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# Constructions unique up to non-unique isomorphism

1) Fields have algebraic closures unique up to a non-unique isomorphism.

2) Nice spaces (without base point) have universal covering spaces unique up to a non-unique isomorphism.

3) Modules have injective hulls unique up to a non-unique isomorphism.

Such situations can lead to interesting groups - the absolute Galois group, the fundamental group, and the "Galois" groups of modules introduced by Sylvia Wiegand in Can. J. Math., Vol. XXIV, No. 4, 1972, pp. 573-579.

I'd appreciate any insight into the abstract features of situations which give rise to this type of phenomenon. And I'd appreciate as many examples from as many parts of mathematics as possible.

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It is possible to functorially define the universal covering space of a non-pointed space X: take the diagram $\Pi_1(X) \to Top$ sending a point $x\in X$ to the (pretty well unique) covering space one gets from the pointed space $(X,x)$, and a homotopy class of paths to the map between covering spaces this induces. The colimit of this diagram is the universal covering space of $X$, and this time this is functorial wrt maps $X\to Y$. I learned this from Todd Trimble, I think. –  David Roberts Jan 30 '11 at 9:59
@ David: "the colimit of this diagram is the universal covering space of X", are you sure this is correct? What you described is essentially the groupoid of universal covering spaces and covering space maps inside Top and you are taking the colimit inside Top. This is the same as taking a single covering space and taking the quotient by all covering automorphisms, so you just get X back. –  Chris Schommer-Pries Jan 30 '11 at 14:46
Doesn't this always happen when the construction has not-trivial automorphisms? –  Nick S Jan 30 '11 at 18:36
@ David - this coequalizer still just gives B back. Try an explicit example like B = circle. The point is that there are many paths and so the different universal covers are identified with eachother in more than one way. This forces you to take a quotient of the universal covers which is too small (namely B itself) and no longer a universal cover. You can see that this has to be the case because the paths from b to b (up to homotopy) are just $pi_1$ and so this colimit factors through the quotient by the action of $pi_1$. Agreed? Where in the n-lab is this written? –  Chris Schommer-Pries Jan 31 '11 at 18:57
There cannot be any such construction. The group of homeomorphisms from the circle to itself has no compatible action on the (or should I say "a") universal covering space. –  Tom Goodwillie May 17 '11 at 4:10

The first two examples can be described more or less uniformly. Associated to a field $F$ is the category $C_F$ of algebraic field extensions of $F$ (whose objects are morphisms $F \to E$ and whose morphisms are commutative triangles). This category has a weak terminal object given by any algebraic closure $F \to \bar{F}$. The full subcategory on the algebraic closures is what one might call the absolute Galois groupoid of $F$ (which is a perfectly canonical construction), and choosing an object in this groupoid (which is not) gives the absolute Galois group.

Similarly, associated to a nice space $X$ is the category $C_X$ of connected covers of $X$ (whose objects are covering maps $Y \to X$ and whose morphisms are commutative triangles). This category has a weak initial object given by any universal cover $\bar{X} \to X$. The full subcategory on the universal covers is (equivalent to?) the fundamental groupoid of $X$ (again, a perfectly canonical construction), and choosing an object in this groupoid (which is not) gives the fundamental group.

So you will get this kind of behavior in any situation where you have a weak universal object instead of a universal one. (This partially covers the third example, since injectivity is also a weak universal property.) A general way to engineer a situation similar to the above two might be to look at something like the category of (epi?)morphisms into an object or (mono?)morphisms out of it in your favorite category and see what happens.

In any case, if you are only interested in these constructions because they produce interesting groups, then I think nowadays the modern thing to do is to produce interesting groups using Tannaka-Krein duality.

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Nice observation! In fact, the Galois group of modules alluded to in example 3) is also of this type: it's the automorphism group of the weak initial object given by the injective hull inside the category of embeddings (monomorphisms) $M \to I$ into an injective, so it's analogous to example 1) and dual to 2). –  Theo Buehler Jan 30 '11 at 14:18
Not to mention it's also the weak terminal object in the category of essential extensions of M (with embeddings for the maps). –  Harry Altman Jan 30 '11 at 14:27
"You will get this kind of behavior in any situation where you have a weak universal object instead of a universal one" --- this is so only when the weak universal objects are all isomorphic, no? –  Steven Landsburg Jan 30 '11 at 15:00
@Steven: I guess that would give a groupoid which is not connected, but it's still possible to restrict attention to each of its connected components. –  Qiaochu Yuan Jan 30 '11 at 15:17
I don't understand your last comment. It seems to me that this is due to the fact that after all I'm not so sure what the good definition of "weak initial/terminal" object is, anyway. –  Theo Buehler Jan 30 '11 at 15:58

Any two injective resolutions (of an object in an abelian category) are homotopy equivalent, but this homotopy equivalence is not unique. This is of course because the lifting property in the definition of "injective" does not require any uniqueness.

The connected sum of oriented manifolds is unique up to homeomorphism, but this homeomorphism is not unique.

A bit silly, but: In a short exact sequence $0 \to A \to B \to C \to 0$ in a semisimple abelian category $B$ is unique up to isomorphism (namely, $B \cong A \oplus C$), but the isomorphism is not unique.

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But the equivalence of injective resolutions is unique in the correct higher sense, i.e. the space of choices is contractible. –  Thomas Nikolaus May 17 '11 at 7:31
@Thomas, is that indeed true? I've never thought of it but it looks like an interesting question. What do you mean by 'space of choices' in this case? –  Fernando Muro Feb 10 '12 at 12:22

For a field $k$ and a natural number $n$, the vector space of dimension $n$ over $k$ is unique up to a non-unique isomorphism, though this somehow feels "less unique" to me than your other examples. I thought at first that this might be due to its not fitting into the class of examples described by Qiaochu, but I suppose you can force it into that class by considering the category of $n$-dimensional vector spaces over $k$. But that in turn feels considerably more ad hoc (at least to me) than considering the category of algebraic field extensions.

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Except when the field is $\mathbf{F}_2$ and the dimensions is $1$:) –  Chandan Singh Dalawat Jan 30 '11 at 15:52
Or any field and the dimension is 0 ;-)) –  Johannes Hahn Jan 31 '11 at 0:15
I attended a talk about some problem in the classification of—well, of something to do with finite fields, I don't remember—at which a colleague was moved to ask “What happens here if $\lambda \ne \mu$?” “Oh, yes,” said the speaker, as if discounting a trivial special case, “the results don't work if there are more than 2 non-$0$ scalars in the field.” –  L Spice May 17 '11 at 5:03

My favourite: the mapping cone of a morphism in a triangulated category is unique up to non-unique isomorphism. This fact has originated a lot of research in this topic, and it still does.

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In recent work in set theory the concept of "canonical structure" has emerged, in connection with combinatorial work on pcf theory. The idea is that there are many constructions that depend on the axiom of choice but, once realized, are actually independent of the specific choices made. Usually, this involves two steps: You construct an object, which is not quite canonical (say, a collection of subsets of a cardinal $\kappa$), but then you recognize that there is a natural ideal (say, the non-stationary ideal on $\kappa$) and the corresponding equivalence classes are canonical. Of course, by switching to a new model of set theory, the "canonical structure" may change, so sometimes one thinks of it as a sort of invariant of the models.

The first papers that explicitly mentioned the name "canonical structure" are by Cummings, Foreman, and Magidor, "Canonical structures in the universe of set theory", Parts I and II, Annals of Pure and Applied Logic 129 (2004), 211-243, and 142 (2006), 55-75.

The following quote is from the beginning of the introduction to Part I:

It is a distinguishing feature of modern set theory that many of the most interesting questions are not decided by ZFC, the theory in which we profess to work; to put it another way, ZFC admits a large variety of models. A natural response to this is to identify invariants which may take different values in different models, and which codify a large amount of information about a model.

Of particular interest are invariants which are canonical, in the sense that the Axiom of Choice is needed to show that they exist, but once shown to exist they are independent of the choices made. For example the uncountable regular cardinals are canonical in this sense.

Shelah discovered a large class of canonical invariants, the study of which he labeled PCF theory. These invariants include two which are central in this paper; Shelah [24, 26] (under some mild cardinal arithmetic assumptions on the singular cardinal $\mu$) defined two stationary subsets of $\mu^+$, the sets of good and approachable points. The definitions of these sets appear to depend on certain arbitrary choices, but (modulo the club filter) are in fact independent of these choices. Other canonical structures we study in this paper include the stationary sets of tight and internally approachable structures, and the collection of good points on a scale.

The two references cited in the quote are S. Shelah, "On successors of singular cardinals", in M. Boffa, D. van Dalen, and K. McAloon, editors, Logic Colloquium ’78, pages 357–380, Amsterdam, 1979. North-Holland; and S. Shelah, "Cardinal Arithmetic". Oxford University Press, Oxford, 1994.

Besides the ongoing work by Cummings-Foreman-Magidor and Shelah, these ideas have been extended by others; Krueger and Ishiu come to mind.

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Sounds like a case for 'the': ncatlab.org/nlab/show/the –  David Roberts Jan 30 '11 at 20:47
Ok, @David, I'm curious. What do you mean? –  Andres Caicedo Jan 31 '11 at 0:47
@Andres: What David is referring to is the fact that universal constructions in category theory have this property, and that this is only a new feature of set theory. For instance, suppose we want to make the product with an object $S$ a functor (i.e. the functor $(-)\times S$. However, while the product is unique up to unique isomorphism, we have to choose a representative of each isomorphism class as well as the connecting morphisms between them. This requires choice or global choice, but after we choose a specific representative of the functor usign choice, it is unique up to unique iso. –  Harry Gindi Jan 31 '11 at 9:16
So when I said above "the functor $(-)\times S$", I was abusing language, since any given construction of "the" functor is only "a functor $(-)\times S$". This might seem like we're being overly cautious, but when we move from isomorphism of objects to equivalence of objects (in a bicategory), this makes some difference. This can be resolved by using Mac Lane's coherence theorem for bicategories, but, when we move up to tricategories, such a coherence theorem is proven not to exist. –  Harry Gindi Jan 31 '11 at 9:21
I see. Thanks, @Harry! –  Andres Caicedo Jan 31 '11 at 14:11

The homology of a differential graded algebra has an $A_\infty$-algebra structure which is unique up to non-unique isomorphism.

See Keller's nice expository paper, for instance. In particular, he states this result in Section 3.3 (as a theorem due to Kadeishvili, among others). It is stated there as a result about the homology of an $A_\infty$-algebra, but any differential graded algebra may be viewed as an $A_\infty$-algebra.

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Hi John, Thanks, I wish you would say more people or give a reference at least. –  David Feldman Jan 30 '11 at 23:22
@David: I added a reference. –  John Palmieri Jan 30 '11 at 23:36

Here are some examples that are less of an algebraic nature (but all seem to be subsumed by Qiaochu's observation in that they are "weakly initial" or "weakly terminal" objects in appropriate categories):

Consider the categories of metric spaces or complete metric spaces and $1$-Lipschitz maps. Isbell has shown that in these categories there are injective hulls, unique up to non-unique isomorphism. A metric space $I$ is injective if for every isometric embedding $A \to B$ and every $1$-Lipschitz map $A \to I$ there exists a $1$-Lipschitz extension $B \to I$. The automorphism groups of the injective hull of a space seems exceedingly hard to determine (even for finite spaces) but there's one case I find interesting. If $M$ happens to be a (real) Banach space and $I(M)$ is its injective hull then $I(M)$ is a Banach space, uniquely determined up to unique linear isometry, and it is of the form $C(K)$ where $K$ is an extremally disconnected Hausdorff space. H. Elton Lacey and co-authors have given a complete (finite!) list of possible injective hulls of separable Banach spaces.

