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- https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/05%3A_Solutions_and_Hints_for_Selected_Exercises/5.01%3A_Section_1-Using \(frac{\varepsilon}{2}\) in part \((2^{\prime})\) of Proposition 1.5.1 applied to the sets \(A\) and \(B\), there exits \(a \in A\) and \(b \in B\) such that \[\sup A-\frac{\varepsilon}{2}<a \te...Using \(frac{\varepsilon}{2}\) in part \((2^{\prime})\) of Proposition 1.5.1 applied to the sets \(A\) and \(B\), there exits \(a \in A\) and \(b \in B\) such that \[\sup A-\frac{\varepsilon}{2}<a \text { and } \sup B-\frac{\varepsilon}{2}<b .\] It follows that \[\sup A+\sup B-\varepsilon<a+b .\] This proves condition \((2^{\prime})\) of Proposition 1.5.1 applied to the set \(A + B\) that \(\sup A + \sup B = \sup (A + B)\) as desired
- https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/An_Introduction_to_Number_Theory_(Veerman)/05%3A_Modular_Arithmetic_and_Primes/5.05%3A_New_PageIt tells us that there is a unique element \(a^{-1}\) such that \(aa^{-1} = _{b} 1\) if and only if \(a\) is in the reduced set of residues (modulo \(b\)). For every \(a \in \mathbb{Z}_{p}\) there is ...It tells us that there is a unique element \(a^{-1}\) such that \(aa^{-1} = _{b} 1\) if and only if \(a\) is in the reduced set of residues (modulo \(b\)). For every \(a \in \mathbb{Z}_{p}\) there is a unique \(a′ = _{p} -a\) such that \(a+a′ = _{p} 0\). For every \(a \in \mathbb{Z}_{p}\) and \(a \ne 0\), there is a unique \(a′ = a^{-1}\) so that \(aa′ = _{p} 1\).
- https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/An_Introduction_to_Number_Theory_(Veerman)/01%3A_A_Quick_Tour_of_Number_Theory/1.02%3A_Rational_and_Irrational_NumbersNote that any non-empty set \(S\) of integers with a lower bound can be transformed by addition of a integer \(b \in N_{0}\) into a non-empty \(S+b\) in \(N_{0}\). The advantage of expressing a ration...Note that any non-empty set \(S\) of integers with a lower bound can be transformed by addition of a integer \(b \in N_{0}\) into a non-empty \(S+b\) in \(N_{0}\). The advantage of expressing a rational number as the solution of a degree 1 polynomial, however, is that it naturally leads to Definition 1.12. The crux of the following proof is that we take an interval and scale it up until we know there is an integer in it, and then scale it back down.
- https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/An_Introduction_to_Number_Theory_(Veerman)/04%3A_Number_Theoretic_Functions/4.06%3A_New_PageShow that \(\varphi (n) = |S-\bigcup_{i=1}^{r} A_{i}\). (Hint: any number that is not co-prime with \(n\) is a multiple of at least one of the \(p_{i}\).) Show that the principle of inclusion-exclusio...Show that \(\varphi (n) = |S-\bigcup_{i=1}^{r} A_{i}\). (Hint: any number that is not co-prime with \(n\) is a multiple of at least one of the \(p_{i}\).) Show that the principle of inclusion-exclusion implies that \(|S-\bigcup_{i=1}^{r} A_{i}| = n+n \sum_{l=1}^{r} (-1)^{l} \sum_{I_{l} \subseteq R} \prod_{i \in I_{l}} \frac{1}{p_{i}}\). Show that \(n+n \sum_{l=1}^{r} (-1)^{l} \sum_{I_{l} \subseteq R} \prod_{i \in I_{l}} \frac{1}{p_{i}} = n \prod_{i=1}^{r} (1-\frac{1}{p_{i}})\).
