) = {xf}\left( x\right) + y \] \n\nfor all \( x, y \in \mathbb{R} \) . | Solution. Before I begin solving the problem, can you guess what the answer is? Clearly, \( f\left( x\right) = + x \) works. But there’s actually a second solution: \( f\left( x\right) = - x \) . In general, a \ | No |
Solve \( f\left( {x + y}\right) = f\left( x\right) + f\left( y}\right) \) over \( \mathbb{Q} \) . | Solution. As before we begin by examining which functions we think the answers are. Trying out the most general \( f\left( x\right) = {kx} + c \), we find that \( c = 0 \) but \( k \) can be anything. So our guess is that the answer is \( f\left( x\right) = {kx} \) .\n\nWe now prove this guess is right. First of all, a... | No |
Solve over \( \mathbb{Q} \) :\n\n\[ f\left( x\right) + f\left( y\right) = {2f}\left( \frac{x + y}{2}\right) . \] | Solution. This time, our preliminary checks reveal that \( f\left( x\right) = {kx} + c \) works for any \( k \) and \( c \) . (In a vague sense, the fact that \( c \) is free to vary is manifested in the fact that plugging in all zeros yields the tautology \( 0 = 0 \) .) So now we do the following trick: we can shift t... | Yes |
Solve over \( \mathbb{R} \) :\n\n\[ f\left( {x + y}\right) = f\left( x\right) + f\left( y\right) \;\text{ and }\; f\left( {xy}\right) = f\left( x\right) f\left( y\right) . | Solution. We claim \( f\left( x\right) = x \) and \( f\left( x\right) = 0 \) are the only solutions (which both work). According to the theorem, to prove \( f \) is linear it suffices to show \( f \) is nonnegative over some nontrivial interval. Now,\n\n\[ f\left( {t}^{2}\right) = f{\left( t\right) }^{2} \geq 0 \]\n\nf... | Yes |
Solve over \( \mathbb{R} \) : \n\n\[ \nf\left( {{x}^{2} + y}\right) = f\left( {{x}^{27} + {2y}}\right) + f\left( {x}^{4}\right) .\n\] | Solution. For this problem, we claim the only answer is the constant function \( f = 0 \) . As usual our first move is to take the all-zero setting, which gives \( f\left( 0\right) = 0 \) .\n\nNow, let's step back: can we do anything that will make lots of terms go away? There's actually a very artificial choice that w... | Yes |
Solve over \( \mathbb{R} \) :\n\n\[ \left( {x - y}\right) f\left( {x + y}\right) - \left( {x + y}\right) f\left( {x - y}\right) = {4xy}\left( {{x}^{2} - {y}^{2}}\right) . \] | Solution. First of all, the \( x - y \) and \( x + y \) everywhere are a mess, so we replace them with \( a = x + y \) and \( b = x - y \) (this doesn’t lose any information). Then \( {x}^{2} - {y}^{2} = {ab} \) , and also \( {2x} = a + b,{2y} = a - b \) . So, the equation is just saying that\n\n\[ {bf}\left( a\right) ... | Yes |
Prove that \( {a}^{2} + {b}^{2} + {c}^{2} \geq {ab} + {bc} + {ca} \) and \( {a}^{4} + {b}^{4} + {c}^{4} \geq {a}^{2}{bc} + {b}^{2}{ca} + {c}^{2}{ab} \) . | Proof. By AM-GM,\n\n\[ \frac{{a}^{2} + {b}^{2}}{2} \geq {ab}\text{ and }\frac{2{a}^{4} + {b}^{4} + {c}^{4}}{4} \geq {a}^{2}{bc}. \]\n\nSimilarly,\n\n\[ \frac{{b}^{2} + {c}^{2}}{2} \geq {bc}\text{ and }\frac{2{b}^{4} + {c}^{4} + {a}^{4}}{4} \geq {b}^{2}{ca}. \]\n\n\[ \frac{{c}^{2} + {a}^{2}}{2} \geq {ca}\text{ and }\fra... | Yes |
Since \( \left( {5,0,0}\right) \succ \left( {3,1,1}\right) \succ \left( {2,2,1}\right) \), prove that \( {a}^{5} + {a}^{5} + {b}^{5} + {b}^{5} + {c}^{5} + {c}^{5} \geq {a}^{3}{bc} + {a}^{3}{bc} + {b}^{3}{ca} + {b}^{3}{ca} + {c}^{3}{ab} + {c}^{3}{ab} \geq {a}^{2}{b}^{2}c + {a}^{2}{b}^{2}c + {b}^{2}{c}^{2}a + {b}^{2}{c}^... | From this we derive \( {a}^{5} + {b}^{5} + {c}^{5} \geq {a}^{3}{bc} + {b}^{3}{ca} + {c}^{3}{ab} \geq {abc}\left( {{ab} + {bc} + {ca}}\right) \). | No |
Prove that if \( {abc} = 1 \) then \( {a}^{2} + {b}^{2} + {c}^{2} \geq a + b + c \) . | Proof. AM-GM alone is hopeless here, because whenever we apply AM-GM, the left and right hand sides of the inequality all have the same degree. So we want to use the condition \( {abc} = 1 \) to force the problem to have the same degree. The trick is to notice that the given inequality can be rewritten as\n\n\[ \n{a}^{... | Yes |
Given \( a + b + c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \), prove that\n\n\[ \frac{1}{{\left( 2a + b + c\right) }^{2}} + \frac{1}{{\left( a + 2b + c\right) }^{2}} + \frac{1}{{\left( a + b + 2c\right) }^{2}} \leq \frac{3}{16}. \] | Proof. First, we want to eliminate the condition. The original problem is equivalent to\n\n\[ \frac{1}{{\left( 2a + b + c\right) }^{2}} + \frac{1}{{\left( a + 2b + c\right) }^{2}} + \frac{1}{{\left( a + b + 2c\right) }^{2}} \leq \frac{3}{16} \cdot \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{a + b + c}. \]\n\nNow the... | Yes |
If \( a, b, c \) are the three sides of a triangle, prove that\n\n\[ \sqrt{a + b - c} + \sqrt{b + c - a} + \sqrt{c + a - b} \leq \sqrt{a} + \sqrt{b} + \sqrt{c}. \] | Proof. Again assume WLOG that \( a \geq b \geq c \) and notice that \( \left( {a + b - c, c + a - b, b + c - a}\right) \succ \) \( \left( {a, b, c}\right) \) . Apply Karamata on \( f\left( x\right) = \sqrt{x} \) . | Yes |
If \( a + b + c = 3 \), prove that\n\n\[ \n{18}\mathop{\sum }\limits_{\text{cyc }}\frac{1}{\left( {3 - c}\right) \left( {4 - c}\right) } + 2\left( {{ab} + {bc} + {ca}}\right) \geq {15}.\n\] | Proof. We can rewrite the given inequality as\n\n\[ \n\mathop{\sum }\limits_{\text{cyc }}\left( {\frac{18}{\left( {3 - c}\right) \left( {4 - c}\right) } - {c}^{2}}\right) \geq 6.\n\]\n\nUsing the tangent line trick lets us obtain the magical inequality\n\n\[ \n\frac{18}{\left( {3 - c}\right) \left( {4 - c}\right) } - {... | Yes |
Prove \( \mathop{\sum }\limits_{\text{cyc }}\frac{{\left( b + c - a\right) }^{2}}{{a}^{2} + {\left( b + c\right) }^{2}} \geq \frac{3}{5} \) . | Proof. Since the inequality is homogeneous, we may assume WLOG that \( a + b + c = 3 \) . So the inequality we wish to prove is\n\n\[ \mathop{\sum }\limits_{\text{cyc }}\frac{{\left( 3 - 2a\right) }^{2}}{{a}^{2} + {\left( 3 - a\right) }^{2}} \geq \frac{3}{5} \]\n\nWith some computation, the tangent line trick gives awa... | Yes |