Closely related are projective covers in the category of compact Hausdorff spaces and continuous maps. There, the projectives are precisely the extremally disconnected spaces (Gleason).

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I should add that there is an abstract version of injective hulls/projective covers considered in Adamek-Rosicky, Locally presentable and accessible categories, that seems to subsume all the examples given so far (except maybe the example involving bases). –  Theo Buehler Jan 30 '11 at 14:48

A compact connected semisimple Lie Group $G$ has an essentially unique maximal torus $T$, a maximal abelian subgroup of maximum dimension (the rank of $G$ is the dimension of this torus). Although $G$ has lots of such torii (in fact any element of $G$ is contained in at least one), any two are conjugate to one another by some element of $G$.

In a similar vein, one can break a given maximal torus $T$ up into congruent pieces (the images of Weyl chambers under the exponential map applied to the Lie algebra of $T$), any two of which are equivalent to one another by an element of the Weyl Group of $G$. The value of any class function on $G$ is then completely determined on all of $G$ by its values on a single one of these pieces.

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The countable dense linear order is unique up to a non-unique isomorphism. The countable atomless Boolean algebra is unique up to a non-unique isomorphism. The random graph is unique up to a non-unique isomorphism. (Algebraically closed fields of a given characteristic and transcendence degree, and vector spaces of given dimension over a given field, are other examples already mentioned above.)

In general, if $T$ is a $\kappa$-categorical first-order theory (in a countable language), then the model $M$ of $T$ of cardinality $\kappa$ is unique up to a non-unique isomorphism.

Even more generally: if $T$ is any complete theory and $\kappa$ an infinite cardinal, then the saturated model $M$ of $T$ of cardinality $\kappa$—if it exists at all—is unique up to a non-unique isomorphism. ($M$ is unique by a standard back-and-forth argument. Non-uniqueness of the isomorphism amounts to saying that $\operatorname{Aut}(M)$ is nontrivial. By homogeneity, it suffices to exhibit two elements of $M$ with the same type. If there exists a nonprincipal parameter-free $1$-type, we can easily find two elements that realize it. If all $1$-types are principal, there are only finitely many, hence two elements of $M$ have to realize the same type by the pigeonhole principle.)

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Let me mention Sullivan's minimal models.

Every commutative differential graded $\mathbb{Q}$-algebra (cdga) $A^*$ concentrated in non-negative degrees and such that $H^0(A^*)=\mathbb{Q}$ admits a minimal Sullivan model $i:M^*\to A^*$ where $M^*$ is a free commutative graded algebra obtained from $\mathbb{Q}$ by adding generators of non-negative degrees so that the differential of each generator is a $\mathbb{Q}$-linear combination of products of length $\geq 2$ of the previous generators, and $i$ is a map of cgda's that induces a cohomology isomorphism (i.e., a quasi-isomorphism).

The minimal model is unique up to a non-unique isomorphism. More generally, if $f:A^*\to B^*$ is a map of cdga's and $j:N^*\to B^*$ is a minimal model of $B^*$, then there is a cdga map $g:M^*\to N^*$, defined up to cdga homotopy, such that $fi=gj$ up to cdga homotopy; moreover, if $f$ is a quasi-isomorphism, then $g$ is an isomorphism.

This reduces the classification of non-negative cdga's up to quasi-isomorphism (and as a consequence, the classification of simply connected topological spaces up to rational homotopy) to the classification of algebras of a certain kind up to isomorphism.

Of course, this example is similar to some mentioned before (in a sense it is the commutative analog of the answer of John Palmieri).

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Vector spaces have a basis that is unique up to a non-unique isomorphism.
Hilbert spaces have an orthonormal basis that is unique up to a non-unique unitary.
(At least, if you accept Zorn's lemma, i.e. the axiom of choice)

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Wait, how do vector spaces have a unique basis? If $V = \langle x, y\rangle$, then $x + y, y$ is also a basis of $V$, but it is not the same basis. –  Simon Rose Jan 30 '11 at 16:04
@Simon: but there is an linear isomorphism that maps the former basis to the latter. So indeed a basis is not unique, but unique up to an isomorphism. Finally, this isomorphism itself is not unique, as it can for example be composed with any isomorphism that leaves the latter basis invariant. –  Chris Heunen Jan 30 '11 at 16:14
@Chris: We might be splitting hairs, but I think that this misses the spirit of the question. There is some notion of canonicality which defines an algebraic closure, for example, but no such notion that defines a basis of a vector space. –  Simon Rose Jan 30 '11 at 16:23
@Simon: I agree this might only answer the letter of the question and perhaps not its spirit. Neverthelesss, OP asked for as many examples from as many parts of mathematics as possible, and this is one. Moreover, these isomorphisms certainly form interesting groups, namely $SL(n)$ and $U(n)$. Anyway, if one is after insight into the abstract features of such situations, isn't it important to also consider examples that fall outside one's initial intuition? –  Chris Heunen Jan 30 '11 at 17:29
I do think an example lurks here, but you haven't nailed it. You don't want a basis, you want only that for which you thought you wanted a basis, namely rigidification. So for one given field and a given dimension fix a model vector space $M$ of that dimension. Define a rigidification of any given vectors space (over the same field, with the same dimension) as in isomorphism $i:V\rightarrow M$. An isomorphism of rigidified vector spaces is a map $j:M\rightarrow M$ that makes the triangle commute. Now you get $GL(n)$ (and other groups if you force more structure on $M$.) –  David Feldman Jan 31 '11 at 5:00

One example that springs to mind is when you are secretly working with the objects of a higher category, and so the choice is not unique up to a unique isomorphism, but the choice of isomorphism is also subject to higher coherence data. In a 2-category this would mean the isomorphisms are unique up to a unique invertible 2-arrow and so on. In $\omega$-categories, you may have such coherence all the way to infinity, and so end up with no uniqueness after all. One place where this emerges is when your $\omega$-category has all duals - is then an $\omega$-groupoid, because all the duals make everything weakly invertible!

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# zbMATH — the first resource for mathematics

Accelerated iterative hard thresholding algorithm for $$l_0$$ regularized regression problem. (English) Zbl 1441.90121
Summary: In this paper, we propose an accelerated iterative hard thresholding algorithm for solving the $$l_0$$ regularized box constrained regression problem. We substantiate that there exists a threshold, if the extrapolation coefficients are chosen below this threshold, the proposed algorithm is equivalent to the accelerated proximal gradient algorithm for solving a corresponding constrained convex problem after finite iterations. Under some proper conditions, we get that the sequence generated by the proposed algorithm is convergent to a local minimizer of the $$l_0$$ regularized problem, which satisfies a desired lower bound. Moreover, when the data fitting function satisfies the error bound condition, we prove that both the iterate sequence and the corresponding sequence of objective function values are R-linearly convergent. Finally, we use several numerical experiments to verify our theoretical results.
##### MSC:
 90C25 Convex programming
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##### References:
 [1] Aubin, JP; Cellina, A., Differential Inclusions: Set-Valued Maps and Viability Theory (1984), Berlin: Springer, Berlin · Zbl 0538.34007 [2] Beck, A.; Teboulle, M., A fast iterative shrinkage-thresholding algorithm for linear inverse problems, SIAM J. Imaging Sci., 2, 1, 183-202 (2009) · Zbl 1175.94009 [3] Bian, W.; Chen, XJ, Optimality and complexity for constrained optimization problems with nonconvex regularization, Math. Oper. Res., 42, 4, 1063-1084 (2017) · Zbl 1386.90167 [4] Blumensath, T.; Davies, ME, Iterative thresholding for sparse approximations, J. Fourier Anal. Appl., 14, 5-6, 629-654 (2008) · Zbl 1175.94060 [5] Blumensath, T.; Davies, ME, Iterative hard thresholding for compressed sensing, Appl. Comput. Harmon. Anal., 27, 3, 265-274 (2009) · Zbl 1174.94008 [6] Blumensath, T.; Davies, ME, Normalized iterative hard thresholding: guaranteed stability and performance, IEEE J. Sel. Top. Signal Process., 4, 2, 298-309 (2010) [7] Candes, EJ; Romberg, J.; Tao, T., Robust uncertainty principles: exact signal reconstruction from highly incomplete frequency information, IEEE Trans. Inf. Theory, 52, 2, 489-509 (2006) · Zbl 1231.94017 [8] Candes, EJ; Tao, T., Decoding by linear programming, IEEE Trans. Inf. Theory, 51, 12, 4203-4215 (2005) · Zbl 1264.94121 [9] Candes, EJ; Wakin, MB; Boyd, SP, Enhancing sparsity by reweighted $$l_1$$ minimization, J. Fourier Anal. Appl., 14, 5-6, 877-905 (2008) · Zbl 1176.94014 [10] Chambolle, A., An algorithm for total variation minimization and applications, J. Math. Imaging Vis., 20, 1-2, 89-97 (2004) · Zbl 1366.94048 [11] Chambolle, A.; DeVore, RA; Lee, NY; Lucier, BJ, Nonlinear wavelet image processing: variational problems, compression, and noise removal through wavelet shrinkage, IEEE Trans. Image Process., 7, 3, 319-335 (1998) · Zbl 0993.94507 [12] Chartrand, R., Exact reconstruction of sparse signals via nonconvex minimization, IEEE Signal Process. Lett., 14, 10, 707-710 (2007) [13] Chartrand, R.; Staneva, V., Restricted isometry properties and nonconvex compressive sensing, Inverse Probl., 24, 3, 1-14 (2008) · Zbl 1143.94004 [14] Chen, S.; Cowan, CFN; Grant, PM, Orthogonal least-squares learning algorithm for radial basis function networks, IEEE Trans. Neural Netw., 2, 2, 302-309 (1991) [15] Chen, XJ; Ng, MK; Zhang, C., Non-Lipschitz $$l_p$$-regularization and box constrained model for image restoration, IEEE Trans. Image Process., 21, 12, 4709-4721 (2012) · Zbl 1373.94080 [16] Donoho, DL, Compressed sensing, IEEE Trans. Inf. Theory, 52, 4, 1289-1306 (2006) · Zbl 1288.94016 [17] Fan, JQ; Li, RZ, Variable selection via nonconcave penalized likelihood and its oracle properties, J. Am. Stat. Assoc., 96, 456, 1348-1360 (2001) · Zbl 1073.62547 [18] Figueiredo, MAT; Nowak, RD, An EM algorithm for wavelet-based image restoration, IEEE Trans. Image Process., 12, 8, 906-916 (2003) · Zbl 1279.94015 [19] Foucart, S.; Lai, MJ, Sparsest solutions of underdetermined linear systems via $$l_q$$-minimization for \(0
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# Kerala Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines

## Kerala State Syllabus 9th Standard Maths Solutions Chapter 6 Parallel Lines

### Kerala Syllabus 9th Standard Maths Parallel Lines Text Book Questions and Answers

Textbook Page No. 88

Parallel Lines Class 9 Kerala Syllabus Question 1.
Draw an 8 cm long line and divide it in the ratio 2 : 3.
Draw AB whose length is 8 cm. Draw AC with length 5 cm. Join BC. Let the point D divides AC in the ratio 2 : 3. Draw a line through D parallel to BC. This line divides AB in the ratio 2 : 3.

