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for $|x|<1$ : $$ \frac{1}{1-x}=\sum_{n\geq 0} x^n $$ put $t=-x$ you get : $$ f(t)=\frac{1}{1+t}=\sum_{n\geq 0} (-1)^n t^n=\sum_{n\geq 0} \frac{(-1)^nn!}{n!} t^n=\sum_{n\geq 0} \frac{f^{(n)}(0)}{n!} t^n $$ so $$ f^{(n)}(0)=(-1)^n n! $$


The result is correct if $\sum_{i=1}^na_i$ is a multiple of $n$. Let $a=\frac1n\sum_{i=1}^na_i$; we want to show that $$\sum_{i=1}^n\binom{a_i}r\ge n\binom{a}r\;.\tag{1}$$ Suppose that $a_i<a<a_k$. Then $$\begin{align*} \binom{a_i}r+\binom{a_k}r&=\binom{a_i}r+\binom{a_k-1}r+\binom{a_k-1}{r-1}\\ &\ge\binom{a_i}r+\binom{a_k-1}r+\binom{a_i}{r-1}...


I think you have your wires crossed; particularly it seems like you're thinking about how the sum of the first $n$ odd numbers gives $n^2$. This is woefully incorrect. Here is how you should frame it: Suppose that $x_1,\ldots, x_n\ge 2$, then you want to show that $x_1\cdots x_n$ is odd if and only if $x_i$ is odd for all $i$. In terms of an induction ...


the result is correct but in the demonstration it is best not to confuse the symbol $\varepsilon$ and $\delta$ Since $f$ is uniformly continuous then $\forall\epsilon>0, \exists\delta>0$ s.t. $\forall x,x'\in X, d(x,x')<\delta\Rightarrow\rho(f(x),f(x'))<\epsilon$ But since $(x_n)$ is Cauchy in $X$, then $\forall\delta>0, \exists N$ s.t. $\...


Suppose $$A=B+C$$ If $$B^T=B, $$ $$C^T=-C,$$ then according to the known property of transposition of sum of matrices $$A^T=(B+C)^T=B^T+C^T=B+(-C)=B-C$$ Now we have $$A=B+C \tag 1\\ $$ $$A^T=B-C\tag 2\\$$ Adding $(1)$ to $(2)$ gives $$B={(A+A^T)\over 2}\\ $$ Subtracting $(2)$ from $(1)$ gives $$C={(A-A^T)\over 2}\\ $$

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