Theorem 2.8 \( \left( {n - 1\mathrm{{EV}}}\right) \)\n\nLet \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) be real numbers, and suppose \( {a}_{1} + {a}_{2} + \cdots + {a}_{n} \) is fixed. Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a function with exactly one inflection point. If\n\n\[ f\left( {a}_{1}\right) + f\left( {... | Proof. See page 15 of Olympiad Inequalities, by Thomas Mildorf. The main idea is to use Karamata to \ | No |
Let \( a, b, c \) be positive reals. Prove \( 1 \leq \mathop{\sum }\limits_{\text{cyc }}\frac{a}{\sqrt{{a}^{2} + {8bc}}} < 2 \) . | Proof. Set \( {e}^{x} = \frac{bc}{{a}^{2}},{e}^{y} = \frac{ca}{{b}^{2}},{e}^{z} = \frac{ab}{{c}^{2}} \) . We have the condition \( x + y + z = 0 \) and want to prove\n\n\[ 1 \leq f\left( x\right) + f\left( y\right) + f\left( z\right) < 2 \]\n\nwhere \( f\left( x\right) = \frac{1}{\sqrt{1 + 8{e}^{x}}} \) . You can compu... | Yes |
Let \( {x}_{1},{x}_{2},\ldots ,{x}_{n} \) be positive reals satisfying \( \mathop{\sum }\limits_{{i = 1}}^{n}\frac{1}{{1998} + {x}_{i}} = \frac{1}{1998} \) . Prove\n\n\[ \frac{\sqrt[n]{{x}_{1}{x}_{2}\ldots {x}_{n}}}{n - 1} \geq {1998} \] | Proof. Let \( {y}_{i} = \frac{1998}{{1998} + {x}_{i}} \) . Since \( {y}_{1} + {y}_{2} + \cdots + {y}_{n} = 1 \), the problem becomes\n\n\[ \mathop{\prod }\limits_{{i = 1}}^{n}\left( {\frac{1}{{y}_{i}} - 1}\right) \geq {\left( n - 1\right) }^{n} \]\n\nSet \( f\left( x\right) = \ln \left( {\frac{1}{x} - 1}\right) \), so ... | Yes |
Prove \( 3\left( {a + b + c}\right) \geq 8\sqrt[3]{abc} + \sqrt[3]{\frac{{a}^{3} + {b}^{3} + {c}^{3}}{3}} \) . | Proof. By Power Mean with \( r = 1, s = \frac{1}{3},{w}_{1} = \frac{1}{9},{w}_{2} = \frac{8}{9} \), we find that\n\n\[ \n{\left( \frac{1}{9}\sqrt[3]{\frac{{a}^{3} + {b}^{3} + {c}^{3}}{3}} + \frac{8}{9}\sqrt[3]{abc}\right) }^{3} \leq \frac{1}{9}\left( \frac{{a}^{3} + {b}^{3} + {c}^{3}}{3}\right) + \frac{8}{9}\left( {abc... | Yes |
Theorem 3.3 (Hölder's Inequality)\n\nLet \( {\lambda }_{a},{\lambda }_{b},\ldots ,{\lambda }_{z} \) be positive reals with \( {\lambda }_{a} + {\lambda }_{b} + \cdots + {\lambda }_{z} = 1 \) . Let \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) , \( {b}_{1},{b}_{2},\ldots ,{b}_{n},\ldots ,{z}_{1},{z}_{2},\ldots ,{z}_{n} \) be po... | Proof. WLOG \( {a}_{1} + \cdots + {a}_{n} = {b}_{1} + \cdots + {b}_{n} = \cdots = 1 \) (note that the degree of the \( {a}_{i} \) on either side is \( {\lambda }_{a} \) ). In that case, the LHS of the inequality is 1, and we just note\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}^{{\lambda }_{a}}{b}_{i}^{{\lambda ... | Yes |
Prove \[ \mathop{\sum }\limits_{\text{cyc }}\frac{a}{\sqrt{{a}^{2} + {8bc}}} \geq 1 \] | Proof. By Holder \[ {\left( \mathop{\sum }\limits_{\text{cyc }}a\left( {a}^{2} + 8bc\right) \right) }^{\frac{1}{3}}{\left( \mathop{\sum }\limits_{\text{cyc }}\frac{a}{\sqrt{{a}^{2} + {8bc}}}\right) }^{\frac{2}{3}} \geq \left( {a + b + c}\right) \] So it suffices to prove \( {\left( a + b + c\right) }^{3} \geq \mathop{\... | Yes |
Prove \( \frac{1}{a\left( {b + c}\right) } + \frac{1}{b\left( {c + a}\right) } + \frac{1}{c\left( {a + b}\right) } \geq \frac{27}{2{\left( a + b + c\right) }^{2}} \) . | Proof. Again by Holder,\n\n\[ \n{\left( \mathop{\sum }\limits_{\mathrm{{cyc}}}a\right) }^{\frac{1}{3}}{\left( \mathop{\sum }\limits_{\mathrm{{cyc}}}b + c\right) }^{\frac{1}{3}}{\left( \mathop{\sum }\limits_{\mathrm{{cyc}}}\frac{1}{a\left( {b + c}\right) }\right) }^{\frac{1}{3}} \geq 1 + 1 + 1 = 3. \n\] | Yes |
Prove \( \mathop{\sum }\limits_{\text{cyc }}\frac{{a}^{3} + 5{b}^{3}}{{3a} + b} \geq \frac{3}{2}\left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) \) . | Proof. We use Cauchy (Titu) to obtain\n\n\[ \mathop{\sum }\limits_{\text{cyc }}\frac{{a}^{3}}{{3a} + b} = \mathop{\sum }\limits_{\text{cyc }}\frac{{\left( {a}^{2}\right) }^{2}}{3{a}^{2} + {ab}} \geq \frac{{\left( {a}^{2} + {b}^{2} + {c}^{2}\right) }^{2}}{\mathop{\sum }\limits_{\text{cyc }}3{a}^{2} + {ab}}. \]\n\nWe can... | No |
If \( {abc} = 1 \), prove \( \frac{1}{{a}^{5}{\left( b + 2c\right) }^{2}} + \frac{1}{{b}^{5}{\left( c + 2a\right) }^{2}} + \frac{1}{{c}^{5}{\left( a + 2b\right) }^{2}} \geq \frac{1}{3} \) . | Proof. We can use Hölder to eliminate the square roots in the denominator:\n\n\[ \n{\left( \mathop{\sum }\limits_{\mathrm{{cyc}}}ab + 2ac\right) }^{2}\left( {\mathop{\sum }\limits_{\mathrm{{cyc}}}\frac{1}{{a}^{5}{\left( b + 2c\right) }^{2}}}\right) \geq {\left( \mathop{\sum }\limits_{\mathrm{{cyc}}}\frac{1}{a}\right) }... | Yes |
Theorem 1.1 (Swapping the order of summation)\n\nLet \( f\left( {a, b}\right) \) be a function. Then\n\n\[ \mathop{\sum }\limits_{{a \in A}}\mathop{\sum }\limits_{{b \in B}}f = \mathop{\sum }\limits_{{b \in B}}\mathop{\sum }\limits_{{a \in A}}f \] | This seemingly obvious fact is a nontrivial step in more than \( {50}\% \) of summation problems (in the same way that cyclic quadrilaterals is used in more than \( {50}\% \) of geometry problems). We will see this key idea again and again in the examples that follow. Thus any time you see a double sum, you should alwa... | No |
Evaluate the sum\n\[ \mathop{\sum }\limits_{{n \geq 1}}\frac{{7n} + {32}}{n\left( {n + 2}\right) } \cdot {\left( \frac{3}{4}\right) }^{n} \] | Solution. Again, since we have a polynomial denominator, our first reflex is to decompose its partial fractions. This gives us\n\n\[ \frac{{7n} + {32}}{n\left( {n + 2}\right) } = \frac{16}{n} - \frac{9}{n + 2} \]\n\nSuddenly we’re done, because the sum telescopes according to the fact that \( 9 = {\left( 3/4\right) }^{... | Yes |