Kerala Syllabus 9th Standard Maths Chapter 6 Question 2.
Draw a rectangle of perimeter 15 cm and sides in the ratio 3 : 4
Perimeter is 15 cm.
So length + breadth = 7.5 cm
Draw AB, a line segment of length 7.5 cm.
Draw AC from A, 4 + 3 = 7cm long.
Mark D on this line such that AD = 4 cm. Draw BC.
Draw a line parallel to BC and passing through D, which meets AB at E. Draw rectangle AEGF whose length as AE and breadth as EB.

Class 9 Maths Parallel Lines Kerala Syllabus Question 3.
Draw rectangles specified below each of perimeter 10 cm.
i. Equilateral triangle.
ii. Sides in the ratio 3 : 4 : 5
iii. Sides in the ratio 2 : 3 : 4
i. Draw a line AB of length 10 cm.
Draw AC with length 9 cm. Join BC.
AD = DE = EC = 3
AF : FG : GB = 3 : 3 : 3 = 1 : 1 : 1
Divide it in the ratio 1 : 1 : 1 and draw a triangle.

Draw ΔFHG with FG as its side.

ii. Draw a line segment AB of length 10 cm. Draw AC = 12 cm, join BC. Divide it in the ratio 3 : 4 : 5.
AD : DE : EC = 3 : 4 : 5
AF : FG : GB = 3 : 4 : 5
Draw ΔGHB with sides AF, FG and GB.

iii. Draw a line segment AB of length 10 cm. Draw AC with length 9 cm, join BC. Divide it in the ratio 2 : 3 : 4.

AD : DE : EC = 2 : 3 : 4
AF : FG : GB = 2 : 3 : 4
Draw ΔGHB with sides AF, FG and GB.

Parallel Lines 9th Standard Kerala Syllabus Question 4.
In the picture below, the diagonals of the trapezium ABCD intersect at P. Prove that PA × PD = PB × PC.

Sides AB, CD are parallel.
Draw EF parallel to AB and passing through P.
Now lines AB, EF and DC are parallel.
These lines cut the lines AC and BD in the same ratio.
So, PA : PC = PB : PD
i.e., $$\frac {PA}{PC} = \frac {PB}{PD}$$
From this by cross multiplication we get PA × PD = PC × PB

Textbook Page No. 93

Parallel Lines Class 9 Kerala Syllabus Question 1.
In the picture, the perpendicular is drawn from the midpoint of the hypotenuse of a right triangle to the base. Calculate the length of the third side of the large right triangle and the lengths of all three sides of small right triangle.

AC = 10 cm
∴ AM = 5 cm

MN, CB are perpendicular to AB
MN, CB are parallel lines.
∴ MN divides AB and AC in the same ratio.
∴ AN = BN = 4cm
BC = $$\sqrt{10^{2} – 8^{2}} = \sqrt {100 – 64} = \sqrt {36}$$ = 6 cm
M is the midpoint of the AC. N is the midpoint of AB.
The length of the line joining the midpoints of two sides of a triangle is half the length of the third side. So MN is half of BC.
∴ MN = 3cm

Parallel Lines Chapter Class 9 Kerala Syllabus Question 2.
Draw a right triangle and the perpendicular from the midpoint of the hypotenuse to the base.

i. Prove that this perpendicular is half the perpendicular side of the large triangle.
ii. Prove that perpendicular bisects the bottom side of the larger triangle.
iii. Prove that in the large triangle the distances from the mid point of the hypotenuse to all the vertices are equal.
iv. Prove that the circumcentre of a right triangle is the mid point of its hypotenuse.
MN is perpendicular to AB.
MN and CB are parallel lines. MN divides AC and AB in the same ratio.

i. AM = $$\frac{1}{2}$$AC, AN = $$\frac{1}{2}$$AB
∴ MN = $$\frac{1}{2}$$CB

ii. We get two right triangles of same base and perpendicular. So their hypotenuses are also equal.
∴ MA = MC = MB
AN = NB

iii. ∠ANM = ∠BNM = 90°
MN = MN
ΔANM ≅ ΔBNM
AM = MB …….(1)
M is the midpoint of AC
AM = MC …….(2)
(1) = (2) ⇒ AM = MB = MC

iv. The points A, B and C are at same distance from M. So a circle can be drawn with M as centre and pass through there three points. So this circle is the circum circle of ΔABC.

9th Standard Maths Notes Kerala Syllabus Question 3.
In the parallelogram ABCD, the line drawn through a point P on AB, parallel to BC, meets AC at Q. The line through Q, parallel to AB meets AD at R.

Prove that $$\frac {AP}{PB} = \frac {AR}{RD}$$
In ΔABC, RQ the parallel line of BC, divides AB and BC in the same ratio When we consider the relation
$$\frac {AR}{RD} = \frac {AQ}{QC}, \frac {AP}{PB} = \frac {AQ}{AC} = \frac {AR}{RD}$$

$$\frac {AR}{RD} = \frac {AQ}{QC}$$
$$\frac {AP}{PB} = \frac {AQ}{AC} = \frac {AR}{RD}$$
∴ $$\frac {AP}{PB} = \frac {AR}{RD}$$

Hsslive Guru 9th Maths Kerala Syllabus Question 4.
In the picture below, two vertices of a parallelogram are joined to the mid points of two sides. Prove that these lines divide the diagonal in the picture into three equal parts.

In the parallelogram ABCD, AB = CD and AB parallel CD.
P is the mid point of AB and Q is the midpoint of DC.
Since P and Q are the midpoints of AB and CD respectively.
PB = ½AB
DQ = ½CD = ½AB
∴ PB = DQ
i.e., PB = DQ and PB parallel DQ.
∴ PBQD is a parallelogram.
Since the parallel lines PD and BQ cut AB into equal parts, they cut AY also into equal parts.
AX = XY
The parallel lines PD and BQ cut CD into equal parts. So they cut CX also into equal parts.
YC = XY
∴ AX = XY = YC

9th Standard Maths Kerala Syllabus Question 5.
Prove that the quadrilateral formed by joining the mid points of a quadrilateral is a parallelogram. What if the original quadrilateral is a rectangle? What if it is a rhombus?

P, Q, R and S are the midpoints of the sides of the quadrilateral ABCD. Draw the diagonals AC and BD.
The length of the line joining the mid-points of two sides of a trianglle is half the length of the third side.
In ΔABC, PQ = ½AC
∴ PQ = SR
In ΔABD, PS = ½BD
In ΔBCD, QR = ½BD
∴ PS = QR
In quadrilateral PQRS, both pairs of opposite sides are equal. So it is a parallelogram.
PQ = ½AC, RS = ½AC
∴ RS = PQ = ½AC
PS = ½BD, QR = ½BD
∴ PS = QR = ½BD = ½AC
∴ PQ = RS = PS = QR
All sides of the quadrilateral PQRS are equal. So it is a rhombus.
Since ABCD is a rhombus. AC perpendicular to BD.
PQ and SP are perpendicular, i.e., ∠SPQ = 90°
One angle of the parallelogram PQRS is 90° and therefore it is a rectangle.

### Kerala Syllabus 9th Standard Maths Parallel Lines Exam Oriented Text Book Questions and Answers

Class 9 Maths Chapter 6 Kerala Syllabus Question 1.
Draw ΔABC with AB = 5cm, BC = 4cm, AC = 6cm, ΔABC. In BC mark P such that BP = 1cm, PC = 3cm. What is the relation between the areas of ΔABP and ΔAPC?
Area of ΔAPC is 3 times the area of ΔABP.

If two triangles have the common vertex and their bases are along the same straight line, the ratio between their areas is equal to the ratio between the lengths of their bases.
⇒ $$\frac{Area of ΔABP}{Area of ΔAPC} = \frac{BP}{PC} = \frac{1}{3}$$

Chapter 6 Maths Class 9 Kerala Syllabus Question 2.
In ΔABC, the line parallel to BC meet AB and AC at D and E respectively. AE = 4.5cm, $$\frac{AD}{DB} = \frac{2}{5}$$ = then find EC.

$$\frac{AD}{DB} = \frac{AE}{EC}$$
$$\frac{2}{5} = \frac{4.5}{EC}$$ = EC = $$\frac{4.5 \times 5}{2} = \frac{22.5}{2}$$
= 11.25 cm

Hsslive Guru Maths 9th Kerala Syllabus Question 3.
In ΔPQR, PQ = 9cm, QR = 14 cm, PR = 12 cm The bisector of ∠PRS meets QR in S. Find QS and RS.

$$\frac{PQ}{PR} = \frac{QS}{SR}$$;
$$\frac{18}{24} = \frac{6}{RS}$$
= RS = $$\frac{24 \times 6}{18}$$ = 8 cm

Hsslive Guru 9 Maths Kerala Syllabus Question 4.
In ΔABC, the point P is on AB such that the length of AP is double the length of PB. The line through P parallel to BC meets AC at Q and the length of AQ is 1 cm more than that of QC, what is the length of AC?

AP = 2 × PB
AP : PB = 2 : 1
$$\frac{AP}{PB} = \frac{AQ}{QC} = \frac{2}{1}$$ ⇒
AQ = 2QC
Consider QC = x cm
x + 1 = 2x ⇒ x = 1 ⇒ QC = 1
AQ = 2 × 1 = 2
AC = AQ + QC = 2 + 1 = 3 cm

Hss Live Guru Maths Kerala Syllabus Question 5.
In ΔABC, a line parallel to BC cuts a B and AC at X and Y

(a) If AB = 3.6cm, AC = 2.4cm, AX = 2.1 cm, what is the length of AY?
(b) If AB = 2cm, AC = 1.5cm, AY = 0.9cm
what is the length of BX?
a. $$\frac{AB}{AX} = \frac{AC}{AY} ⇒ \frac{3.6}{2.1} = \frac{2.4}{AY}$$;
⇒ 3.6 × AY = 2.4 × 2.1 ⇒ AY = 1.4 cm

b. Let length of AX = xcm
$$\frac{AX}{AB} = \frac{AY}{AC} ⇒ \frac{x}{2} = \frac{0.9}{1.5}$$;
1.5 × x = 1.8 ⇒ x = 1.2;
BX = AB – AX; BX = 2 – 1.2 = 0.8 cm

Hss Live Guru Maths 9 Kerala Syllabus Question 6.
In the figure, ABCD is a trapezium, and AB || CD. P is the midpoint of AD. If PQ || AB, then prove that Q is the midpoint of BC.

Given PQ || AB,
therefore
PQ || DC also. Line segments AD and BC are dividing 3 parallel lines EC, PQ and AB. Given P is the midpoint of AD
⇒ DP = PA ∴ $$\frac{DP}{PA}$$ = 1, ∴ $$\frac{CQ}{QB}$$ = 1
∴ CQ = QB ∴ Q is the midpoint of BC.

