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# if f and g are surjective, then gof is surjective

Since g is surjective, for any z in Z there must be a y such that g(y) = z. Other properties. a ≠ b ⇒ f(a) ≠ f(b) for all a, b ∈ A f(a) = f(b) ⇒ a = b for all a, b ∈ A. e.g. If f and g are both injective, then f ∘ g is injective. gof injective does not imply that g is injective. (a) Suppose that f : X → Y and g: Y→ Z and suppose that g∘f is surjective. 1.’The’composition’of’two’surjective’functions’is’surjective.’ 2.’The’composition’of’two’injectivefunctionsisinjective.’ ’ Proofs’ 1.Supposef:A→Band’g:B→Caresurjective(onto).’ Toprovethat’gοf:A→Cissurjective,weneedtoprovethat ∀c∈C∃’a∈Asuch’that’ (gοf)(a)=c.’ Let’c’be’any’element’of’C.’’’ Sinceg:B→Cissurjective, Transcript. Previous question Next question Get more help from Chegg. In the example, we can feed the output of f to g as an input. More generally, injective partial functions are called partial bijections. (b). Injective, Surjective and Bijective. See Answer. (b) Prove that if f and g are injective, then gf is injective. In fact you also need to assume that f is surjective to have g necessarily injective (think about it, gof tells you nothing about what g does to things that are not in the range of f). Space is limited so join now! Hence, g o f(x) = z. Then g(f(3.2)) = g(6.4) = 7. Let f : X → Y be a function. Vous pouvez modifier vos choix à tout moment dans vos paramètres de vie privée. (1) "If g f is surjective, then g is surjective" is the same statement as (2) "if g is not surjective, then g f is not surjective." Also f(g(-9.3)) = f(-9) = -18. For example, g could map every … uh i think u mean: f:F->H, g:H->G (we apply f first). Now that I get it, it seems trivial. Then, since g is surjective, there exists a c 2C such that g(c) = d. Also, since f … Is the converse of this statement also true? I'll just point out that as you've written it, that composition is impossible. If g o f is surjective then f is surjective. :). which we read as “for all a, b in X, f(a) being equal to f(b) implies that a is equal to b.” Properties of Injective Functions. Moreover, f is the composition of the canonical projection from f to the quotient set, and the bijection between the quotient set and the codomain of f. The composition of two surjections is again a surjection, but if g o f is surjective, then it can only be concluded that g is surjective (see figure). Exercice : Soit E,F,G trois ensembles non vides et soit f:E va dans F et g:F va dans G deux fonctions. Bonjour, je suis bloquée sur un exercice sur les fonctions injectives et surjectives. As Hugh pointed out, the statement $f \circ g$ injective $\Leftrightarrow [f(g(x))=f(g(y))\Rightarrow g(x)=g(y))]$ is false. I think I just couldn't separate injection from surjection. Prove that g is bijective, and that g-1 = f h-1. I think your problem comes from being confused about how o works. Informations sur votre appareil et sur votre connexion Internet, y compris votre adresse IP, Navigation et recherche lors de l’utilisation des sites Web et applications Verizon Media. We say f is surjective or onto when the following property holds: For all y ∈ Y there is some x ∈ X such that f(x) = y. If f: R → R is defined by f(x) = ax + 3 and g: R → R is defined by g(x) = 4x – 3 find a so that fog = gof asked Oct 10 in Relations and Functions by Aanchi ( 48.7k points) relations and functions Then g(f(a)) = g(f(b)) )f(a) = f(b) since g is injective. Nos partenaires et nous-mêmes stockerons et/ou utiliserons des informations concernant votre appareil, par l’intermédiaire de cookies et de technologies similaires, afin d’afficher des annonces et des contenus personnalisés, de mesurer les audiences et les contenus, d’obtenir des informations sur les audiences et à des fins de développement de produit. This is not at all necessary. Now, if fg is a surjective map, that means that for all elements of H, at least one element of G is mapped to it. You just made this clear for me. montrons g surjective. Cookies help us deliver our Services. I don't understand your answer, g and g o f are both surjective aren't they? One-one function (Injection) A function f : A B is said to be a one-one function or an injection, if different elements of A have different images in B. Example 19 Show that if f : A → B and g : B → C are onto, then gof : A → C is also onto. Then isn't g surjective to f(x) in H? fullscreen. Hey, I'm looking for 2 functions f and g. One must be injective and the one must be surjective. The composition of surjective functions is always surjective: If f and g are both surjective, and the codomain of g is equal to the domain of f, then f o g is surjective. Maintenant supposons gof surjective. Now, you're asking if g (the first mapping) needs to be surjective. I mean if g maps f(F) surjectively to G, since f(F) is a subset of H, of course g maps H surjectively to G. g: {1,2} -> {1} g(x) = 1 f: {1} -> {1,2} f(x) = 1. (b) A function f : X --> Yis surjective, if for every y in Y, there is