A random permutation of \( \{ 1,\ldots, n\} \) has an average of one fixed point. | To see this, let’s look at the case \( n = 4 \) :\n\n<table><thead><tr><th></th><th>W</th><th>\( \mathrm{X} \)</th><th>Y</th><th>Z</th><th>\( \sum \)</th></tr></thead><tr><td>1</td><td>W</td><td>X</td><td>Y</td><td>Z</td><td>4</td></tr><tr><td>2</td><td>W</td><td>X</td><td>Z</td><td>Y</td><td>2</td></tr><tr><td>3</td><... | Yes |
Let \( X \) and \( Y \) be random variables (not necessarily independent). Then \( \mathbb{E}\left\lbrack {X + Y}\right\rbrack = \) \( \mathbb{E}\left\lbrack X\right\rbrack + \mathbb{E}\left\lbrack Y\right\rbrack \) | Proof. For simplicity we do the case where \( X, Y \) take nonnegative integer values; for the general case one should use \( \int \) rather than \( \sum \) . By definition, we have\n\n\[ \mathbb{E}\left\lbrack {X + Y}\right\rbrack = \mathop{\sum }\limits_{{n \geq 0}}{nP}\left( {X + Y = n}\right) = \mathop{\sum }\limit... | Yes |
Lemma 2.4 \( \left( {\varphi * 1 = \mathrm{{id}}}\right) \)\n\nLet \( n \geq 1 \) be an integer. Then\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\varphi \left( d\right) = n \] | Proof. We will give a more pedestrian proof later on in \ | No |
Compute\n\n\\[ \n\\mathop{\\sum }\\limits_{{k \\geq 0}}\\left( \\begin{matrix} {1000} \\\\ {3k} \\end{matrix}\\right) \n\\] | Solution. We can rewrite the sum as\n\n\\[ \n\\mathop{\\sum }\\limits_{{n \\geq 0}}\\left( \\begin{matrix} {1000} \\\\ n \\end{matrix}\\right) f\\left( n\\right) \n\\]\n\nwhere\n\\[ \nf\\left( n\\right) = \\left\\{ \\begin{array}{ll} 1 & n \\equiv 0\\;\\left( {\\;\\operatorname{mod}\\;3}\\right) \\\\ 0 & \\text{ otherw... | No |
Lemma 3.1 (Fermat's little theorem)\n\nLet \( p \) be a prime. Then \( {a}^{p - 1} \equiv 1\\left( {\\;\\operatorname{mod}\\;p}\\right) \) whenever \( \\gcd \\left( {a, p}\\right) = 1 \) . | Proof. Multiplication by \( a \) exhibits a bijection from \( \{ 1,\\ldots, p - 1\} \) to itself. Now take the products to get \( {a}^{p - 1}\\left( {p - 1}\\right) ! \equiv \\left( {p - 1}\\right) !\\left( {\\;\\operatorname{mod}\\;p}\\right) \), and cancel the resulting \( \\left( {p - 1}\\right) ! \) . | Yes |
Lemma 3.2 (Wilson's theorem)\n\nFor any prime \( p \) ,\n\n\[ \left( {p - 1}\right) ! \equiv - 1\;\left( {\;\operatorname{mod}\;p}\right) \] | Exercise 3.3. Prove this theorem if you don't already know how. | No |
Lemma 3.4 (Sums of powers modulo \( p \) )\n\nLet \( p \) be a prime and \( m \) an integer. Then\n\n\[ \n{1}^{m} + {2}^{m} + \cdots + {\left( p - 1\right) }^{m} \equiv \left\{ \begin{array}{ll} 0\;\left( {\;\operatorname{mod}\;p}\right) & \text{ if }p - 1 \nmid m \\ - 1\;\left( {\;\operatorname{mod}\;p}\right) & \text... | Proof. Suffices to show the case \( p - 1 \nmid m \) since the other is obvious. Let \( g \) be a primitive root modulo \( p \) (see [1]), then the above sum equals\n\n\[ \n1 + {g}^{m} + \cdots + {g}^{\left( {p - 2}\right) m} \equiv \frac{{g}^{\left( {p - 1}\right) m} - 1}{{g}^{m} - 1}\;\left( {\;\operatorname{mod}\;p}... | Yes |
Let \( p > 3 \) be a prime. Then\n\n\[ \left( {p - 1}\right) !\left( {\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{p - 1}}\right) \equiv 0\;\left( {\;\operatorname{mod}\;{p}^{2}}\right) . | Proof. Let \( S = {1}^{-1} + {2}^{-1} + \cdots + {\left( p - 1\right) }^{-1}\left( {\;\operatorname{mod}\;{p}^{2}}\right) \) . It’s already clear (by say Lemma 3.4 with \( m = - 1 \) ) that \( p \mid S \), but we in fact want \( {p}^{2} \mid S \) . We exploit the symmetry by pairing up the opposite terms:\n\n\[ {2S} = ... | Yes |
Example 4.6 \( \left( {\varphi * \mathbf{1} = \mathrm{{id}}}\right) \)\n\nLet \( n \geq 1 \) be an integer. Then\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\varphi \left( d\right) = n \] | Proof. Rephrasing the problem, we wish to show that\n\n\[ \varphi * \mathbf{1} = \mathrm{{id}}. \]\n\nIt is true for prime powers \( n = {p}^{e} \), because the left-hand side is \( 1 + \left( {p - 1}\right) + \left( {{p}^{2} - p}\right) + \) \( \cdots + \left( {{p}^{e} - {p}^{e - 1}}\right) = {p}^{e} \) . But since bo... | No |
Theorem 4.9 (Möbius inversion formula)\n\nLet \( f \) and \( g \) be any arithmetic functions (possibly not multiplicative). Then\n\n\[ g\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}f\left( d\right) \Leftrightarrow f\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) g\left( {n/d}\right) . \]... | Proof. Assume \( g = f * \mathbf{1} \) . Then\n\n\[ g * \mu = \left( {f * \mathbf{1}}\right) * \mu = f * \left( {\mathbf{1} * \mu }\right) = f * \delta = f. \] | Yes |
Define a sequence \( {\left( {a}_{n}\right) }_{n \geq 1} \) by \( \mathop{\sum }\limits_{{d \mid n}}{a}_{d} = {2}^{n} \) . Show that \( n \) divides \( {a}_{n} \) . | Solution. By Möbius inversion, \( {a}_{n} = \mathop{\sum }\limits_{{d \mid n}}\mu \left( {n/d}\right) {2}^{d} \) . Now just let \( n = {p}_{1}^{{e}_{1}}\ldots {p}_{r}^{{e}_{r}} \) and bash. Details omitted. | No |
We have\n\[ \mathop{\sum }\limits_{{n \geq 0}}\left( \begin{matrix} {1000} \\ n \end{matrix}\right) = {2}^{1000} \] | Proof. Let \( {a}_{n} = \left( \begin{matrix} {1000} \\ n \end{matrix}\right) \), for example. Of course, by the binomial theorem we have attached generating function\n\n\[ A\left( x\right) = \mathop{\sum }\limits_{{n \geq 0}}{a}_{n}{x}^{n} = \mathop{\sum }\limits_{{n \geq 0}}\left( \begin{matrix} {1000} \\ n \end{matr... | Yes |
\[ \mathop{\sum }\limits_{{n \geq 0}}n\left( \begin{matrix} {1000} \\ n \end{matrix}\right) = {1000} \cdot {2}^{999} \] | Proof. Take the derivative of the previous example \[ \mathop{\sum }\limits_{{n \geq 1}}\left( \begin{matrix} {1000} \\ n \end{matrix}\right) n{x}^{n - 1} = {1000}{\left( 1 + x\right) }^{999}. \] This time, the we obtain \( \mathop{\sum }\limits_{{n \geq 0}}n\left( \begin{matrix} {1000} \\ n \end{matrix}\right) = {1000... | No |