Ch 6 Maths Class 9 Kerala Syllabus Question 7.
In the figure AB || EH || DC, Answer the following questions given below.

a. If DE = 3cm, EA = 4cm and DG = 5cm, find GB?
b. If BH = 6cm,CB = 16cm and DB = 20cm, find DG?
c. If DA = 36cm, EA = 22cm and DG = 15cm find DB?
d. If DE = 20cm, EA = 25cm and HB = 22cm, find CH?
a. In ΔDAB, AB || EG
∴ $$\frac{DE}{EA} = \frac{DG}{GB}; \frac{3}{4} = \frac{5}{GB} \Rightarrow GB = \frac{20}{3} = 6 \frac{2}{3}$$cm

b. In ΔBDC, DC || GH
∴ $$\frac{BH}{BC} = \frac{BG}{BD}; \frac{6}{16} = \frac{BG}{20}$$;
BG = $$\frac{120}{16} = 7\frac{1}{2}$$cm
DG = BD – BG = 20 – $$7\frac{1}{2}$$ = $$12\frac{1}{2}$$cm

c. In ΔABD, AB || EG
$$\frac{DA}{EA} = \frac{DB}{BG}; \frac{36}{22} = \frac{DB}{BG}$$;
$$\frac{36}{22} \frac{DB}{DB – DG}$$
⇒ 36D(DB – DG) = DB × 22
14 DB = 540 DB = $$\frac{540}{14}$$ = 38$$\frac{4}{7}$$cm

d. AB, EH and DC are 3 parallel lines.
$$\frac{DE}{EA} = \frac{CH}{HB}; \frac{20}{25} = \frac{CH}{22}$$ ⇒ CH = $$\frac{88}{5}$$ = 17$$\frac{3}{5}$$cm

Hss Live Guru Class 9 Maths Kerala Syllabus Question 8.
In ABC, a line parallel to BC cuts AB and AC at P and Q. Show that $$\frac{AP}{AB} = \frac{AQ}{AC}$$

$$\frac{AB}{PB} = \frac{AQ}{QC}$$ (BC || PQ)
$$\frac{AP}{PB}+1=\frac{AQ}{QC}+1; \frac{AP+PB}{PB} = \frac{AQ + QC}{QC}$$
$$\frac{AB}{PB} = \frac{AQ}{QC}$$;
∴ $$\frac{AP}{AB} = \frac{AQ}{AC}$$

Question 9.
In the figure below, ABC is a right angled triangle. AB = 10cm, AC = 6cm, BC = 8cm, The midpoint of AB is M. Compute the lengths of the sides of ΔMBN

(MN || AC)
∴ $$\frac{AM}{MB} = \frac{NC}{NB}$$
1 = $$\frac{NC}{NB}$$ ∴ CN = NB = 4
∴ MB = 5, NB = 4, MN = 3

Question 10.
In the figure ABC is a right angled triangle and M is the midpoint of AB. Prove that MN = V2AC and also prove that MN = MA = MB.
MN = $$\sqrt{x^{2} – y^{2}}$$
AC = $$\sqrt{4x^{2} – 4y^{2}} = 2\sqrt {x^{2} – y^{2}}$$
AC = 2 × MN or MN = $$\frac{1}{2}$$AC
MC = $$\sqrt{x^{2} – 4^{2} + 4^{2}}$$; MC = X


text:
## Solution – Find the values of a and b that make the function differentiable everywhere.

Find all values of $a$ and $b$ that make the following function differentiable for all values of $x$.

$$f(x) = \begin{cases} bx^2-3 & \mbox{if } x\leq -1 \\ ax+b & \mbox{if } x>-1 \end{cases}$$

When trying to solve a problem like this, there are actually two things you will need to consider for our function $f(x)$.  Obviously, we need to make sure that it’s differentiable everywhere, but this actually implies something else that we will want to consider as well.

Since a function being differentiable implies that it is also continuous, we also want to show that it is continuous.  The reason for this is that any function that is not continuous everywhere cannot be differentiable everywhere.  Once we make sure it’s continuous, then we can worry about whether it’s also differentiable.

## Making sure f(x) is continuous everywhere

I’m not going to go into quite as much detail to show the part about making sure the function is continuous because I have already done this, which you can see by clicking here.

To make sure $f(x)$ is continuous at $x=-1$ we need to make sure that $$\lim_{x \to -1} f(x) = f(-1).$$  Since we have a piecewise function, we will need to consider each one-sided limit, but in this case only the right sided limit will tell us something useful.

$$\lim_{x \to -1^{+}} f(x) = f(-1)$$

$$\lim_{x \to -1^{+}} ax+b = b(-1)^2-3$$

$$a(-1)+b=b-3$$

$$-a+b=b-3$$

$$-a=-3$$

$$a=3$$

So now we know that $f(x)$ will be continuous everywhere as long as $a=3$.  However, this doesn’t really tell us that $f(x)$ is differentiable everywhere as well.

## Making sure f(x) is differentiable everywhere

We now know that we will need to let $a=3$ in order for this function to be continuous and to have a chance of being differentiable.  As a result, we can say that we are now trying to make this function differentiable everywhere:

$$f(x) = \begin{cases} bx^2-3 & \mbox{if } x\leq -1 \\ 3x+b & \mbox{if } x>-1 \end{cases}$$

We can see that the only place this function would possibly not be differentiable would be at $x=-1$.  The reason for this is that each function that makes up this piecewise function is a polynomial and is therefore continuous and differentiable on its entire domain.  The only place we may have a problem is when we have to switch between the two functions.

#### What does it mean for a function to be differentiable?

It means that its derivative exists for all values of $x$.  In other words, we need to be able to find its derivative no matter what $x$ is.

However, as I mentioned above, in this case we really only need to make sure that we can find the derivative of $f(x)$ when $x=-1$ since we know it would exist for all other values of $x$.  By using the definition of a derivative, we need to make sure the following limit exists at $x=-1$.

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Since we need to check this when $x=-1$, we can plug in $-1$ for $x$.  Therefore, we need to make sure this limit exists:

$$\lim_{h \to 0} \frac{f(-1+h)-f(-1)}{h}$$

I went over this limit definition in greater detail previously.  If you want a refresher on where this is coming from you can find that by clicking here.

Just like when we had to find the limit to make sure that $f(x)$ was continuous, we will need to consider each one sided limit separately in order to find this limit.  And also like when we checked for continuity, each one sided limit is going to require the use of a different section of our piecewise function.

#### Setting up the limits

When $h$ is slightly less than $0$, and we are considering the left sided limit, $f(-1+h)$ would need to be found using the $y=bx^2-3$ because this would involve inputting $x$ values which are less than $-1$.

By the same reasoning, when $h$ is slightly greater than $0$, and we are considering the right sided limit, $f(-1+h)$ would need to be found using the $y=3x+b$ because this would involve inputting $x$ values which are greater than $-1$.  Therefore, we need to consider the following one sided limits:

$$\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)^{2}-3\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$\lim_{h \to 0^{+}} \frac{\Big[3(-1+h)+b\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

#### Now what do we do with these limits?

Now remember, as I discussed in the lesson about one-sided limits, in order for a limit to exist we need both of its one-sided limits to exist and they need to be equal.  Therefore, in order to show that the derivative of $f(x)$ exists at $x=-1$, these two one-sided limits need to be equal to each other.  Before setting them equal to each other, first we’ll simplify them a bit.  First the left side limit.

$$\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)^{2}-3\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)(-1+h)-3\Big]-\Big[b(1)-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b(1-2h+h^2)-3\Big]-\Big[b-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b-2bh+bh^2-3\Big]-\Big[b-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{b-2bh+bh^2-3-b+3}{h}$$

$$=\lim_{h \to 0^{-}} \frac{bh^2-2bh}{h}$$

$$=\lim_{h \to 0^{-}} \frac{h(bh-2b)}{h}$$

$$=\lim_{h \to 0^{-}} bh-2b$$

$$=-2b$$

And now the right sided limit.

$$=\lim_{h \to 0^{+}} \frac{\Big[3(-1+h)+b\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$=\lim_{h \to 0^{+}} \frac{\Big[-3+3h+b\Big]-\Big[b(1)-3\Big]}{h}$$

$$=\lim_{h \to 0^{+}} \frac{-3+3h+b-b+3}{h}$$

$$=\lim_{h \to 0^{+}} \frac{3h}{h}$$

$$=\lim_{h \to 0^{+}} 3$$

$$=3$$

Now if we set these two simplified versions of the one-sided limits equal to each other, we get

$$-2b=3$$

$$b=-\frac{3}{2}$$

#### What does this tell us?

So now if we put both pieces together, we know that $a=3$ will ensure that $f(x)$ is continuous and then making $b=-\frac{3}{2}$ will also make sure $f(x)$ is differentiable at $x=-1$.  This would in turn make $f(x)$ differentiable for all values of $x$, or make it differentiable everywhere.

As always, I want to hear your questions!  Go check out my other lessons about derivatives and if you can’t get your question answered, I’d love to hear from you.  Leave a comment below or email me at jakesmathlessons@gmail.com.  If you have questions on this problem and solution or if you have another question you would like to see me answer, just ask it.  Or if you have an entire topic you would like to see me write a lesson about, just let me know.

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## Solution – Find the values of a and b that make f continuous everywhere

Consider the following piecewise function: $$f(x) = \begin{cases} \frac{x^2-4}{x-2} & \mbox{if } x<2 \\ ax^2-bx+3 & \mbox{if } 2\leq x<3 \\ 2x-a+b & \mbox{if } x\geq 3 \end{cases}$$

Problem credit: Chapter 2.4 #36 in Single Variable Calculus: Concepts and Contexts by James Stewart.

## How to find a and b for a piecewise function to be continuous everywhere.

When we see piecewise functions like this and our goal is to make sure it is continuous everywhere, there are two cases we want to consider separately:

1. making each individual piece continuous on its reduced domain,
2. and making the different pieces line up when our function switches between them.

First let’s consider each piece individually on the reduced domain where $f$ is actually defined by that function.

Piece 1: $y=\frac{x^2-4}{x-2}$ when $x<2$

When considering a function to find x values where the function is not continuous, two very common things to look for would be taking the square root of a negative number or dividing by zero.  Obviously we don’t need to worry about taking the square root of a negative number here since we don’t have any square roots.  However, we do have a fraction, and therefore we should consider if there is an x value that would cause us to divide by zero.

Looking at the denominator of our fraction, we see that we would divide by zero when we plug in some number for x that causes

$$x-2=0.$$

Clearly, this would happen when $x=2$.  And in fact, this is the only value for x where $y=\frac{x^2-4}{x-2}$ is not continuous.  However, remember that I said we want to see if there are any x values that cause the function not to be continuous within our limited domain for the specific piece of the function.  Since $x=2$ is NOT in the domain $x<2$, we can say that this piece of our function is continuous on its entire domain and won’t contribute any discontinuities to f.

Piece 2: $y=ax^2-bx+3$ when $2\leq x<3$

Looking at this piece of our piecewise function, clearly we need to consider our constants a and b.  Since our function f is a function of x (indicated by f(x)), we can consider the other letters in this piece of our function (a and b) to be constants.  I discussed this in a bit more detail here, but it basically means that a and b are some set number, they do not change.

The important thing to realize though, is that regardless of what values we choose for a and b, this piece of our function will just be a polynomial.  More specifically, it will be a quadratic function.  No matter what our a and b end up being we will have an $x^2$ term, an x term, and a constant term.  Since we know this piece of our function will be a polynomial no matter what our a and b are we can say that this piece of our function will be continuous for any a and b we select.