- https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/05%3A_Solutions_and_Hints_for_Selected_Exercises/5.02%3A_Section_2-Define \[\alpha_{n}=\sup _{k \geq n}\left(a_{n}+b_{n}\right), \beta_{n}=\sup _{k \geq n} a_{k}, \gamma_{n}=\sup _{k \geq n} b_{k} .\] By the definition, \[\limsup _{n \rightarrow \infty}\left(a_{n}+b_...Define \[\alpha_{n}=\sup _{k \geq n}\left(a_{n}+b_{n}\right), \beta_{n}=\sup _{k \geq n} a_{k}, \gamma_{n}=\sup _{k \geq n} b_{k} .\] By the definition, \[\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=\lim _{n \rightarrow \infty} \alpha_{n}, \limsup _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} \beta_{n}, \limsup _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} \gamma_{n} .\] By Exercise 2.5.3, \[\alpha_{n} \leq \beta_{n}+\gamma_{n} \text { for all } n \in \math…
- https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/04%3A_Differentiation/4.05%3A_Section_5-Consider the function \[g(t)=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(t)}{k !}(x-t)^{k}-\frac{\lambda}{(n+1) !}(x-t)^{n+1} .\] Then \[g(\bar{x})=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(\bar{x})}{k !}(x-\bar{x})^{k...Consider the function \[g(t)=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(t)}{k !}(x-t)^{k}-\frac{\lambda}{(n+1) !}(x-t)^{n+1} .\] Then \[g(\bar{x})=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(\bar{x})}{k !}(x-\bar{x})^{k}-\frac{\lambda}{(n+1) !}(x-\bar{x})^{n+1}=f(x)-P_{n}(x)-\frac{\lambda}{(n+1) !}(x-\bar{x})^{n+1}=0 .\] and \[g(x)=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(x)}{k !}(x-x)^{k}-\frac{\lambda}{(n+1) !}(x-x)^{n+1}=f(x)-f(x)=0 .\] By Rolle's theorem, there exists \(c\) in between \(\bar{x}\) and \(x\) such that…
- https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/02%3A_SequencesMore precisely, a sequence of elements of a set \(A\) is a function \(f: \mathbb{N} \rightarrow A\). We will denote the image of \(n\) under the function with subscripted variables, for example, \(a_{...More precisely, a sequence of elements of a set \(A\) is a function \(f: \mathbb{N} \rightarrow A\). We will denote the image of \(n\) under the function with subscripted variables, for example, \(a_{n}=f(n)\). We will also denote sequences by \(\left\{a_{n}\right\}_{n=1}^{\infty}\), \(\left\{a_{n}\right\}_{n}\), or even \(\left\{a_{n}\right\}\). Each value \(a_{n}\) is called a term of the sequence, more precisely, the \(n\)-th term of the sequence.
- https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/An_Introduction_to_Number_Theory_(Veerman)/02%3A_The_Fundamental_Theorem_of_Arithmetic/2.04%3A_New_Page\[\begin{array}{ccc} {a = \prod_{i=1}^{s} p_{i}^{k_{i}}}&{and}&{b = \prod_{i=1}^{s} p_{i}^{l_{i}}} \end{array} \nonumber\] On the other hand, any integer greater than \(m\) has a unique factorization ...\[\begin{array}{ccc} {a = \prod_{i=1}^{s} p_{i}^{k_{i}}}&{and}&{b = \prod_{i=1}^{s} p_{i}^{l_{i}}} \end{array} \nonumber\] On the other hand, any integer greater than \(m\) has a unique factorization that either contains a prime not in the list \(P\) and therefore divides neither \(a\) nor \(b\), or, if not, at least one of the primes in \(P\) in its factorization has a power greater than \(m_{i}\).
- https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/zz%3A_Back_Matter
- https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/An_Introduction_to_Number_Theory_(Veerman)/03%3A_Linear_Diophantine_Equations/3.6_ExercisesDefine \(\gcd(p_{1}, p_{2})\) as the highest degree polynomial that is a divisor of both \(p_{1}\) and \(p_{2}\). (Remark: if you do this very carefully, you will realize that in this problem you real...Define \(\gcd(p_{1}, p_{2})\) as the highest degree polynomial that is a divisor of both \(p_{1}\) and \(p_{2}\). (Remark: if you do this very carefully, you will realize that in this problem you really need to consider equivalence classes of polynomials, in the sense that two polynomials \(p\) and \(q\) are equivalent if there is a non-zero constant \(c \in \mathbb{R}\) such that \(p = cq\).
- https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/04%3A_Differentiation/4.01%3A_Definition_and_Basic_Properties_of_the_DerivativeFor example, given functions \(f: G_{1} \rightarrow \mathbb{R}\), \(g: G_{2} \rightarrow \mathbb{R}\), and \(h: G_{3} \rightarrow \mathbb{R}\) such that \(f\left(G_{1}\right) \subset G_{2}\), \(g\left...For example, given functions \(f: G_{1} \rightarrow \mathbb{R}\), \(g: G_{2} \rightarrow \mathbb{R}\), and \(h: G_{3} \rightarrow \mathbb{R}\) such that \(f\left(G_{1}\right) \subset G_{2}\), \(g\left(G_{2}\right) \subset G_{3}\), \(f\) is differentiable at \(a\), \(g\) is differentiable at \(f(a)\), and \(h\) is differentiable at \(g(f(a))\), we obtain that \(h \circ g \circ f\) is differentiable at \(a\) and \((h \circ g \circ f)^{\prime}(a)=h^{\prime}(g(f(a))) g^{\prime}(f(a)) f^{\prime}(a)\)