Theorem 5.4 (Explicit form for Lucas numbers)\n\nLet \( \alpha = \frac{1}{2}\left( {1 + \sqrt{5}}\right) \) and \( \beta = \frac{1}{2}\left( {1 - \sqrt{5}}\right) \) . Then\n\n\[ \n{L}_{n} = {\alpha }^{n} + {\beta }^{n} \n\] | Proof. We consider the generating function\n\n\[ \nL\left( x\right) = 2 + x + 3{x}^{2} + 4{x}^{3} + 7{x}^{4} + \ldots \n\]\n\nWe aim to find its compact form. Write\n\n\[ \nL\left( x\right) = 2 + x + 3{x}^{2} + 4{x}^{3} + 7{x}^{4} + \ldots \n\]\n\n\[ \n{xL}\left( x\right) = {2x} + {x}^{2} + 3{x}^{3} + 4{x}^{4} + \ldots... | Yes |
Theorem 5.6 (Generalized binomial theorem)\n\nLet \( r \) be any real number (not necessarily an integer). Then\n\n\[{\left( 1 + x\right) }^{r} = \mathop{\sum }\limits_{{n \geq 0}}\left( \begin{array}{l} r \\ n \end{array}\right) {x}^{n}\;\text{ where }\;\left( \begin{array}{l} r \\ n \end{array}\right) = \frac{r\left(... | Proof. For concreteness, let’s show the coefficient of \( {x}^{3} \) is \( \frac{r\left( {r - 1}\right) \left( {r - 2}\right) }{3!} \) . Suppose\n\n\[{\left( 1 + x\right) }^{r} = \mathop{\sum }\limits_{{k \geq 0}}{a}_{k}{x}^{k} = {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} + {a}_{3}{x}^{3} + {a}_{4}{x}^{4} + \ldots\n\nTake the... | Yes |
Let \( S \) be the set of triples \( \\left( {i, j, k}\\right) \) of positive integers which satisfy \( i + j + k = {17} \) . Compute \[ \\mathop{\\sum }\\limits_{{\\left( {i, j, k}\\right) \\in S}}{ijk} \] | Solution. The point is to notice that one can consider the quantity \[ F\\left( x\\right) = \\left( {\\mathop{\\sum }\\limits_{{i \\geq 0}}i{x}^{i}}\\right) \\left( {\\mathop{\\sum }\\limits_{{j \\geq 0}}j{x}^{j}}\\right) \\left( {\\mathop{\\sum }\\limits_{{k \\geq 0}}k{x}^{k}}\\right) \] and that we merely want the co... | Yes |
For \( n \geq 0 \), compute\n\n\[ \mathop{\sum }\limits_{{k \geq 0}}\left( \begin{matrix} n + k \\ {2k} \end{matrix}\right) {2}^{n - k} \] | Proof. As described above, we let\n\n\[ A\left( x\right) = \mathop{\sum }\limits_{{n \geq 0}}\left\lbrack {\mathop{\sum }\limits_{{k \geq 0}}\left( \begin{matrix} n + k \\ {2k} \end{matrix}\right) {2}^{n - k}}\right\rbrack {x}^{n} = \mathop{\sum }\limits_{{k \geq 0}}\mathop{\sum }\limits_{{n \geq 0}}\left( \begin{matri... | Yes |
For \( n \geq 0 \), compute\n\n\[ \mathop{\sum }\limits_{{k \geq 0}}\left( \begin{matrix} k \\ n - k \end{matrix}\right) \] | Solution. Snake Oil:\n\n\[ \mathop{\sum }\limits_{{n \geq 0}}\left\lbrack {\mathop{\sum }\limits_{{k \geq 0}}\left( \begin{matrix} k \\ n - k \end{matrix}\right) }\right\rbrack {x}^{n} = \mathop{\sum }\limits_{{k \geq 0}}\mathop{\sum }\limits_{{n \geq 0}}\left( \begin{matrix} k \\ n - k \end{matrix}\right) {x}^{n} \]\n... | Yes |
Problem 6.2 (USAMO 2010/5, Titu Andreescu). Let \( q = \frac{{3p} - 5}{2} \) where \( p \) is an odd prime, and let \[ {S}_{q} = \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{5 \cdot 6 \cdot 7} + \cdots + \frac{1}{q\left( {q + 1}\right) \left( {q + 2}\right) }.\] | Prove that if \( \frac{1}{p} - 2{S}_{q} = \frac{m}{n} \) for integers \( m \) and \( n \), then \( m - n \) is divisible by \( p \) . | No |
Problem 6.3 (Princeton Individual Finals 2015, Xiaoyu Xu). Let \( p \) be an odd prime. Prove that \( {p}^{2} \mid {2}^{p} - 2 \) if and only if | \[ \frac{1}{1 \cdot 2} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{\left( {p - 2}\right) \left( {p - 1}\right) } \equiv 0\;\left( {\;\operatorname{mod}\;p}\right) . \] | No |
Ryan is learning number theory. He reads about the Möbius function \( \mu : \mathbb{N} \rightarrow \mathbb{Z} \), defined by \( \mu \left( 1\right) = 1 \) and \( \mu \left( n\right) = \) \( - \mathop{\sum }\limits_{{d \mid n, d \neq n}}\mu \left( d\right) \) for \( n > 1 \) . However, Ryan doesn’t like negative numbers... | Answer: \( \frac{11}{7} \) . Let \( f\left( {i, j}\right) = \Delta \left( {{3}^{i}{5}^{j}}\right) \) . Find a recurrence for \( f \), and deduce that the corresponding two-variable generating function obeys \( F\left( {x, y}\right) = \frac{1}{2}\left( {1 + \frac{1}{1 - {2x} - {2y} + {2xy}}}\right) \) . Compute the \( {... | Yes |
Theorem 3.4 (Triangles Sum to \( {180}^{ \circ } \) )\n\nFor any lines \( \ell, m, n \) we have\n\n\[ \measuredangle \left( {\ell, m}\right) + \measuredangle \left( {m, n}\right) + \measuredangle \left( {n,\ell }\right) = 0. \]\n\nIn particular, for any points \( A, B, C \) we have\n\n\[ \measuredangle {ABC} + \measure... | Proof. Check it yourself. | No |
Theorem 3.5 (Collinearity Criteria)\n\nLet \( X \) be any point. Points \( A, B, C \) are collinear if and only if\n\n\[ \measuredangle {XBC} = \measuredangle {XBA}\text{.} \] | Proof. Prove this yourself. (Show that the assertion is equivalent to \( \measuredangle {ABC} = 0 \) .) | No |
For any positive integers \( n \) and \( k \) ,\n\n\[ \left( \begin{array}{l} n \\ k \end{array}\right) < \frac{1}{e}{\left( \frac{en}{k}\right) }^{k} \]\n\nHere \( e \approx {2.718}\ldots \) is Euler’s constant. | Proof. Do \( \left( \begin{array}{l} n \\ k \end{array}\right) < \frac{{n}^{k}}{k!} \) and then use calculus to prove that \( k! \geq e{\left( k/e\right) }^{k} \) . Specifically,\n\n\[ \ln 1 + \ln 2 + \cdots + \ln k \geq {\int }_{x = 1}^{k}\ln {xdx} = k\ln k - k + 1 \]\n\nwhence exponentiating works. | No |
Let \( n \) and \( k \) be integers with \( n \leq {2}^{k/2} \) and \( k \geq 3 \) . Then it is possible to color the edges of the complete graph on \( n \) vertices with the following property: one cannot find \( k \) vertices for which the \( \left( \begin{array}{l} k \\ 2 \end{array}\right) \) edges among them are m... | Solution. Again we just randomly color the edges and hope for the best. We use a coin flip to determine the color of each of the \( \left( \begin{array}{l} n \\ 2 \end{array}\right) \) edges. Let’s call a collection of \( k \) vertices bad if all \( \left( \begin{array}{l} k \\ 2 \end{array}\right) \) edges are the sam... | Yes |