This is the case because polynomials are always continuous everywhere.  If this section of our function would be continuous everywhere, it would certainly be continuous on our limited domain of $2\leq x<3$.  Therefore, we don’t need to worry about our a and b causing any discontinuities here.

Piece 3: $y=2x-a+b$ when $x\geq 3$

This is actually going to be similar to piece 2 above.  We can see by looking at this function that no matter what a and b values we select here, they will both be constants which will combine to a single constant when we combine like terms.  Once we do this, we will be left with an x term and a constant term.  Therefore, this is also a polynomial (in this case it would also be a linear function), and would be continuous everywhere as a result.  Because of this, it must be continuous on our limited domain of $x\geq 3$.

Now we have shown that we will not get any discontinuities within any of our restricted domains that make up our piecewise function, but what about on their edges.  We also have to make sure we are selecting an a and a b value so that we can switch from one part of our function to the other without jumping up or down and leaving a hole in our function.  If we do not select the correct a and b we may get discontinuities at the x values where we switch from one function to the other.

In order to prevent this, we need to make sure that our first two pieces of f(x) transition smoothly when $x=2$ (this is the place we switch between these two functions) and we need to make sure that the second and third piece of our function have a smooth transition between them at $x=3$.

Continuity at $x=2$

Remember by definition, a function f(x) is continuous at $x=a$ if $$\lim_{x \to a} f(x) = f(a).$$

We need to apply this rule here to make sure that our function f is continuous at $x=2$. Therefore, we need to pick an a and a b so that $$\lim_{x \to 2} f(x) = f(2).$$

As I explained here, a limit will only exist if both of its one-sided limits exist and are equal to each other.  Therefore, rather than finding the above limit, we actually want to find these two one-sided limits:

$$\lim_{x \to 2^{-}} f(x) = f(2)$$

$$\lim_{x \to 2^{+}} f(x) = f(2).$$

Let’s start with the left-sided limit.  Since we are approaching $x=2$ from the left, we are looking at x values that are slightly less than 2.  Therefore, we need to use the piece of our function that is defined for $x<2$.  So the limit we really want to find, and the equation we want to solve is:

$$\lim_{x \to 2^{-}} \frac{x^2-4}{x-2} = f(2).$$

When attempting to solve this limit, we see that we can not just plug in 2 for x because this would cause us to divide by zero.  As a result of this, I would first suggest simplifying the function whose limit we are trying to find.  Using the difference of squares rule on the top of our fraction we can rewrite this limit as

$$\lim_{x \to 2^{-}} \frac{(x+2)(x-2)}{x-2} = f(2).$$

Once we do this, we can see that the top and bottom of our fraction contain a $(x-2)$ term so we can use them to cancel each other out.  This leaves us with

$$\lim_{x \to 2^{-}} (x+2) = f(2).$$

$y=x+2$ is a linear function and is continuous everywhere, so we can simply plug in $x=2$ to find this limit.

$$4=f(2)$$

Now we need to figure out the right side of the equation.  To find f(2), we just need to plug in $x=2$ into our function f(x), which is our original piecewise function.  The first thing we need to do is decide which piece defines our function when $x=2$.  We know that $f(x)=ax^2-bx+3$ when $2\leq x<3$.  We can use this part to find f(2) because this is the piece that defines f when $x=2$.  Plugging in 2 for x we get

$$4=ax^2-bx+3$$

$$4=a(2)^2-b(2)+3$$

$$4=4a-2b+3$$

$$1=4a-2b$$

Notice we have only one equation at this point which relates two unknown constants, making it impossible to find one unique solution for a and b.  To do this we will need another equation relating these two constants.

I would also like to point out that this equation came from using the equation involving only the left sided limit, and we still need to look at the right sided limit.

$$\lim_{x \to 2^{+}} f(x) = f(2)$$

If we look at this limit, we see that we would be approaching $x=2$ from the right side, which would mean we are looking at x values slightly larger than 2.  For x values slightly larger than 2, but infinitely close to 2, we would use the $y=ax^2-bx+3$ piece to define f.  We already found f(2) above, so putting these two facts together we see

$$\lim_{x \to 2^{+}} ax^2-bx+3 = a(2)^2-b(2)+3$$

Since the function whose limit we are trying to find is continuous everywhere (since it’s a polynomial) we can just plug in 2 for x to find this limit.

$$a(2)^2-b(2)+3=a(2)^2-b(2)+3$$

$$4a-2b+3=4a-2b+3$$

Notice both sides of this equation are the same.  Because of this, this equation will actually be true no matter what we put in for a and b.  This doesn’t really help us at all in this case, but it was important to test it out and see what it told us.  So we know that f will be continuous at $x=2$ as long as

$$1=4a-2b.$$

At this point we have found a set of a and b values that make f continuous at $x=2$.  But we need to find just one a and one b that will accomplish this for all x values.

But this does not tell us anything about whether it would also be continuous at $x=3$.  Checking that may give us another relationship between a and b that we can use to find the single unique solution for the two constants that will make f continuous everywhere.

Continuity at $x=3$

Making sure that f is continuous at $x=3$ will be an extremely similar process to what we just did around $x=2$.  Similar to above, we will need to make sure the following equations are true:

$$\lim_{x \to 3^{-}} f(x) = f(3)$$

$$\lim_{x \to 3^{+}} f(x) = f(3).$$

Let’s start with the right sided limit.  This means that we are getting closer and closer to $x=3$ and we are coming from x values that are slightly larger than 3.  Therefore, we will need to use the piece of our function that defines f for $x\geq 3$.

$$\lim_{x \to 3^{+}} 2x-a+b = f(3)$$

Since a and b are both constants, $y=2x-a+b$ is a linear function, and is continuous everywhere as a result.  Because of this, we can just plug 3 in for x to find this limit.

$$2(3)-a+b = f(3)$$

$$6-a+b = f(3)$$

To find f(3) we just need to plug 3 in for x into the piece of our function that defines it when $x=3$, which is the third piece of f.

$$6-a+b = 2(3)-a+b$$

$$6-a+b = 6-a+b$$

We see here another case where this equation gives us no useful information.  This is because this equality will be true no matter what we plug in for a and b.  Let’s move onto the left sided limit and see what we get there.

$$\lim_{x \to 3^{-}} f(x) = f(3)$$

Now that our x value is approaching 3 from the left side, it is slightly smaller than 3 and slowly increasing.  Since we are considering what our function is doing around $x=3$ for x values slightly smaller than 3, we will want to use the piece of f that is defined for $2\leq x<3$.  We already found f(3) above, so we will also plug that in.

$$\lim_{x \to 3^{-}} ax^2-bx+3 = 6-a+b$$

Just like before, since we know $y=ax^2-bx+3$ is a polynomial, it is continuous everywhere and we can find this limit by plugging in $x=3$.

$$a(3)^2-b(3)+3 = 6-a+b$$

$$9a-3b+3 = 6-a+b$$

$$10a-4b=3$$

Putting it all together

Now we have another relationship that relates a and b.  Alone, it isn’t very useful, but we can take the previous equation we found from ensuring f is continuous at $x=2$ and solve the following system of equations:

$$\mbox{(1): }4a-2b=1$$

$$\mbox{(2): }10a-4b=3.$$

Now that we have two equations that relate these two unknown constants, we have enough information to solve for them.  You always need at least as many equations as you have variables (or unknown constants in this case).  There are a couple different ways to solve a system of equations like this, but I will use substitution.  First we will solve for b in equation (1), then plug that into equation (2).

$$4a-2b=1$$

$$-2b=-4a+1$$

$$\mbox{(3): }b=2a-\frac{1}{2}$$

Now we will plug this into equation (2) and solve for a.

$$10a-4\Big(2a-\frac{1}{2}\Big)=3$$

$$10a-8a+2=3$$

$$2a+2=3$$

$$2a=1$$

$$a=\frac{1}{2}$$

Now we can plug this back into our equation for b, equation (3).

$$b=2\Big(\frac{1}{2}\Big)-\frac{1}{2}$$

$$b=1-\frac{1}{2}$$

$$b=\frac{1}{2}$$

Now we have shown that f(x) will be continuous at $x=2$ and at $x=3$ if $a=\frac{1}{2}$ and $b=\frac{1}{2}$.  We already showed that f is continuous for $x<2$, $2, and $x>3$ for all values of a and b.  Now we also know that f is continuous at $x=2$ and $x=3$ if $a=\frac{1}{2}$ and $b=\frac{1}{2}$.  Therefore, we can say that f would be continuous everywhere if $a=\frac{1}{2}$ and $b=\frac{1}{2}$.

In fact, we can see that f is continuous by plugging in the values we found for a and b and graphing it using Desmos.

This type of problem is one that you will likely run into a few times.  It actually comes in handy when you are trying to solve a similar type of problem that requires you to make a function differentiable everywhere.  If you would like to see this I have worked through one of these and you can see that by clicking this link: Solution – Find the values of a and b that make the function differentiable everywhere.

## SHORTCUT – More examples

Check out the other topics I’ve covered and the problems I’ve worked through.  You can see a list of related lessons on my limits page.

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text:
Math and Arithmetic

# What is the square root of 1481?

###### Wiki User

Rounded to two decimal places, sqrt(1481) = Â±38.48

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There are infinitely many of them. They include square root of (4.41) square root of (4.42) square root of (4.43) square root of (4.44) square root of (4.45) square root of (5.3) square root of (5.762) square root of (6) square root of (6.1) square root of (6.2)

### What is the square root of a square that has no root?

It's not a square if it has no root. If a number is a square then, by definition, it MUST have a square root. If it did not it would not be a square.

### What is the square root of 20 in radical form?

square root of 20 = square root of 4X5=square root of 4 X square root of 5=2*square root of 5.

### Square root 2 times square root 3 times square root 8?

square root 2 times square root 3 times square root 8

### What is the square root of 4x?

We use the property of square roots that says the square root of (ab)=square root (a) multiplied by square root of b So square root (4x)=square root (4) mutiplies by square root of x =2(square root (x)) 2sqrt(x)

### What is a principal square root?

The principal square root is the non-negative square root.

### What is the difference between principal square root and perfect square root?

A principal square root is any square root that's answer is positive, and a perfect square root is a square root that's answer is an integer.

### What is the square root of 20 in simplest form?

square root of 20 = square root of 4 * square root of 5. square root of 4 = 2, so your answer is 2 square root of 5.

### What is the square root of -24?

The square root of -24 is the same thing as the square root of -1, times the square root of 24. So, the square root of -1 is called 'i'. and the square root of 24 is equal to 2 times the square root of 6. Therefore, the square root of -24 is equal to: 2i* square root of 6.

### Simplify square root of 5 Times Square root of 6?

Square root of 5 time square root of 6 is the same as square root of 5 x 6, that is, square root of 30.

### What numbers are always irrational?

the square root of 3, the square root of 5, the square root of 6, the square root of 7, the square root of 8 etc

### What is the square root of -28?

the square root of -28 is the square root of -1 time the square root of 28. The square root of -1 is not a real number, it is represented by i. The square root of 28 is the square root of 4 times the square root of 7 (4*7=28) The square root 4 is equal to 2, and the square root of 7 cannot be simplified.The answer is 2iâˆš7.

