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binominal
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binominal
binominal, a. (baɪˈnɒmɪnəl) [f. L. binōmin-is, f. bi- two + nōmin- (nom. nōmen) name + -al1.] Having or characterized by two names, esp. those of genus and species in scientific nomenclature.1880 Gunther Fishes 10 Applying binominal terms to the species. 1881 Trans. Vict. Inst. 24 In this way the bi...
Oxford English Dictionary
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Intiornis
Intiornis (meaning "Inti bird", the binominal naming means "Unexpected Sun bird") is an extinct genus of avisaurid enantiornithean birds which existed
wikipedia.org
en.wikipedia.org
Is binominal the same as boolean? In the case of describing an attribute type which has two values (yes or no) is binominal the same as boolean? A source would be great.
_Binominal_ stresses that the item on hand has two names, rather than two values. It is not so appropriate.
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Find sum of binominal formula and prove it I have hard time with finding sum of this: $$ \sum_{k=1}^{n}k{n\choose k} $$ Please help! Prferably with some good hints.
**Hint**. Recall that $\binom nk$ counts the $k$-subsets of an $n$-set. So $$ \sum_{k=1}^n k \binom nk = \left|\left\\{(A,x) : x \in A, A \subseteq \\{0,\ldots, n-1\right\\}\right| $$ Counting in another way (first choosing $x$, then the set $A$), we have $$ \sum_{k=1}^n k\binom nk = n \sum_{k=1}^{n...
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Squaring a binominal I have a really simple question that I can't find the answer to. In a algebra test, I was asked to simplify $(5 + $$\sqrt{3}$)$^2$. What I did was to square each term individually: $5^2$+$\sqrt{3...
$(5+\sqrt{3})^2=(5+\sqrt{3})(5+\sqrt{3})$ then FOIL.
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Using Binomial Theorem to solve a question The question is this: > "We assume that $a$,$b$ are positive numbers(they might not be integers) but $a+b$ and $a*b$ are integers, prove that for every $n$ $\in$$\mathbb{...
You don't need the Binomial Theorem. Use induction. Note that $a^n+b^n$ holds for $n=0,1$ from the conditions provided. Now asssume it is true for $n=m$, $n=m+1$. Then using Newton's Identity $$a^{m+2}+b^{m+2}=(a+b)(a^{m+1}+b^{m+1})-ab(a^m+b^m)$$ We find it holds for $m+2$ as well. Thus, our proof i...
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Prove that the nth root converges to 1 with Binominal Theorem Prove with the use of the Binominal Theorem that: > **$$\sqrt{x}=(\sqrt[n] {\sqrt{x}})^x=(\sqrt[x] {\sqrt{x}}-1)+1)^x≥1+x(\sqrt[n] {\sqrt{x}}-1) \quad \f...
It's a little simpler than you are making it.... $$ \begin{align} ((\sqrt[x] {\sqrt{x}}-1)+1)^x &= \sum_{\ell=0}^x \binom x\ell(\sqrt[x] {\sqrt{x}}-1)^{\ell} \\\ &\ge \binom x0(\sqrt[x] {\sqrt{x}}-1)^0 + \binom x1(\sqrt[x] {\sqrt{x}}-1)^1 \\\ &=1+ x(\sqrt[x] {\sqrt{x}}-1) \end{align} $$
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可合并堆1：二项堆（Binominal Heap）-CSDN博客
斐波那契堆(Fibonacci heap)是堆中一种，它和二项堆一样，也是一种可合并堆；可用于实现合并优先队列。斐波那契堆比二项堆具有更好的平摊分析性能，它的合并操作的时间复杂度是O(1)。 与二项堆一样，它也是由一组堆最小有序树组成，并且是一种可合并堆。
blog.csdn.net
dionymal
dionymal, a. (daɪˈɒnɪməl) [f. as prec. + -al1.] Of or pertaining to a dionym; = binominal.1656 Blount Glossogr., Dionymal, that hath two names. 1884 J. A. Allen On Zoöl. Nomen. in The Auk Oct. 352 The binomial (or dionymal) system.