Given a triangle-free graph \( G \) with average degree \( d \) and \( N \) vertices, we can find an independent set with size at least \( {0.01}\frac{N}{d}\log d \) . | Our first move is to try and replace the \ | No |
A graph \( G \) has \( n \) vertices and average degree \( d \) . Prove that it is possible to select an independent set of size at least \( \frac{n}{2d} \) . | Proof. Rather than selecting \( \frac{n}{2d} \) vertices randomly and hoping the number of edges is 1, we’ll instead select each vertex with probability \( p \) . (We will pick a good choice of \( p \) later.)\n\nThat means the expected number of vertices we will take is \( {np} \) . Now there are \( \frac{1}{2}{nd} \)... | No |
Given a hypergraph \( G \) with \( N \) vertices and with edges all of size 3, suppose that for any two vertices at most one 3-edge joins them. Then we can find an independent set with size at least \( c \cdot \frac{N}{\sqrt{d}}\sqrt{\log d} \) . | In the context of the IMO problem, suppose we consider each of the \( n \) lines as a vertex and each finite region as a hyper-edge. Like in the previous solution, we treat pentagons, hexagons, . . . as just quadrilaterals; hence we can assume all edges have size either 3 or 4. Once again we use a coin flip weighted wi... | No |
Given circles \( {\omega }_{1} \) and \( {\omega }_{2} \) intersecting at points \( X \) and \( Y \), let \( {\ell }_{1} \) be a line through the center of \( {\omega }_{1} \) intersecting \( {\omega }_{2} \) at points \( P \) and \( Q \) and let \( {\ell }_{2} \) be a line through the center of \( {\omega }_{2} \) int... | If you try to draw the points in the order listed, you are sunk: the chance that \( P, Q, R \) , \( S \) are cyclic if you let \( {\ell }_{1} \) and \( {\ell }_{2} \) be arbitrary is basically zero.\n\nInstead, let’s draw \( {\omega }_{1} \) and \( {\omega }_{2} \) (with centers \( {O}_{1} \) and \( {O}_{2} \) ), so th... | Yes |
Theorem 1 (Line). The equation of a line is \( {ux} + {vy} + {wz} = 0 \) where \( u, v, w \) are reals. (These \( u \) , \( v \) and \( w \) are unique up to scaling.) | This is a corollary of the area formula, Theorem 10. | No |
Corollary 2 (Line through a vertex). The equation of a line passing through \( A \) is simply of the form \( y = {kz} \) for some constant \( k \) . | In particular, the equation for the line \( {AB} \) is simply \( z = 0 \), by substituting \( \left( {1,0,0}\right) \) and \( \left( {0,1,0}\right) \) into \( {ux} + {vy} + {wz} = 0 \) . | No |
Corollary 3 (Ceva’s Theorem). Let \( {AD},{BE} \) and \( {CF} \) be cevians of a triangle \( {ABC} \) . Then the cevians concur if and only if\n\n\[ \frac{BD}{DC}\frac{CE}{EA}\frac{AF}{FB} = 1 \] | Proof. Since \( D \) lies on \( {BC} \), the point \( D \) has the form \( D = \left( {0, d,1 - d}\right) \) . So the equation of line \( {AD} \) is simply\n\n\[ z = \frac{1 - d}{d}y \]\n\nSimilarly, if we let \( E = \left( {1 - e,0, e}\right) \) and \( F = \left( {f,1 - f,0}\right) \) then the lines \( {BE} \) and \( ... | Yes |
Theorem 4 (Evan’s Favorite Forgotten Trick). Consider displacement vectors \( \overrightarrow{MN} = \left( {{x}_{1},{y}_{1},{z}_{1}}\right) \) and \( \overrightarrow{PQ} = \left( {{x}_{2},{y}_{2},{z}_{2}}\right) \) . Then \( {MN} \bot {PQ} \) if and only if\n\n\[ 0 = {a}^{2}\left( {{z}_{1}{y}_{2} + {y}_{1}{z}_{2}}\righ... | Proof. Translate \( \overrightarrow{O} \) to \( \overrightarrow{0} \) . It is necessary and sufficient that \( \left( {{x}_{1}\overrightarrow{A} + {y}_{1}\overrightarrow{B} + {z}_{1}\overrightarrow{C}}\right) \cdot \left( {{x}_{2}\overrightarrow{A} + {y}_{2}\overrightarrow{B} + {z}_{2}\overrightarrow{C}}\right) = 0 \) ... | Yes |
Corollary 5. Consider a displacement vector \( \overrightarrow{PQ} = \left( {{x}_{1},{y}_{1},{z}_{1}}\right) \) . Then \( {PQ} \bot {BC} \) if and only if | \[ 0 = {a}^{2}\left( {{z}_{1} - {y}_{1}}\right) + {x}_{1}\left( {{c}^{2} - {b}^{2}}\right) \] | Yes |
Theorem 7 (Distance Formula). Consider a displacement vector \( \overrightarrow{PQ} = \left( {x, y, z}\right) \) . Then\n\n\[{\left| PQ\right| }^{2} = - {a}^{2}{yz} - {b}^{2}{zx} - {c}^{2}{xy}\] | Exercise. Prove the distance formula by using\n\n\[P{Q}^{2} = \left( {x\overrightarrow{A} + y\overrightarrow{B} + z\overrightarrow{C}}\right) \cdot \left( {x\overrightarrow{A} + y\overrightarrow{B} + z\overrightarrow{C}}\right)\]\n\nand the fact that \( x + y + z = 0 \) . | No |
Theorem 8. The general equation of a circle is\n\n\[ \n- {a}^{2}{yz} - {b}^{2}{zx} - {c}^{2}{xy} + \left( {{ux} + {vy} + {wz}}\right) \left( {x + y + z}\right) = 0 \n\] \n\nfor reals \( u, v, w \) . | Proof. Assume the circle has center \( \left( {j, k, l}\right) \) and radius \( r \) . Then this is just \n\n\[ \n- {a}^{2}\left( {y - k}\right) \left( {z - l}\right) - {b}^{2}\left( {z - l}\right) \left( {x - j}\right) - {c}^{2}\left( {x - j}\right) \left( {y - k}\right) = {r}^{2} \n\] \n\nExpand everything, and colle... | Yes |
Theorem 10 (Area Formula). The area of a triangle with vertices \( P = \left( {{x}_{1},{y}_{1},{z}_{1}}\right), Q = \left( {{x}_{2},{y}_{2},{z}_{2}}\right) \) and \( R = \left( {{x}_{3},{y}_{3},{z}_{3}}\right) \) is\n\n\[ \left\lbrack {PQR}\right\rbrack = \left\lbrack {ABC}\right\rbrack \cdot \left| \begin{array}{lll} ... | A proof of the area formula can be found in [2]. | No |
Corollary 11 (First Collinearity Criteria). The points \( P = \left( {{x}_{1} : {y}_{1} : {z}_{1}}\right), Q = \left( {{x}_{2} : {y}_{2} : {z}_{2}}\right) \) and \( R = \left( {{x}_{3} : {y}_{3} : {z}_{3}}\right) \) are collinear if and only if\n\n\[ \left| \begin{array}{lll} {x}_{1} & {y}_{1} & {z}_{1} \\ {x}_{2} & {y... | Proof. The points are collinear iff the area of the triangle they determine is zero, and this doesn't change under scaling \( \left( {x, y, z}\right) \mapsto \left( {{kx} : {ky} : {kz}}\right) \) . | Yes |