### Square root of 5 and square root of 11?

Square root of 5 = ± 2.236068Square root of 11 = ± 3.316625

### Square root of 18 in radical form?

the square root of 18 can be simplified to 3 * the square root of 2.The perfect square 9 is a factor of 18. That means the square root of 9 * the square root of 2 = the square root of 18.the square root of 9 can be simplified to 3.That leaves 3 * the square root of 2.

### Can the square root of x be added to the square root of 3x?

No. The Square root of x is not the value of x. So it can not be simplified beyond: Root X + root 3x Yes. The square root of 3x equals the square root of 3 times the square root of x, so when you add another square root of x, you can factor out the square root of x, thereby simplifying the expression to the square root of x times the sum of one plus the square root of three.

### Square root of 2 times the square root of 46?

Square root of 2 times the square root of 46 = Â±9.59166305

### What does the square root of 27 plus the square root of 5?

The square root of 27 plus the square root of 5 = 7.4322204

### What is the square root of 348203489028490285409?

Square root of 348203489028490285409 is 1,394,284,239.13463 Square root of 348203489028490285409 is 1,394,284,239.13463 Square root of 348203489028490285409 is 1,394,284,239.13463

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# nLab plethory

A plethory is a monoid object in the monoidal category of birings. The most famous example of such a creature is $\Lambda$, the ring of symmetric polynomials. See:

• J. Borger, B. Wieland, Plethystic algebra, Advances in Mathematics 194 (2005), 246–283. (web)
Created on June 29, 2010 00:31:03 by John Baez (99.11.157.15)


text:
Future major features

Hello everybody

In this thread, I would like to discuss a new feature for Moodle - so called "Automatic updates deployment". This is a follow-up project on the recently introduced "Available updates notification" feature. In short, Moodle administrators should be able to download and deploy (install) updates purely via the web interface. We believe that this feature would encourage more admins to keep their sites up-to-date which is considered as the key protection against security attacks.

In the first phase, we plan to focus on updating and installing contributed extensions from the Plugins directory. Later, we plan to add a support for updating (and even installing!) Moodle core itself via this new tool.

The use cases and a rough implementation plan is described at Automatic updates deployment project page. I admit it's a bit late for significant changes in the spec for the first version of the tool (planned for 2.4 release). However, any idea, comment and feedback will be still useful as it can be considered in later phases of the development.

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From the wiki:

A standalone tool (script) that is able to fetch a ZIP package from a given URL, compare its MD5 content hash with a given value, unpack it, and deploy the content to a given location inside wwwroot.

Isn't there something like this for the lang downloads? It seems like that would be a similar funcitonality - maybe we can re-use other code or handle those two together?

I agree that the downloader/installer code needs to be separate code as handling includes will be very problematic.

We should probably learn a bit from other projects - have you seen drush?

cheers,
Tomek

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Yes, the language installer does a very similar job. Just note that the language installer has a lot of underlying Moodle environment ready to be used - starting with parameters handling from displaying the output. In this case, we must not include anything from Moodle. Thence direct re-using of any code is a bit tricky...

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I'm interested how hosted, very secure installations will cope with this.

We currently have to go through code check etc before thuings are installed, often quite detailed.  Can this be made secure enough for the providers to be happy with?

I think it is a great idea.  Regularly it is a small fix that needs to be sorted.  I'll head off to the docs to see if this new feature involves working for themes and course layouts as well.

Cheers

-Derek

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I am not sure I understand what you mean by "be made secure enough". Strict environments will actually want to have this feature locked to prevent accidental installation of a plugin that has not been reviewed by their own.

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OK, that's the answer to my question: it's not secure enough for strict environments.  So it's a speed thing for those who host things themselves, and a speed thing for hosted environments to do it for us.  But no advantage to give more power to users in a hosted enviroment.  OK, I can live with that.

-Derek

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In your plan, you indicate that the downloaded zip file's MD5 hash is compared with an "expected value".

• You don't indicate how the expected MD5 is obtained.  In order to have any security value, the expected MD5 must be obtained via a separate channel from the zip file -- otherwise, anyone who can tamper with the zip file can tamper with the MD5.  I would suggest using a public key digital signature scheme, instead of just a simple hash.  Alternatively, it can ensure that the zip file is downloaded via HTTPS (if you trust the SSL certificate authorities).
• MD5 is no longer considered secure.  Is there any reason for not using SHA-512?

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Good point Hubert. Yes, we are aware of this. We are planning to use HTTPS for obtaining the MD5 hash and maybe even for downloading the ZIP. I am also aware of MD5's weakness from the cryptography point of view. But in this case, it actually serves as a control checksum, not a cypher. Of course, hackers might be able to reconstruct the original contents of the ZIP file knowing the MD5 hash - but I don't think they would waste their time on it really And I am not aware of any simple way how to produce infected ZIP with malicious contents having the same MD5 hash and reasonable size. So, to paraphrase yourself - is there a reason for using SHA-512?

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DISCLAIMER: know nothing about cryptography, I just read Bruce Schneier. (In fact, is this even under the term 'cryptography'?)

So, to paraphrase yourself - is there a reason for using SHA-512?

Well, MD5 has had collision exploits, see: Microsoft Technet: Flame malware collision attack explained. So its not beyond the realms of possibility that such a zip could be created, (though I agree, this wouldn't keep me up at night).

These cryptography algorithms seemingly an arms race, SHA-512 is the best runner at the moment and widely available, so, why not?

(Lets not mention using two different algorithms to generate the signature to help protect against weakness in one).

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Why use SHA-512?  Because it is the currently generally-believed-to-be most secure hash algorithm.

IMHO, when designing a new protocol/system/etc., one of the goals of the design should be to build for the future as much as possible and as much as practical.  That means, for example, making things extensible, and avoiding having to change things in the future.  Even though there may not be any currently known attack for generating a malicious Zip file, the known weaknesses in MD5 indicate that it can't be trusted for much longer.  In fact, US-CERT says that you "should avoid using the MD5 algorithm in any capacity." (emphasis mine)

Also, regarding building for the future, even though you say that its intended  purpose is as a control checksum rather than a cipher, unless you are absolutely certain that nobody will think that it's a good idea to also have a cipher, if you're choosing to use a cryptographic hash function anyways, why not use a stronger one that can be used for both purposes?  You say that you don't think that hackers would waste their time trying to build a malicious Zip file, but with Moodle being one of the most popular VLEs, and with more companies and institutions investing in VLEs, and with things such as people using the same password for multiple sites, Moodle could very well be a tempting target.

IMHO, when someone says "hash", one's first thought, barring any other constraints or considerations, should be the current best hash algorithm, which is currently generally agreed to be SHA-512.  Some possible constraints or considerations are:

• hash size: in some cases (e.g., if many hashes are used and/or if the data is small), a large hash size may cause too great of an overhead.  But this isn't the case in this situation, since the hash size is many, many times smaller than the ZIP file.
• speed: one hash algorithm may be faster than another, and if many hashes are calculated, this may be an issue.  But in this case, a hash is only calculated occasionally, so this is not an issue.
• availability of implementation: in some cases, the current-best hash algorithm might not be readily available in the target system.  But in this case, PHP 5.3 already includes SHA-512 support, so this is not an issue.

Given that no other issues are, IMHO, relevant, the remaining distinguishing feature for hashes is the security, and in this regard, SHA-512 is the clear winner.  Even if you think that SHA-512 is overkill, at least use SHA-256, or even SHA-1.

Another thing that I would like to point out is that by using MD5, you may be encouraging others, who might not know better, to do the same.  "If it's good enough for Moodle, it's good enough for me."  We should be setting an example of migrating away from MD5 wherever possible, rather than encouraging its use.

As I see it, there is simply no reason to use MD5 here, especially since it is just a one-line change.  (Well, OK, two lines -- one line on the server, and one line on the client.)

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So, I think what we are saying is that the hash-checking system should be able to cope with any type of hash that PHP supports. In other words, when you download https://download.moodle.org/qtype_stack.2012092600.hash the the file contents will include both the name of the hash algorithm, and the value to check against.

Looking at http://php.net/manual/en/function.crypt.php is seems this could be achieved if the contents of the .hash file looked like

$6$rounds=5000$somehashstringXX$D4IrlXatmP7rx3P3InaxBeoomnAihCKRVQP22JZ6EY47Wc6BkroIuUUBOov1i.S5KPgErtP/EN5mcO.ChWQW21

which is the kind of thing PHP's crypt function needs to tell it to use SHA512. That gets us future-proof.

This is not really overengineering. Compared to the time to download and install a plugin, the time to compute the has of the zip using any algorithm is trivial.

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Yes, specifying the hash type is a good way to make things more flexible.  I think it would also be good to allow for multiple hash types (like what Debian did) to be downloaded, which would make it easier to transition from one hash type to another, if the need should ever arise -- if a weakness is found in hash algorithm foo, then you can provide both hash values: the new one for security, and the old one for code that is not yet able to support the new one.

I don't know if we'd want to use the crypt format, because I don't know if we want to be stuck with the predefined numbers that it uses.  We probably don't want to be using the crypt function for hashing, anyways, since that would require us to read the whole file into memory.  It would be better to use the hash family of functions, like hash_file http://php.net/manual/en/function.hash-file.php

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Oh yes, that is a much better function.

So, a sensible format for the .hash file would be something like

sha512|abc123...
md5|abc123...
...

where the algorithm name is anything that is a valid first arugment to hash_file. I am just noting the hash_algos method, to get a list of supported hash algorithms, since I had to look that up.

The implemenations could be to tests the first, or all, hashes from the .hash file where this PHP version supports that algorithm, with an error if none of the hashes use a supported algorithm.

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Yes, something like that would probably work well enough.  The implementation could make up its own mind about what hash algorithms to trust/use, possibly even allowing the administrator to configure this.

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I think some settings and checks for how it's handled would be very important and in no particular order:

- ignore this update (once, always) / view ignored updates

- check compatibility with moodle before updating plugins

- check compatibility with plugins before updating moodle

- try to activate plugins that aren't marked as compatible after core update (maybe it'll still work?)

- backup database before performing update

- set site in maintenance more while updating w/ custom message

- schedule auto-update time via cron

-----

on a deeper level: integrate a way to mark a plugin as working/not working with a specific version of moodle if it's not explicitly marked as compatible (similar to wordpress)

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Hello David

some suggestions and feature requests:

• option to hide/disable the feature completely by CFG seeting in config.php
• option to hide/disable the feature for installing new plugins

As Moodle Parnter we host hundreds of systems and keep the system up to date by our own services. So the support all the time knows which packages are installed for a client.  There are lots of systems with same code basis but seperate databases. If one of the admins updates the system its updated for all other systems.
In some environments several packages are not allowed, mainly due to data protection regulations. If admins who are not aware of the information and install additional packages it could be critical.

• Reset the system to former version.

During last version updates lots of APIs have changed. Connected systems often got problems. If we simplify the update process lots of system connections  to external systems will crash. So it should be possible to set the system after a quick test back to the former version.

I know there are a lot of admins who like such a feature in Moodle. I also saw a lot of wordpress users who crashed their systems by adding plugins.  We should implement it carefull and with some barriers that it is not so simple and could be disabled. Often teachers are admins and most of the teachers are collectors of features and gamers.  They really like to try new features but often they are not aware about the problems if plugins are not updated and supported continously. I think its different to host a personal blog or to admin a school/corporate leraning platform.