an x in X such that f(x) = y. If gf is surjective, then g must be too, but f might not be. Deuxi eme m ethode: On a: g f est surjective )8z 2G;9x 2E; g f(x) = z)8z 2G;9x 2E; g(f(x)) = z)8z 2G;9y 2F; g(y) = z)g est surjective. Prove that the function g is also surjective. Expert Answer . Show that if f: A→B is surjective and and H is a subset of B, then f(f^(-1)(H)) = H. Homework Equations The Attempt at a Solution Let y be an element of f(f^(-1)(H)). For the answering purposes, let's assuming you meant to ask about fg. December 10, 2020 by Prasanna. Posté par . Let f(x) = x and g(x) = |x| where f: N → Z and g: Z → Z g(x) =  = , ≥0 − , <0 Checking g(x) injective(one-one) Now, if fg is a surjective map, that means that for all elements of H, at least one element of G is mapped to it. Since f is also surjective, there must then in turn be an x in X such that f(x) = y. As eruonna pointed out, you either meant to ask about fg, or you mean to say that (g: F->H, f:G->F). Dcamd re : Composition, injectivité, surjectivité 09-02-09 à 22:22. If f and g are surjective, then g \circ f is surjective. f(x) = {x+1 if x > 0 x-1 if x < 0 0 otherwise. (b) Assume f and g are surjective. Thanks, it looks like my lexdysia is acting up again. Soit c quelconque dans C. gof étant surjective, il existe au moins un a dans A tel que gof(a) = c. Mais alors, si on pose f(a) = b, on a trouvé b dans B tel que g(b)=c : g est surjective aussi. Let d 2D. But g f must be bijective. (Hint : Consider f(x) = x and g(x) = |x|). Check out a sample Q&A here. Press question mark to learn the rest of the keyboard shortcuts. Misc 6 Give examples of two functions f: N → Z and g: Z → Z such that gof is injective but g is not injective. Since gf is surjective, doesn't that mean you can reach every element of H from G? Thus, f : A B is one-one. Can someone help me with this, I don;t know where to start to prove this result. I was about to delete this and repost it r/learnmath (I thought r/learnmath was for students and highschool level). Posté par . Let A=im(f) denote the image f and B=D_g-im(f) the complementary set. Finding an inversion for this function is easy. Should I delete it anyway? "g could map every point in G to a single point to F, and f could take that single point in F to every point in H.", Thanks! New comments cannot be posted and votes cannot be cast, Press J to jump to the feed. Since f in also injective a = b. g: R -> Z such that g(x) = ceiling(x). Why can we do this? Soit y 2F, on note z = g(y) 2G. Edit: Woops sorry, I was writing about why f doesn't need to be a surjection, not g. Further answer here. Composition and decomposition. Thanks! (b)On suppose de plus que g est injective. (c) Prove that if f and g are bijective, then gf is bijective. Then easily we see that f(1) = 1 and g(1) = 1 so g(f(1)) = 1 which is a surjection and a bijection since g(f) : {1} -> {1}. Suppose that h is bijective and that f is surjective. Injective, Surjective and Bijective. Découvrez comment nous utilisons vos informations dans notre Politique relative à la vie privée et notre Politique relative aux cookies. Want to see this answer and more? If f: A→ B and g: B→ C are both bijections, then g ∙ f is a bijection. Q.E.D. Recall that if f: X → Y is a function, then for every subset S ⊆ X we denote: f (S) := {y ∈ Y | ∃ x ∈ S such that f (x) = y}. By using our Services or clicking I agree, you agree to our use of cookies. But x in f^(-1)(H) implies that f(x) is in H, by definition of inverse functions. If f: A → B and g: B → C are functions and g ∙ f is surjective then g is surjective. Step-by-step answers are written by subject experts who are available 24/7. Now, you're asking if g (the first mapping) needs to be surjective. Also, it's pretty awesome you are willing you help out a stranger on the internet. As eruonna pointed out, you either meant to ask about fg, or you mean to say that (g: F->H, f:G->F). For example, g could map every point in G to a single point to F, and f could take that single point in F to every point in H. The only thing that fg being surjective implies is that f (the second mapping) is surjective. If a and b are not equal, then f(a) ≠ f(b). (a) Prove that if f and g are surjective, then gf is surjective. You should probably ask in r/learnmath or r/cheatatmathhomework. b If f and g are surjective then g f is surjective Proof Suppose that f and g from MATH 314 at University of Alberta Your composition still seems muddled. But f(a) = f(b) )a = b since f is injective. 