Corollary 12 (Line Through 2 Points). The equation of a line through the points \( P = \left( {{x}_{1} : {y}_{1} : {z}_{1}}\right) \) and \( Q = \left( {{x}_{2} : {y}_{2} : {z}_{2}}\right) \) is\n\n\[ \left| \begin{matrix} {x}_{1} & {y}_{1} & {z}_{1} \\ {x}_{2} & {y}_{2} & {z}_{2} \\ x & y & z \end{matrix}\right| = 0 \... | The previous corollary also implies the line formula. | No |
Corollary 13 (Second Collinearity Criteria). The points \( P = \left( {{x}_{1},{y}_{1},{z}_{1}}\right), Q = \left( {{x}_{2},{y}_{2},{z}_{2}}\right) \) and \( R = \left( {{x}_{3},{y}_{3},{z}_{3}}\right) \), are collinear if and only if\n\n\[ \left| \begin{array}{lll} {x}_{1} & {y}_{1} & 1 \\ {x}_{2} & {y}_{2} & 1 \\ {x}... | Proof. It is a property of the determinant that\n\n\[ \left| \begin{array}{lll} {x}_{1} & {y}_{1} & {z}_{1} \\ {x}_{2} & {y}_{2} & {z}_{2} \\ {x}_{3} & {y}_{3} & {z}_{3} \end{array}\right| = \left| \begin{array}{lll} {x}_{1} & {y}_{1} & {x}_{1} + {y}_{1} + {z}_{1} \\ {x}_{2} & {y}_{2} & {x}_{2} + {y}_{2} + {z}_{2} \\ {... | No |
Theorem 16 (Conway’s Formula). Given a point \( P \) with counter-clockwise directed angles \( \measuredangle {PBC} = \) \( \theta \) and \( \measuredangle {BCP} = \phi \), we have \( P = \left( {-{a}^{2} : {S}_{C} + {S}_{\phi } : {S}_{B} + {S}_{\theta }}\right) \) . | Proof. By the Law of Sines on triangle \( {PBC},{BP} = \frac{a\sin \phi }{\sin \left( {\theta + \phi }\right) } \), and \( {CP} = \frac{a\sin \theta }{\sin \left( {\theta + \phi }\right) } \), so the area of triangle \( {PBC} \) is\n\n\[ \n\frac{1}{2} \cdot {BC} \cdot {BP}\sin \theta = \frac{{a}^{2}\sin \theta \sin \ph... | Yes |
Theorem 4 (Evan’s Favorite Forgotten Trick). Consider displacement vectors \( \overrightarrow{MN} = \left( {{x}_{1},{y}_{1},{z}_{1}}\right) \) and \( \overrightarrow{PQ} = \left( {{x}_{2},{y}_{2},{z}_{2}}\right) \) . Then \( {MN} \bot {PQ} \) if and only if | \[ 0 = {a}^{2}\left( {{z}_{1}{y}_{2} + {y}_{1}{z}_{2}}\right) + {b}^{2}\left( {{x}_{1}{z}_{2} + {z}_{1}{x}_{2}}\right) + {c}^{2}\left( {{y}_{1}{x}_{2} + {x}_{1}{y}_{2}}\right) \] | Yes |
Corollary 5. Consider a displacement vector \( \overrightarrow{PQ} = \left( {{x}_{1},{y}_{1},{z}_{1}}\right) \) . Then \( {PQ} \bot {BC} \) if and only if | \[ 0 = {a}^{2}\left( {{z}_{1} - {y}_{1}}\right) + {x}_{1}\left( {{c}^{2} - {b}^{2}}\right) \] | Yes |
Corollary 6. The perpendicular bisector of \( {BC} \) has equation | \[ 0 = {a}^{2}\left( {z - y}\right) + x\left( {{c}^{2} - {b}^{2}}\right) \] | Yes |
Corollary 9. The circumcircle has equation | \[ {a}^{2}{yz} + {b}^{2}{zx} + {c}^{2}{xy} = 0 \] | Yes |
Corollary 11 (First Collinearity Criteria). The points \( P = \left( {{x}_{1} : {y}_{1} : {z}_{1}}\right), Q = \left( {{x}_{2} : {y}_{2} : {z}_{2}}\right) \) and \( R = \left( {{x}_{3} : {y}_{3} : {z}_{3}}\right) \) are collinear if and only if | \[ \left| \begin{array}{lll} {x}_{1} & {y}_{1} & {z}_{1} \\ {x}_{2} & {y}_{2} & {z}_{2} \\ {x}_{3} & {y}_{3} & {z}_{3} \end{array}\right| = 0 \] | Yes |
Theorem 1.1 (Linearity) If \( {F}_{1}\left( t\right) \) and \( {F}_{2}\left( t\right) \) are two functions whose Laplace transform exists, then\n\n\[ \mathcal{L}\left\{ {a{F}_{1}\left( t\right) + b{F}_{2}\left( t\right) }\right\} = a\mathcal{L}\left\{ {{F}_{1}\left( t\right) }\right\} + b\mathcal{L}\left\{ {{F}_{2}\lef... | Proof\n\n\[ \mathcal{L}\left\{ {a{F}_{1}\left( t\right) + b{F}_{2}\left( t\right) }\right\} = {\int }_{0}^{\infty }\left( {a{F}_{1} + b{F}_{2}}\right) {e}^{-{st}}{dt} \]\n\n\[ = {\int }_{0}^{\infty }\left( {a{F}_{1}{e}^{-{st}} + b{F}_{2}{e}^{-{st}}}\right) {dt} \]\n\n\[ = * - a{\int }_{0}^{\infty }{F}_{1}{e}^{-{st}}{dt... | Yes |
Theorem 1.2 (First Shift Theorem) If it is possible to choose constants \( M \) and \( \alpha \) such that \( \left| {F\left( t\right) }\right| \leq M{e}^{\alpha t} \), that is \( F\left( t\right) \) is of exponential order, then\n\n\[ \mathcal{L}\left\{ {{e}^{-{bt}}F\left( t\right) }\right\} = f\left( {s + b}\right) \... | Proof The proof is straightforward and runs as follows:-\n\n\[ \mathcal{L}\left\{ {{e}^{-{bt}}F\left( t\right) }\right\} = \mathop{\lim }\limits_{{T \rightarrow \infty }}{\int }_{0}^{T}{e}^{-{st}}{e}^{-{bt}}F\left( t\right) {dt} \]\n\n\[ = {\int }_{0}^{\infty }{e}^{-{st}}{e}^{-{bt}}F\left( t\right) {dt}\text{(as the li... | Yes |
Example 1.1 Find the Laplace transform of the function \( F\left( t\right) = t \) . | Solution Using the definition of Laplace transform,\n\n\[ \mathcal{L}\left( t\right) = \mathop{\lim }\limits_{{T \rightarrow \infty }}{\int }_{0}^{T}t{e}^{-{st}}{dt} \]\n\nNow, we have that\n\n\[ {\int }_{0}^{T}t{e}^{-{st}}{dt} = {\left\lbrack -\frac{t}{s}{e}^{-{st}}\right\rbrack }_{0}^{T} - {\int }_{0}^{T} - \frac{1}{... | Yes |
Find the Laplace transform of \( \mathcal{L}\left\{ {t{e}^{at}}\right\} \) and deduce the value of \( \mathcal{L}\left\{ {{t}^{n}{e}^{at}}\right\} \), where \( a \) is a real constant and \( n \) a positive integer. | Solution Using the first shift theorem with \( b = - a \) gives\n\n\[ \mathcal{L}\left\{ {F\left( t\right) {e}^{at}}\right\} = f\left( {s - a}\right) \]\n\nso with\n\n\[ F\left( t\right) = t\text{ and }f = \frac{1}{{s}^{2}} \]\n\nwe get\n\n\[ \mathcal{L}\left\{ {t{e}^{at}}\right\} = \frac{1}{{\left( s - a\right) }^{2}}... | Yes |
Find the Laplace transform of the function represented by \( F\\left( t\\right) \) where\n\n\[ \nF\\left( t\\right) = \\left\\{ \\begin{array}{ll} t & 0 \\leq t < {t}_{0} \\\\ 2{t}_{0} - t & {t}_{0} \\leq t \\leq 2{t}_{0} \\\\ 0 & t > 2{t}_{0} \\end{array}\\right. \n\] | Solution This function is of the \ | No |