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Is this "Automatic updates deployment" feature actually meant to be working in Moodle 2.4? I am currently conducting tests and have come upon a problem with updating the hotpotatoes mod, described in this discussion.

Using Moodle 2.4beta+ (Build: 20121123)

Joseph

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Yes, it should work for most sites. See my post in the other thread you mentioned for details.

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I'm lost - where is the post you say gives more detail in the other thread that Joseph mentioned? When I click on Joseph's link above I get "coding error " in red etc.

EDIT: Response is here https://moodle.org/mod/forum/discuss.php?d=216584&parent=943037#p943408 but still don't understand the coding error -errro.

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@Mary,

Mea culpa. The link to that discussion I started in the HotPot module forum should have been: see this discussion.

Joseph

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Anyone:

Just tryng to get this sorted in one of the sites I work with, but getting errors, and will switch deugging on in the night when I think not too many people will be around.

Just curious: is this working as planned?  Are there people out there using this?

-Derek

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text:
location:  Publications → journals
Search results

Search: All articles in the CMB digital archive with keyword real C*-algebras

 Expand all        Collapse all Results 1 - 1 of 1

1. CMB 2011 (vol 54 pp. 593)

Boersema, Jeffrey L.; Ruiz, Efren
 Stability of Real $C^*$-Algebras We will give a characterization of stable real $C^*$-algebras analogous to the one given for complex $C^*$-algebras by Hjelmborg and RÃ¸rdam. Using this result, we will prove that any real $C^*$-algebra satisfying the corona factorization property is stable if and only if its complexification is stable. Real $C^*$-algebras satisfying the corona factorization property include AF-algebras and purely infinite $C^*$-algebras. We will also provide an example of a simple unstable $C^*$-algebra, the complexification of which is stable. Keywords:stability, real C*-algebrasCategory:46L05
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Paul's Online Notes
Home / Calculus II / Series & Sequences / Alternating Series Test
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### Section 4-8 : Alternating Series Test

The last two tests that we looked at for series convergence have required that all the terms in the series be positive. Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series.

The test that we are going to look into in this section will be a test for alternating series. An alternating series is any series, $$\sum {{a_n}}$$, for which the series terms can be written in one of the following two forms.

\begin{align*}{a_n} &= {\left( { - 1} \right)^n}{b_n}\hspace{0.5in}{b_n} \ge 0\\ {a_n} &= {\left( { - 1} \right)^{n + 1}}{b_n}\hspace{0.5in}{b_n} \ge 0\end{align*}

There are many other ways to deal with the alternating sign, but they can all be written as one of the two forms above. For instance,

\begin{align*}{\left( { - 1} \right)^{n + 2}} & = {\left( { - 1} \right)^n}{\left( { - 1} \right)^2} = {\left( { - 1} \right)^n}\\ {\left( { - 1} \right)^{n - 1}} & = {\left( { - 1} \right)^{n + 1}}{\left( { - 1} \right)^{ - 2}} = {\left( { - 1} \right)^{n + 1}}\end{align*}

There are of course many others, but they all follow the same basic pattern of reducing to one of the first two forms given. If you should happen to run into a different form than the first two, don’t worry about converting it to one of those forms, just be aware that it can be and so the test from this section can be used.

#### Alternating Series Test

Suppose that we have a series $$\sum {{a_n}}$$ and either $${a_n} = {\left( { - 1} \right)^n}{b_n}$$ or $${a_n} = {\left( { - 1} \right)^{n + 1}}{b_n}$$ where $${b_n} \ge 0$$ for all $$n$$. Then if,

1. $$\mathop {\lim }\limits_{n \to \infty } {b_n} = 0$$ and,

2. $$\left\{ {{b_n}} \right\}$$ is a decreasing sequence

the series $$\sum {{a_n}}$$ is convergent.

A proof of this test is at the end of the section.

There are a couple of things to note about this test. First, unlike the Integral Test and the Comparison/Limit Comparison Test, this test will only tell us when a series converges and not if a series will diverge.

Secondly, in the second condition all that we need to require is that the series terms, $${b_n}$$ will be eventually decreasing. It is possible for the first few terms of a series to increase and still have the test be valid. All that is required is that eventually we will have $${b_n} \ge {b_{n + 1}}$$ for all $$n$$ after some point.

To see why this is consider the following series,

$\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}{b_n}}$

Let’s suppose that for $$1 \le n \le N$$ $$\left\{ {{b_n}} \right\}$$ is not decreasing and that for $$n \ge N + 1$$ $$\left\{ {{b_n}} \right\}$$ is decreasing. The series can then be written as,

$\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}{b_n}} = \sum\limits_{n = 1}^N {{{\left( { - 1} \right)}^n}{b_n}} + \sum\limits_{n = N + 1}^\infty {{{\left( { - 1} \right)}^n}{b_n}}$

The first series is a finite sum (no matter how large $$N$$ is) of finite terms and so we can compute its value and it will be finite. The convergence of the series will depend solely on the convergence of the second (infinite) series. If the second series has a finite value then the sum of two finite values is also finite and so the original series will converge to a finite value. On the other hand, if the second series is divergent either because its value is infinite or it doesn’t have a value then adding a finite number onto this will not change that fact and so the original series will be divergent.

The point of all this is that we don’t need to require that the series terms be decreasing for all $$n$$. We only need to require that the series terms will eventually be decreasing since we can always strip out the first few terms that aren’t actually decreasing and look only at the terms that are actually decreasing.

Note that, in practice, we don’t actually strip out the terms that aren’t decreasing. All we do is check that eventually the series terms are decreasing and then apply the test.

Let’s work a couple of examples.

Example 1 Determine if the following series is convergent or divergent.

$\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{n}}$
Show Solution

First, identify the $${b_n}$$ for the test.

$\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{n}} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\frac{1}{n}} \hspace{0.5in}{b_n} = \frac{1}{n}$

Now, all that we need to do is run through the two conditions in the test.

$\mathop {\lim }\limits_{n \to \infty } {b_{n}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0$ ${b_n} = \frac{1}{n} > \frac{1}{{n + 1}} = {b_{n + 1}}$

Both conditions are met and so by the Alternating Series Test the series must converge.

The series from the previous example is sometimes called the Alternating Harmonic Series. Also, the $${\left( { - 1} \right)^{n + 1}}$$ could be $${\left( { - 1} \right)^n}$$ or any other form of alternating sign and we’d still call it an Alternating Harmonic Series.

In the previous example it was easy to see that the series terms decreased since increasing $$n$$ only increased the denominator for the term and hence made the term smaller. In general however, we will need to resort to Calculus I techniques to prove the series terms decrease. We’ll see an example of this in a bit.

Example 2 Determine if the following series is convergent or divergent.

$\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}{n^2}}}{{{n^2} + 5}}}$
Show Solution

First, identify the $${b_n}$$ for the test.

$\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}{n^2}}}{{{n^2} + 5}}} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{{n^2}}}{{{n^2} + 5}}} \hspace{0.5in} \Rightarrow \hspace{0.5in}{b_n} = \frac{{{n^2}}}{{{n^2} + 5}}$

Let’s check the conditions.

$\mathop {\lim }\limits_{n \to \infty } {b_{n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{{n^2} + 5}} = 1 \ne 0$

So, the first condition isn’t met and so there is no reason to check the second. Since this condition isn’t met we’ll need to use another test to check convergence. In these cases where the first condition isn’t met it is usually best to use the divergence test.

So, the divergence test requires us to compute the following limit.

$\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^n}{n^2}}}{{{n^2} + 5}}$

This limit can be somewhat tricky to evaluate. For a second let’s consider the following,

$\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^n}{n^2}}}{{{n^2} + 5}} = \left( {\mathop {\lim }\limits_{n \to \infty } {{\left( { - 1} \right)}^n}} \right)\left( {\mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{{n^2} + 5}}} \right)$

Splitting this limit like this can’t be done because this operation requires that both limits exist and while the second one does the first clearly does not. However, it does show us how we can at least convince ourselves that the overall limit does not exist (even if it won’t be a direct proof of that fact).

$\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^n}{n^2}}}{{{n^2} + 5}} = \mathop {\lim }\limits_{n \to \infty } \left[ {{{\left( { - 1} \right)}^n}\frac{{{n^2}}}{{{n^2} + 5}}} \right]$

Now, the second part of this clearly is going to 1 as $$n \to \infty$$ while the first part just alternates between 1 and -1. So, as $$n \to \infty$$ the terms are alternating between positive and negative values that are getting closer and closer to 1 and -1 respectively.

In order for limits to exist we know that the terms need to settle down to a single number and since these clearly don’t this limit doesn’t exist and so by the Divergence Test this series diverges.

Example 3 Determine if the following series is convergent or divergent.

$\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 3}}\sqrt n }}{{n + 4}}}$
Show Solution

Notice that in this case the exponent on the “-1” isn’t $$n$$ or $$n + 1$$. That won’t change how the test works however so we won’t worry about that. In this case we have,

${b_n} = \frac{{\sqrt n }}{{n + 4}}$

so let’s check the conditions.

The first is easy enough to check.

$\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt n }}{{n + 4}} = 0$

The second condition requires some work however. It is not immediately clear that these terms will decrease. Increasing $$n$$ to $$n + 1$$ will increase both the numerator and the denominator. Increasing the numerator says the term should also increase while increasing the denominator says that the term should decrease. Since it’s not clear which of these will win out we will need to resort to Calculus I techniques to show that the terms decrease.

$f\left( x \right) = \frac{{\sqrt x }}{{x + 4}}\hspace{0.5in}\hspace{0.25in}f'\left( x \right) = \frac{{4 - x}}{{2\sqrt x {{\left( {x + 4} \right)}^2}}}$

Now, there are two critical points for this function, $$x = 0$$, and $$x = 4$$. Note that $$x = - 4$$ is not a critical point because the function is not defined at $$x = - 4$$. The first is outside the bound of our series so we won’t need to worry about that one. Using the test points,

$f'\left( 1 \right) = \frac{3}{{50}}\hspace{0.5in}f'\left( 5 \right) = - \frac{{\sqrt 5 }}{{810}}$

and so we can see that the function in increasing on $$0 \le x \le 4$$ and decreasing on $$x \ge 4$$. Therefore, since $$f\left( n \right) = {b_n}$$ we know as well that the $${b_n}$$ are also increasing on $$0 \le n \le 4$$ and decreasing on $$n \ge 4$$.

The $${b_n}$$ are then eventually decreasing and so the second condition is met.

Both conditions are met and so by the Alternating Series Test the series must be converging.

As the previous example has shown, we sometimes need to do a fair amount of work to show that the terms are decreasing. Do not just make the assumption that the terms will be decreasing and let it go at that.

Let’s do one more example just to make a point.

Example 4 Determine if the following series is convergent or divergent.

$\sum\limits_{n = 2}^\infty {\frac{{\cos \left( {n\pi } \right)}}{{\sqrt n }}}$
Show Solution

The point of this problem is really just to acknowledge that it is in fact an alternating series. To see this we need to acknowledge that,

$\cos \left( {n\pi } \right) = {\left( { - 1} \right)^n}$

If you aren’t sure of this you can easily convince yourself that this is correct by plugging in a few values of $$n$$ and checking.

