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Start with convergence criterion: A necessary and sufficient condition for the Fourier series T(x) of f to converge pointwise to c(x) on E is that there exists a fixed $\delta$ such that $ 0 < \delta < \pi$ and $\int_0^\delta {{g_{c(x)}}(u)\frac{{\sin \left( {nu} \right)}}{u}du \to 0} $ pointwise on E. Here ${g_{c(x)}}(u) = \frac{1}{2}\left( ...


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@goblin has already given you a good list of books that you could have a look at. I second this recommendation. Another recommendation is to take a math class on something like abstract algebra or discrete mathematics. In studying, for example, abstract algebra you learn how to think abstractly about concepts. You are given definitions of certain things and ...


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Note that proof theory isn't the study of how to write an informal proof, its the study of certain logical calculi and the proofs they accept/reject. If I understand correctly, this isn't what you're looking for. You want something more like: How To Prove It How To Think Like A Mathematician Mathematical Proofs: A Transition To Advanced Mathematics Good ...


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Basis step: It's clear that: $F(0)$ is an integer and $F(0)=\frac{3^0+(-1)^0}{2}$ $F(1)$ is an aninteger and $F(1)$=\frac{3^1+(-1)^1}{2}$ Induction steps Given an integer $n$, Assume that for every $0\leq k\leq n$ we have: $F(k)$ is an integer and $F(k)=\frac{3^k+(-1)^k}{2}$ And let's prove this assertion for $n+1$, we have $n\leq n$ and $n-1\leq ...


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You verified the base step and you have the induction assumption. Here's how you can use it: $$ a_{k+1}=\underbrace{2a_{k}}_{\text{even times odd is even}}+\underbrace{3a_{k-1}}_{\text{odd times odd is odd}} $$ Since an even number plus an odd number is an odd number, this proves your result. Note that strong induction was necessary to conclude that ...


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For example, $\,a_0 = 1 = a_1\,$ are both odd so $\,a_2 = 2a_1 + 3a_0\,$ = even + odd = odd. That is precisely how the general induction step is proved, i.e. for $\,k\ge 2\,$ prove in the same way that $\,a_k\,$ is odd, given that, by (strong) induction hypothesis, both $\,a_{k-1}\,$ and $\,a_{k-2}\,$ are odd. Remark $\ $ Insight is gained by viewing things ...


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IH the induction hypothesis is $p(i)$ is true for any :$0< i< c$ Goal we want to prove that $P(c)$ is true We can use : $$a_c=\underbrace{2a_{c-1}}_{\text{even}}+\underbrace{3a_{c-2}}_{\text{odd}} $$ and IH implies that both $a_{c-2}$ and $a_{c-1}$ are odd so $a_{c}$ is odd hence $P(c)$ is true.


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If you want to prove that every polynomial map $\mathbb R^n\to\mathbb R$ is continuous, it is not a good idea to consider a general polynomial and give an $\varepsilon$-$\delta$-proof. Instead, you should start with the building blocks, i.e. the projections $x\mapsto x_i$. Use an $\varepsilon$-$\delta$-proof to show that all the projections are continiuous. ...


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Coq runs on type theory. Then "false" is a type representative of a false statement. Thus Coq proved a false statement. You don't want math software that might accidentally prove false things. If Coq can prove "false" then it might somehow prove $1+1=3$. It could have a snowball effect if it proves false things in the middle of a bigger proof.


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You may observe that, as $x \to 0$, $$ f(x)=a_1x+\sum_{n=2}^{\infty}a_nx^n $$ $$ g(x)=b_1x+\sum_{n=2}^{\infty}b_nx^n $$ giving $$ \frac{f(x)}{g(x)}=\frac{a_1+\sum_{n=2}^{\infty}a_nx^{n-1}}{b_1+\sum_{n=2}^{\infty}b_nx^{n-1}} \to \frac{a_1}{b_1} $$ Similarly, since $$ f'(x)=a_1+\sum_{n=2}^{\infty}na_nx^{n-1} $$ $$ g'(x)=b_1+\sum_{n=2}^{\infty}nb_nx^{n-1} $$ ...


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Hint. First note, that $$ 0 = \lim_{x\to 0} f(x) = \lim_{x\to 0}\sum_n a_nx^n = a_0 $$ and likewise $b_0 = 0$. Along the same line of thought, we have $$ a_1 = \lim_{x\to 0} f'(x), \quad b_1 = \lim_{x\to 0} g'(x) $$ Now $$ \frac{f(x)}{g(x)} = \frac{\sum_{n\ge 0} a_n x^n}{\sum_{n\ge 0} b_n x^n} = \frac{\sum_{n\ge 1} a_n x^n}{\sum_{n\ge 1} b_n x^n} = ...


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$n$ is not prime $\implies n$ has a prime factor $p\in(1,n)$. If $p\leq\sqrt{n}$ then we're done. If $p>\sqrt{n}$ then let $m$ denote $\frac{n}{p}$: $m=\frac{n}{p}\implies{mp=n}\implies\color{red}{m|n}$ $p>\sqrt{n}\implies\frac{n}{p}<\frac{n}{\sqrt{n}}=\sqrt{n}\implies\color{red}{m<\sqrt{n}}$ Let $q$ be a prime factor of $m$: ...



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