Example 1.4 Determine the Laplace transform of the step function \( F\\left( t\\right) \) defined by\n\n\[ \nF\\left( t\\right) = \\left\\{ \\begin{array}{ll} 0 & 0 \\leq t < {t}_{0} \\\\ {at} & \\geq {t}_{0} \\end{array}\\right.\n\] | Solution \( F\\left( t\\right) \) itself is bounded, so there is no question that it is also of exponentially order. The Laplace transform of \( F\\left( t\\right) \) is therefore\n\n\[ \n\\mathcal{L}\\left( {F\\left( t\\right) }\\right) = {\\int }_{0}^{\\infty }{e}^{-{st}}F\\left( t\\right) {dt}\n\]\n\n\[ \n= {\\int }... | Yes |
Theorem 1.3 If \( \mathcal{L}\left( {F\left( t\right) }\right) = f\left( s\right) \) then \( \mathcal{L}\left( {{tF}\left( t\right) }\right) = - \frac{d}{ds}f\left( s\right) \) | Proof Let us start with the definition of Laplace transform\n\n\[ \mathcal{L}\left( {F\left( t\right) }\right) = {\int }_{0}^{\infty }{e}^{-{st}}F\left( t\right) {dt} \]\n\nand differentiate this with respect to \( s \) to give\n\n\[ \frac{df}{ds} = \frac{d}{ds}{\int }_{0}^{\infty }{e}^{-{st}}F\left( t\right) {dt} \]\n... | Yes |
Example 1.5 Determine the Laplace transform of the function \( t\sin t \) . | Solution To evaluate this Laplace transform we use Theorem 1.3 with \( f\left( t\right) = \sin t \) . This gives\n\n\[ \mathcal{L}\{ t\sin t\} = - \frac{d}{ds}\left\{ \frac{1}{1 + {s}^{2}}\right\} = \frac{2s}{{\left( 1 + {s}^{2}\right) }^{2}} \]\n\nwhich is the required result. | Yes |
Theorem 2.1\n\n\[ \n\mathcal{L}\left\{ {{F}^{\prime }\left( t\right) }\right\} = {\int }_{0}^{\infty }{e}^{-{st}}{F}^{\prime }\left( t\right) {dt} = - F\left( 0\right) + {sf}\left( s\right) .\n\] | Proof Integrating by parts once gives\n\n\[ \n\mathcal{L}\left\{ {{F}^{\prime }\left( t\right) }\right\} = {\left\lbrack F\left( t\right) {e}^{-{st}}\right\rbrack }_{0}^{\infty } + {\int }_{0}^{\infty }s{e}^{-{st}}F\left( t\right) {dt}\n\]\n\n\[ \n= - F\left( 0\right) + {sf}\left( s\right)\n\]\n\nwhere \( F\left( 0\rig... | Yes |
Theorem 2.2 If \( F\\left( t\\right) \) is a twice differentiable function of \( t \) then\n\n\\[ \n\\mathcal{L}\\left\\{ {{F}^{\\prime \\prime }\\left( t\\right) }\\right\\} = {s}^{2}f\\left( s\\right) - {sF}\\left( 0\\right) - {F}^{\\prime }\\left( 0\\right) .\n\\] | Proof The proof is unremarkable and involves integrating by parts twice. Here are the details.\n\n\\[\n\\mathcal{L}\\left\\{ {{F}^{\\prime \\prime }\\left( t\\right) }\\right\\} = {\\int }_{0}^{\\infty }{e}^{-{st}}{F}^{\\prime \\prime }\\left( t\\right) {dt}\n\\]\n\n\\[\n= {\\left\\lbrack {F}^{\\prime }\\left( t\\right... | Yes |
Theorem 2.3 If \( \mathcal{L}\left( {F\left( t\right) }\right) = f\left( s\right) \) then \( \mathcal{L}\left\{ \frac{F\left( t\right) }{t}\right\} = {\int }_{s}^{\infty }f\left( u\right) {du} \), assuming that\n\n\[ \mathcal{L}\left\{ \frac{F\left( t\right) }{t}\right\} \rightarrow 0\text{ as }s \rightarrow \infty . \... | Proof Let \( G\left( t\right) \) be the function \( F\left( t\right) /t \), so that \( F\left( t\right) = {tG}\left( t\right) \) . Using the property\n\n\[ \mathcal{L}\{ {tG}\left( t\right) \} = - \frac{d}{ds}\mathcal{L}\{ G\left( t\right) \} \]\n\nwe deduce that\n\n\[ f\left( s\right) = \mathcal{L}\{ F\left( t\right) ... | Yes |
Theorem 2.4 (Second Shift Theorem) If \( F\\left( t\\right) \) is a function of exponential order in\n\n\( t \) then\n\n\\[ \n\\mathcal{L}\\left\\{ {H\\left( {t - {t}_{0}}\\right) F\\left( {t - {t}_{0}}\\right) }\\right\\} = {e}^{-s{t}_{0}}f\\left( s\\right)\n\\]\n\nwhere \( f\\left( s\\right) \) is the Laplace transfo... | Proof This result is proved by direct integration.\n\n\\[ \n\\mathcal{L}\\left\\{ {H\\left( {t - {t}_{0}}\\right) F\\left( {t - {t}_{0}}\\right) }\\right\\} = {\\int }_{0}^{\\infty }H\\left( {t - {t}_{0}}\\right) F\\left( {t - {t}_{0}}\\right) {e}^{-{st}}{dt}\n\\]\n\n\\[ \n= {\\int }_{{t}_{0}}^{\\infty }F\\left( {t - {... | Yes |
Example 2.1 Determine the Laplace transform of the sine function switched on at time \( t = 3 \) . | Solution The sine function required that starts at \( t = 3 \) is \( S\left( t\right) \) where\n\n\[ S\left( t\right) = \left\{ \begin{array}{ll} \sin {tt} & t \geq 3 \\ 0 & t < 3 \end{array}\right. \]\n\nWe can use the Heaviside step function to write\n\n\[ S\left( t\right) = H\left( {t - 3}\right) \sin t. \]\n\nThe s... | Yes |
Theorem 2.5 The inverse Laplace transform is linear, i.e.\n\n\[ \n{\mathcal{L}}^{-1}\left\{ {a{f}_{1}\left( s\right) + b{f}_{2}\left( s\right) }\right\} = a{\mathcal{L}}^{-1}\left\{ {{f}_{1}\left( s\right) }\right\} + b{\mathcal{L}}^{-1}\left\{ {{f}_{2}\left( s\right) }\right\} .\n\] | Proof Linearity is easily established as follows. Since the Laplace transform is linear, we have for suitably well behaved functions \( {F}_{1}\left( t\right) \) and \( {F}_{2}\left( t\right) \) :\n\n\[ \n\mathcal{L}\left\{ {a{F}_{1}\left( t\right) + b{F}_{2}\left( t\right) }\right\} = a\mathcal{L}\left\{ {{F}_{1}\left... | Yes |
Use partial fractions to determine\n\n\[ \n{\mathcal{L}}^{-1}\left\{ \frac{a}{{s}^{2} - {a}^{2}}\right\} \n\] | Solution Noting that\n\n\[ \n\frac{a}{{s}^{2} - {a}^{2}} = \frac{1}{2}\left\lbrack {\frac{1}{s - a} - \frac{1}{s + a}}\right\rbrack \n\]\n\ngives straight away that\n\n\[ \n{\mathcal{L}}^{-1}\left\{ \frac{a}{{s}^{2} - {a}^{2}}\right\} = \frac{1}{2}\left( {{e}^{at} - {e}^{-{at}}}\right) = \sinh \left( {at}\right) . \n\]... | Yes |
Example 2.3 Determine the value of\n\n\[ \n{\mathcal{L}}^{-1}\left\{ \frac{{s}^{2}}{{\left( s + 3\right) }^{3}}\right\} \n\] | Solution Noting the standard partial fraction decomposition\n\n\[ \n\frac{{s}^{2}}{{\left( s + 3\right) }^{3}} = \frac{1}{s + 3} - \frac{6}{{\left( s + 3\right) }^{2}} + \frac{9}{{\left( s + 3\right) }^{3}} \n\]\n\nwe use the first shift theorem on each of the three terms in turn to give\n\n\[ \n{\mathcal{L}}^{-1}\left... | Yes |