Oxford English Dictionary
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黄花瓣螺属
rumphii Blainville, 1819 ≡ Dolabella auricularia (Lightfoot, 1786)
Dolabella scapula O'Donoghue, 1929 ≡ Dolabella auricularia (Lightfoot, 1786) (non-binominal
wikipedia.org
zh.wikipedia.org
Problem using binominal theorem? I tried to solve this problem but I couldn't so i'm looking for an help. > Is there any two digit natural number $n$ which fits with following statement? $$n \mid (4^n - 3^n - 1)$$ ...
Note that by the binomial theorem $$4^n=(3+1)^n=3^n+\sum_{k=1}^{n-1}\binom{n}{k}\cdot 3^k\cdot 1^{n-k}+1^n.$$ Hence $$4^n-3^n-1=\sum_{k=1}^{n-1}\binom{n}{k}\cdot 3^k.$$ Show that for $n=11$ (any prime will work) then the binomial coefficients $\binom{11}{k}$ for $k=1,\dots,10$ are all multiple of $1...
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Proving a Binomial Identity **Problem $\boldsymbol{25}$ [$\boldsymbol{5}$ Points]:** Show that $$ \sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}=2^n $$ _Hint: Denote the left hand side by $f(n)$ and prove that $f(n+1)=2f(n)$._...
Showing the hint is just a straightforward calculation. If $f(n) = \sum \binom{n+k}{k} \frac{1}{2^k}$, then \begin{align*} f(n+1) &= \sum_{k=0}^{n+1} \binom{n+k+1}{k} \frac{1}{2^k} \\\ &= \sum_{k=1}^{n+1} \binom{n+k}{k-1} \frac{1}{2^k} + \sum_{k=0}^{n+1} \binom{n+k}{k} \frac{1}{2^k} \\\ &= \frac{1}{...
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Express dissolved multi-summand binominal equation After dissolving an multi summand binomial equation $$f(x)=(x_0+x_1+x_2+x_3+\ ...)^2$$ I got $n*(n-1)/2$ of these terms: $$x_0x_1+x_0x_2+x_0x_3+x_1x_2+x_1x_3+x_2x_3+...
I suppose that what you are looking for is $$f=\sum _{i=1}^n x_i^2+2 \sum _{i=1}^n x_i \sum _{j=i+1}^n x_j$$
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coefficient of $x^9$ in binominal expansion Find the co efficient of $x^9$ in the expansion of $$\left[\frac{(1+x)^3}{(1-x)^3}\right]^\frac{1}{2}$$ I get answer as $\frac{3}{128}$ But in my book the answer is given ...
You could notice that $$\left(\frac{(1+x)^3}{(1-x)^3}\right)^\frac{1}{2}=\left(\frac{1+x}{1-x}\right)^\frac{3}{2}=\left(1+2\sum_{i=1}^\infty x^i\right)^\frac{3}{2}$$ Now, being patient, you could use the generalized binomial theorem. If you are not, Taylor series (tedious too) would lead to $$\left(...
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Prove that an expression is divisible by a polynomial Question: _Show that the polynomial $f(x)=(x+1)^{2n} +(x+2)^n - 1$ is divisible by $g(x) = x^2+3x+2$, where $n$ is an integer._ I have tried to use mathematical i...
Putting the zeros of $x^2+3x+2=0$ i.e., $-1,-2$ one by one, in $f(x)=(x+1)^{2n}+(x+2)^n-1$ we get, $f(-1)=(-1+1)^{2n}+(-1+2)^n-1=0$ and $f(-2)=(-2+1)^{2n}+(-2+2)^n-1=0$ (i)So, using Remainder Theorem, $(x+1)\mid f(x)$ and $(x+2)\mid f(x)\implies lcm(x+1,x+2)\mid f(x)$ (ii) Alternatively, $\frac{(x+1...
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