So the series is really,

$\sum\limits_{n = 2}^\infty {\frac{{\cos \left( {n\pi } \right)}}{{\sqrt n }}} = \sum\limits_{n = 2}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt n }}\hspace{0.5in} \Rightarrow \hspace{0.25in}\,\,\,\,\,{b_n} = \frac{1}{{\sqrt n }}}$

Checking the two condition gives,

$\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\sqrt n }} = 0$ ${b_n} = \frac{1}{{\sqrt n }} > \frac{1}{{\sqrt {n + 1} }} = {b_{n + 1}}$

The two conditions of the test are met and so by the Alternating Series Test the series is convergent.

It should be pointed out that the rewrite we did in previous example only works because $$n$$ is an integer and because of the presence of the $$\pi$$. Without the $$\pi$$ we couldn’t do this and if $$n$$ wasn’t guaranteed to be an integer we couldn’t do this.

Let’s close this section out with a proof of the Alternating Series Test.

#### Proof of Alternating Series Test

Without loss of generality we can assume that the series starts at $$n = 1$$. If not we could modify the proof below to meet the new starting place or we could do an index shift to get the series to start at $$n = 1$$.

Also note that the assumption here is that we have $${a_n} = {\left( { - 1} \right)^{n+1}}{b_n}$$. To get the proof for $${a_n} = {\left( { - 1} \right)^{n}}{b_n}$$ we only need to make minor modifications of the proof and so will not give that proof.

Finally, in the examples all we really needed was for the $${b_n}$$ to be positive and decreasing eventually but for this proof to work we really do need them to be positive and decreasing for all $$n$$.

First, notice that because the terms of the sequence are decreasing for any two successive terms we can say,

$b_{n} - {b_{n + 1}} \ge 0$

Now, let’s take a look at the even partial sums.

\begin{align*}{s_2}& = {b_1} - {b_2} \ge 0\\ {s_4} & = {b_1} - {b_2} + {b_3} - {b_4} = {s_2} + {b_3} - {b_4} \ge {s_2} & \hspace{0.5in} & {\mbox{because }}{b_3} - {b_4} \ge 0\\ {s_6} & = {s_4} + {b_5} - {b_6} \ge {s_4} & \hspace{0.5in} & {\mbox{because }}{b_5} - {b_6} \ge 0\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \vdots \\ {s_{2n}} & = {s_{2n - 2}} + {b_{2n - 1}} - {b_{2n}} \ge {s_{2n - 2}} & \hspace{0.5in} & {\mbox{because }}{b_{2n - 1}} - {b_{2n}} \ge 0\end{align*}

So, $$\left\{ {{s_{2n}}} \right\}$$ is an increasing sequence.

Next, we can also write the general term as,

\begin{align*}{s_{2n}} & = {b_1} - {b_2} + {b_3} - {b_4} + {b_5} + \cdots - {b_{2n - 2}} + {b_{2n - 1}} - {b_{2n}}\\ & = {b_1} - \left( {{b_2} - {b_3}} \right) - \left( {{b_4} - {b_5}} \right) + \cdots - \left( {{b_{2n - 2}} - {b_{2n - 1}}} \right) - {b_{2n}}\end{align*}

Each of the quantities in parenthesis are positive and by assumption we know that $${b_{2n}}$$ is also positive. So, this tells us that $${s_{2n}} \le {b_1}$$ for all $$n$$.

We now know that $$\left\{ {{s_{2n}}} \right\}$$ is an increasing sequence that is bounded above and so we know that it must also converge. So, let’s assume that its limit is $$s$$ or,

$\mathop {\lim }\limits_{n \to \infty } {s_{2n}} = s$

Next, we can quickly determine the limit of the sequence of odd partial sums, $$\left\{ {{s_{2n + 1}}} \right\}$$, as follows,

$\mathop {\lim }\limits_{n \to \infty } {s_{2n + 1}} = \mathop {\lim }\limits_{n \to \infty } \left( {{s_{2n}} + {b_{2n + 1}}} \right) = \mathop {\lim }\limits_{n \to \infty } {s_{2n}} + \mathop {\lim }\limits_{n \to \infty } {b_{2n + 1}} = s + 0 = s$

So, we now know that both $$\left\{ {{s_{2n}}} \right\}$$ and $$\left\{ {{s_{2n + 1}}} \right\}$$ are convergent sequences and they both have the same limit and so we also know that $$\left\{ {{s_n}} \right\}$$ is a convergent sequence with a limit of $$s$$. This in turn tells us that $$\sum {{a_n}}$$ is convergent.


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# Is the study of algebraic curve is techniquely equal to the advanced division of analytic geometry, if not, what is the difference?

Is the study of algebraic curve is techniquely equal to the advanced division of analytic geometry, if not, what is the difference?

And what is other branch of advanced analytic geometry called?

in addition, what is the differece between algebraic geometry and algebraic curves?

-


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# Complex polygon

The term complex polygon can mean two different things:

## Computer graphics

A complex (self-intersecting) pentagon

In the world of computer graphics, a complex polygon is a polygon which is neither convex nor concave. This includes any polygon which:

• Has a boundary comprising discrete circuits, such as a polygon with a hole in it.

Therefore, unlike simple polygons, a complex polygon may not always be interpreted as a simple polygonal region. Vertices are only counted at the ends of edges, not where edges intersect in space.

A formula relating an integral over a bounded region to a closed line integral may still apply when the "inside-out" parts of the region are counted negatively.

Moving around the polygon, the total amount one "turns" at the vertices can be any integer times 360°, e.g. 720° for a pentagram and 0° for an angular "eight".

## Geometry

In geometry, a complex polygon is a polygon in the complex Hilbert plane, which has two complex dimensions.

A complex number may be represented in the form $(a + ib)$, where $a$ and $b$ are real numbers, and $i$ is the square root of $-1$. A complex number lies in a complex plane having one real and one imaginary dimension, which may be represented as an Argand diagram. So a single complex dimension is really two dimensions, but of different kinds.

The unitary plane comprises two such complex planes, which are orthogonal to each other. Thus it has two real dimensions $x$ and $y,$ and two imaginary dimensions $ix$ and $iy$.

A complex polygon is a two-dimensional example of the more general complex polytope in higher dimensions.

In a real plane, a visible figure can be constructed as the real conjugate of some complex polygon.


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# Tag Info

In signal processing, the Nyquist–Shannon sampling theorem says you need at least 2 samples of a frequency to be able to perfectly reconstruct it. So in your question, a sampling rate of $200\: \mathrm{MHz}$ means you can perfectly reconstruct frequencies in the range of $0 - 100\: \mathrm{MHz}$. So what happens when frequencies above $100\: \mathrm{MHz}$ ...


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## Reflection and a mixed boundary value problem concerning analytic functions.(English)Zbl 0903.30028

The authors study a mixed boundary value problem in complex function theory connected with hydrodynamics: Given a bounded domain $$D$$ with the boundary $$C$$ containing continuum $$A$$ with the boundary $$B$$ such that $$G=D\setminus A$$ is connected it is asked to find a holomorphic function $$f$$ on $$G$$ with prescribed continuous $$\text{Re } f=f_{C}$$ on $$C$$ and $$\text{Im } f=f_{B}+k$$ on $$B$$ ($$k$$ is a constant function). If $$B, C$$ are rectifiable Jordan curves satisfying some regularity conditions the solution $$f$$ can be expressed as the Cauchy type integral with continuous real densities. If $$D$$ admits reflection mapping the problem can be simplified as shown in the article. The solution is found with the help of Fredholm type equation under the “minimal” assumptions on $$B$$ and it is of interest also from the point of view of numerical methods.
Reviewer: J.Veselý (Praha)

### MSC:

 30E25 Boundary value problems in the complex plane 31A25 Boundary value and inverse problems for harmonic functions in two dimensions
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# Any easier way to write this?

Zerio920 285
7 years ago

The Wiki says I can record key presses in a table to determine whether the key is being held or not, so something like

local keys = {}

mouse.KeyDown:connect(function(key)
keys[key] = true
end)
mouse.KeyUp:connect(function(key)
keys[key] = nil
end)

while wait(0.1) do
if keys["q"] and Tool.Val.Value > 0 then
Tool.Val.Value = Tool.Val.Value - 5
elseif keys["e"] and Tool.Val.Value < 200 then
Tool.Val.Value = Tool.Val.Value + 5
end
end

However, I'd also like to differentiate between a key press and someone holding down the key, and while the script is effective enough at doing that, it only checks for key presses every 1/10th of a second, so anytime someone pushes a key in that 10th of a second it won't count. It's minor, yeah, but I'd really like to find a more concise method of determining keystrokes from key presses.

You can make the loop go faster if you want. If you leave out the argument in wait() it will delay the script the minimum amount of time it can, about a 30th of a second.


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In verilog, what effect does the not (!) operator have on high impedance and don't care conditions?

I'm writing some verilog and simulating it using modelsim. I have a block that looks like this:

        if(wr_req & !cam_busy & !lookup_latched & !cam_match_found & !cam_match_found_d1) begin
cam_we           <= 1;
cam_din          <= wr_cmp_data ;
wr_ack           <= 1;
lut_wr_data      <= wr_data;
end

lookup_latched is in high impedence. cam_match_found and cam_match_found_d1 are both in don't care states.

It seems to me like the statement should just ignore those signals.

The one issue I noticed was that the signals were being bitwise anded instead of logically anded (I didn't write the code). Will this have an affect on the result? (I imagine it will)

Thanks!

In verilog, a 'z' at the input of a gate transforms to an 'x' on the output of the gate. An 'x' at the input of the gate also translates to an 'x' on the output of the gate.

Therefore your code reduces to if(wr_req & !cam_busy & X & X & X)

The value of (something & X) can be either 0, or X, and in both cases the if statement will evaluate to false (if(X) is the same as if(0)).

The one issue I noticed was that the signals were being bitwise anded instead of logically anded (I didn't write the code). Will this have an affect on the result? (I imagine it will)

If those are all 1-bit signals, than logical AND (&&) is the same result as bitwise AND (&).

• Ah thank you! I just figured it out. I wasn't forcing a signal that I should have been on the input. I was still uncertain of the behavior – Caustic May 2 '13 at 4:56
• Also, 1-bit signals makes sense for the logical vs bitwise and, but for example, what would the behavior of 8'hXF && 8'hFF be? I'm assuming that this evaluates to true since there are non-zero bits in the first operand. Is this correct? – Caustic May 2 '13 at 5:02
• I believe that 8'hXF will reduce to true for purposes of logical operation, but it's probably not a good idea to use multi-bit values in a logical operation (will probably give lint or synthesis warnings). – Tim May 2 '13 at 5:16


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Question

A parallel beam of monochromatic light falls on a combination of a convex lens and concave lens of focal lengths 15 cm and 5 cm respectively. What is the distance between the two lenses to obtain a parallel beam of light from the concave lens?

A
20 cm
B
3 cm
C
10 cm
D
45 cm

Solution

The correct option is A 20 cmFocal length of convex lens  $$f_1 = 15 \ cm$$Focal length of concave lens   $$f_2 = -5 \ cm$$Since the convex lens forms the image at its focus for the incident parallel beam. Thus image is formed at a distance of $$15 \ cm$$ from the convex lens.Now the rays emerge parallel from the concave lens, thus the object has to be at the focus of the concave lens. So the object is at a distance $$5 \ cm$$ in front of concave lens.Hence, separation between the lenses   $$d = 15+5 = 20 \ cm$$Physics

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