Example 2.4 Determine the following inverse Laplace transforms\n\n\[ \n\\text{(a)}{\\mathcal{L}}^{-1}\\frac{\\left( s + 3\\right) }{s\\left( {s - 1}\\right) \\left( {s + 2}\\right) }\\text{; (b)}{\\mathcal{L}}^{-1}\\frac{\\left( s - 1\\right) }{{s}^{2} + {2s} - 8}\\text{; (c)}{\\mathcal{L}}^{-1}\\frac{{3s} + 7}{{s}^{2}... | Solution All of these problems are tackled in a similar way, by decomposing the expression into partial fractions, using shift theorems, then identifying the simplified expressions with various standard forms.\n\n(a) Using partial fraction decomposition and not dwelling on the detail we get\n\n\[ \n\\frac{s + 3}{s\\lef... | No |
Theorem 2.6 (Initial Value) If the indicated limits exist then\n\n\[ \mathop{\lim }\limits_{{t \rightarrow 0}}F\left( t\right) = \mathop{\lim }\limits_{{s \rightarrow \infty }}{sf}\left( s\right) \]\n\n(The left hand side is \( F\left( 0\right) \) of course, or \( F\left( {0 + }\right) \) if \( \mathop{\lim }\limits_{{... | Proof We have already established that\n\n\[ \mathcal{L}\left\{ {{F}^{\prime }\left( t\right) }\right\} = {sf}\left( s\right) - F\left( 0\right) . \]\n\n(2.1)\n\nHowever, if \( {F}^{\prime }\left( t\right) \) obeys the usual criteria for the existence of the Laplace transform, that is \( {F}^{\prime }\left( t\right) \)... | Yes |
Theorem 2.7 (Final Value) If the limits indicated exist, then\n\n\[ \mathop{\lim }\limits_{{t \rightarrow \infty }}F\left( t\right) = \mathop{\lim }\limits_{{s \rightarrow 0}}{sf}\left( s\right) \]\n | Proof Again we start with the formula for the Laplace transform of the derivative of \( F\left( t\right) \)\n\n\[ \mathcal{L}\left\{ {{F}^{\prime }\left( t\right) }\right\} = {\int }_{0}^{\infty }{e}^{-{st}}{F}^{\prime }\left( t\right) {dt} = {sf}\left( s\right) - F\left( 0\right) \]\n\n(2.2)\n\nthis time writing the i... | Yes |
Demonstrate the initial and final value theorems using the function \( F\left( t\right) = {e}^{-t} \) . Expand \( {e}^{-t} \) as a power series, evaluate term by term and confirm the legitimacy of term by term evaluation. | \[ \mathcal{L}\left\{ {e}^{-t}\right\} = \frac{1}{s + 1} \] \[ \mathop{\lim }\limits_{{t \rightarrow 0}}F\left( t\right) = F\left( 0\right) = {e}^{-0} = 1 \] \[ \mathop{\lim }\limits_{{s \rightarrow \infty }}{sf}\left( s\right) = \mathop{\lim }\limits_{{s \rightarrow \infty }}\frac{s}{s + 1} = 1. \] This confirms the i... | Yes |
Example 2.6 Determine the inverse Laplace transform\n\n\[ \n{\\mathcal{L}}^{-1}\\left\\{ \\frac{{s}^{2}}{{s}^{2} + 1}\\right\\} \n\]\n\nand interpret the \( F\\left( t\\right) \) obtained. | Solution Writing\n\n\[ \n\\frac{{s}^{2}}{{s}^{2} + 1} = 1 - \\frac{1}{{s}^{2} + 1} \n\]\n\nand using the linearity property of the inverse Laplace transform gives\n\n\[ \n{\\mathcal{L}}^{-1}\\left\\{ \\frac{{s}^{2}}{{s}^{2} + 1}\\right\\} = {\\mathcal{L}}^{-1}\\{ 1\\} - {\\mathcal{L}}^{-1}\\left\\{ \\frac{1}{{s}^{2} + ... | Yes |
Find the value of \( {\mathcal{L}}^{-1}\left\{ \frac{{s}^{3}}{{s}^{2} + 1}\right\} \) . | Solution Using a similar technique to the previous example we first see that\n\n\[ \frac{{s}^{3}}{{s}^{2} + 1} = s - \frac{s}{{s}^{2} + 1} \]\n\nso taking inverse Laplace transforms using the linearity property once more yields\n\n\[ {\mathcal{L}}^{-1}\left\{ \frac{{s}^{3}}{{s}^{2} + 1}\right\} = {\mathcal{L}}^{-1}\{ s... | Yes |
Theorem 2.8 Let \( F\left( t\right) \) have period \( T > 0 \) so that \( F\left( t\right) = F\left( {t + T}\right) \) . Then\n\n\[ \mathcal{L}\{ F\left( t\right) \} = \frac{{\int }_{0}^{T}{e}^{-{st}}F\left( t\right) {dt}}{1 - {e}^{-{sT}}}. \] | Proof Like many proofs of properties of Laplace transforms, this one begins with its definition then evaluates the integral by using the periodicity of \( F\left( t\right) \)\n\n\[ \mathcal{L}\{ F\left( t\right) \} = {\int }_{0}^{\infty }{e}^{-{st}}F\left( t\right) {dt} \]\n\n\[ = {\int }_{0}^{T}{e}^{-{st}}F\left( t\ri... | Yes |
determine \( \mathcal{L}\{ F\left( t\right) \} \) . | Solution The graph of \( F\left( t\right) \) is shown in Fig. 2.5. The function \( F\left( t\right) \) actually has period \( \pi \), but it is easier to carry out the calculation as if the period was \( {2\pi } \) . Additionally we can check the answer by using the theorem with \( T = \pi \) . With \( T = {2\pi } \) w... | Yes |
Theorem 3.2 (Convolution) If \( f\left( t\right) \) and \( g\left( t\right) \) are two functions of exponential order (so that their Laplace transforms exist), and writing \( \mathcal{L}\{ f\} = \bar{f}\left( s\right) \) and \( \mathcal{L}\{ g\} = \bar{g}\left( s\right) \) as the two Laplace transforms then \( {\mathca... | Proof In order to prove this theorem, we in fact show that\n\n\[ \bar{f}\bar{g} = \mathcal{L}\{ f\left( t\right) * g\left( t\right) \} \]\n\nby direct integration of the right hand side. In turn, this involves its interpretation in terms of a repeated integral. Now,\n\n\[ \mathcal{L}\{ f\left( t\right) * g\left( t\righ... | Yes |
Example 3.1 Find the value of \( \cos t * \sin t \) . | Solution Using the definition of convolution we have\n\n\[ \cos t * \sin t = {\int }_{0}^{t}\cos \left( \tau \right) \sin \left( {t - \tau }\right) {d\tau }.\]\n\nTo evaluate this we could of course resort to computer algebra: alternatively we use the identity\n\n\[ \sin \left( A\right) \cos \left( B\right) = \frac{1}{... | Yes |
Example 3.2 Find the value of \( \sin t * {t}^{2} \) . | Solution We progress as before by using the definition\n\n\[ \sin t * {t}^{2} = {\int }_{0}^{t}\left( {\sin \tau }\right) {\left( t - \tau \right) }^{2}{d\tau }.\]\n\nIt is up to us to choose the order as from Theorem \( {3.2f} * g = g * f \) . Of course we choose the order that gives the easier integral to evaluate. I... | Yes |
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