isomorph

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isomorph
isomorph (ˈaɪsəʊmɔːf) [mod. f. Gr. type *ἰσόµορϕ-ος of equal form, f. ἰσό-, iso- + µορϕή form: in mod.F. isomorphe.] 1. Chem. and Min. A substance or organism isomorphous with another.1864 Webster, Isomorph, a substance which has the same crystalline form with another. 1885 E. R. Lankester in Encycl... Oxford English Dictionary
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Isomorph
An isomorph is an organism that does not change in shape during growth. importance is the part that is involved in substrate uptake (e.g. the gut surface), which is typically a fixed fraction of the total surface area in an isomorph wikipedia.org
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isomorph
isomorph/ˈaɪsəmɔ:f; `aɪsəˌmɔrf/ nsubstance or organism with the same form or structure as another 同形体; 同晶体. 牛津英汉双解词典
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Isomorph (gene)
After Muller's classification was described as an isomorph, or gene mutation that expresses a nonsense point mutant with expression identical to the original wikipedia.org
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Isomorphism (disambiguation)
Isomorphism or isomorph may refer to: Isomorphism, in mathematics, logic, philosophy, and information theory, a mapping that preserves the structure (gene), a classification of gene mutation See also Isomorph Records, a British music label wikipedia.org
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Isomorph vector space such that X complete and Y not Let $X,Y$ be some isomorph vector spaces and let $X $ be a Banach space. If this isomorphism is isometric $Y $ is complete, too. Could someone provide an example su...
Let $X=C\bigl([0,1]\bigr)$ and $Y=\\{f\in F\mid f\text{ is differentiable and }f(0)=0\\}$. Then$$\begin{array}{rccc}D\colon&Y&\longrightarrow&X\\\&f&\mapsto&f'\end{array}$$is a vector space isomorphism. However, $X$ is complete, whereas $Y$ isn't (in both cases, with respect to the $\sup$-norm).
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Symmetric groups isomorph to dihedral groups. I've noticed, that $S_2 \cong D_1$ and $S_3 \cong D_3$. Is every symmetric group $S_n$ (no including $S_1$) isomorph to the dihedral group $D_{n!/2}$?
No. The symmetric groups $S_n$ for $n\geq 3$ have trivial centre, but the dihedral groups $D_{n!/2}$ all have a centre of order $2$, since $n!/2$ is even.
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Riesz Representation Theorem: isomorph Riesz' Representation Theorem states that every linear functional can be represented by a vector. This shows that the Dual can be ANTILINEARLY and norm preserving identified with...
Yes, it is possible. Fix a _total orthogonal system_ $(e_i)_i$ on your Hilbert space $H$ and consider the conjugation of coordinations: $$\varphi:\sum_i\alpha_ie_i \ \mapsto \ \sum_i\bar{\alpha_i}e_i$$ this is an antilinear isomorphism of $H$ to itself, so composing it with the Riesz presentation $H...
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Shape correction function
The shape correction function is a ratio of the surface area of a growing organism and that of an isomorph as function of the volume. The shape of the isomorph is taken to be equal to that of the organism for a given reference volume, so for that particular volume the surface areas are wikipedia.org
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Non-Isomorph trees of a graph Please consider this graph !enter image description here How many non-Isomorph trees with 4 vertex has this graph? Is there any formula that show number of non-Isomorph trees with $n$ ...
There are only two 4-node trees up to isomorphism: the straight line and the Y-shaped graph. The above graph contains both of them. For the number of trees with $n$ vertices, Wikipedia links here : < .
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$\mathbb C$ isomorphism to $\mathbb{R} \times \mathbb{R}$ under multiplication How can I show, that $(\mathbb C,\cdot)$ is not isomorphic to $(\mathbb{R},\cdot) \times (\mathbb{R},\cdot)$ under multiplication? I tri...
To show that $(\mathbb{C}, \cdot)$ and $(\mathbb{R}, \cdot)^2$ are not isomorphic as semigroups, observe that both semigroups have a (necessarily unique) annihilator, an element $a$ such that $ax = a$ for any $x$. Both semigroups have a unique element $e$ such that $ex = x$ for any $x \not= a$ (in f...
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Prove that all cyclic groups of the same order are isomorphic I want to prove that 2 cyclic groups of the same order are always isomorph. So, let $G,H$ be cyclic groups with $|G| = |H|$. Then: I suspect the isomorphis...
Split into two cases, based on whether the groups are finite. For functions between finite sets of the same size (especially homomorphisms between finite groups of the same size), surjectivity implies bijectivity. For the infinite case, it should be easier to show that the kernel is trivial.
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Draw all non-isomorph graphs with 7 knots in our course we had to draw all non-isomorph graphs with 7 knots without loops. Since this is hard to explain I'll add an Image. !My notes The blue written part is accepted...
The intuition is that two graphs are isomorphic if one can redraw one to get the other. In your case, your left-most graph is isomorphic to graph $1$, since you could "straighten" that horizontal edge to a vertical edge to get graph $1$. Similarly, your middle graph is isomorphic to graph $3$ and yo...
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Why isn't the composition equal to the product? Let $G_1, G_2, H$ be three isomorph graphs. We are given the permutation $\phi:G_1\rightarrow G_2$ with $\phi=(1,2,4,3)$ and the permutation $\psi:G_2\rightarrow H$ with...
There are two different conventions: the product of permutations can be defined either as composition or as composition in the reverse order. With the second convention, $(1\ 2)\cdot (1\ 2\ 4\ 3)$ means the permuation that is obtained by first doing $(1\ 2)$ and then doing $(1\ 2\ 4\ 3)$ (note that ...
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