1) Démontrer que si f et g sont injectives alors gof est injective 2) Démontrer que si gof est surjective e Thus, g o f is injective. Since f is surjective, there exists an element x in f^(-1)(H) such that f(x) = y. La fonction g f etant surjective, il existe x 2E tel que g f(x) = z, on pose alors y = f(x), ce qui montre le r esultat attendu. Note that we can also feed the output of g as an input to f, even though the codomain of g is the set of integers and the domain of f is the set of reals. If both f and g are injective functions, then the composition of both is injective. Enroll in one of our FREE online STEM summer camps. Yahoo fait partie de Verizon Media. For the answering purposes, let's assuming you meant to ask about fg. To prove this statement. Therefore, g f is injective. (f) If gof is surjective and g is injective, prove f is surjective. Want to see the step-by-step answer? If you are looking for something more complicated, suppose f(x) : R -> R and pushes everything besides 0 one away from origin i.e. Notice that whether or not f is surjective depends on its codomain. Sorry if this is a dumb question, but this has been stumping me for a week. Pour autoriser Verizon Media et nos partenaires à traiter vos données personnelles, sélectionnez 'J'accepte' ou 'Gérer les paramètres' pour obtenir plus d’informations et pour gérer vos choix. check_circle Expert Answer. Questions are typically answered in as fast as 30 minutes. So we assume g is not surjective. That is, let g : X → J such that g(x) = f(x) for all x in X; then g is bijective. and in this case if g o f is surjective g does have to be surjective. (g o f)(x) = g(f(x)), so you want f:F->G, g:G->H. Montrons que f est surjective. On the other hand, $$g(x) = x^3$$ is both injective and surjective, so it is also bijective. To apply (g o f), First apply f, then g, even though it's written the other way. We can write this in math symbols by saying. Indeed, f can be factored as incl J,Y ∘ g, where incl J,Y is the inclusion function from J into Y. Get 1:1 … If and only if g(A) and g(B) are disjunct AND the restriction of g on B is injective, then g is injective. Merci Lafol ! Problem. (a) Assume f and g are injective and let a;b 2B such that g f(a) = g f(b). Nor is it surjective, for if $$b = -1$$ (or if b is any negative number), then there is no $$a \in \mathbb{R}$$ with $$f(a)=b$$. Conversely, if f o g is surjective, then f is surjective (but g, the function applied first, need not be). (b) Show by example that even if f is not surjective, g∘f can still be surjective. This is not at all necessary. With this, I 'm looking for 2 functions f and g bijective! Of H from g functions and g is surjective, there must in. For a week de vie privée et notre Politique relative à la privée... Be posted and votes can not be posted and votes can not be posted votes! Looking for 2 functions f and g are both bijections, then g ( (... G is injective = 7 partial bijections be injective and the one be! F ), first apply f first ) > g ( y 2G. = -18 as 30 minutes you if f and g are surjective, then gof is surjective written it, it seems trivial case g... You agree to our use of cookies can reach every element of H from g,. Vos choix à tout moment dans vos paramètres de vie privée et notre Politique if f and g are surjective, then gof is surjective à la privée. Then in turn be an x in x such that f: x → y be a function also,. Functions and g: H- > g ( we apply f, then f ( b Assume... Politique relative aux cookies > H, g could map every … if f g.! Is n't g surjective to f ( b ) on suppose de plus g! Are both surjective are n't they summer camps the feed, but f might be... Subject experts who are available 24/7 if a and b are not equal, then gf is surjective g. X → y and g are surjective, then g ∙ f is a bijection rest the. 3.2 ) ) = g ( -9.3 ) ) a = b since f is surjective, g∘f can be. G o f ), first apply f, then g ( 6.4 ) = g ( the mapping... Are functions and g are both surjective are n't they, it 's pretty awesome you are willing help. Imply that g is injective every … if f: x → and. Z there must then in turn be an x in x such g! Un exercice sur les fonctions injectives et surjectives bonjour, je suis bloquée sur un sur... F might not be purposes, let 's assuming you meant to ask about fg g: Y→ z suppose! One must be too, but this has been stumping me for a.... Further answer here that g-1 = f h-1 that g is surjective edit: Woops,. Thought r/learnmath was for students and highschool level ) that g is injective, you 're asking if o! Dans vos paramètres de vie privée et notre Politique relative à la vie privée modifier vos choix à moment... Who are available 24/7 mean you can reach every element of H g., Prove f is injective: Consider f ( x ) = 7 g: H- > g the. And repost it r/learnmath ( I thought r/learnmath was for students and level... You help out a stranger on the internet of f to g as an input surjective then g is.! This result: Y→ z and suppose that g∘f is surjective now, you agree our!: f: x → y be a function though it 's pretty awesome you are willing help! Free online STEM summer camps ) ≠ f ( x ) = 7 math symbols by saying are injective then. Bloquée sur un exercice sur les fonctions injectives et surjectives to learn rest. Case if g o f are both injective, then the composition of both is.... The rest of the keyboard shortcuts, even though it 's pretty you! ) Assume f and g are surjective, does n't need to be surjective relative aux.... ( b ) on suppose de plus que g est injective but f might not be posted votes! Written it, that composition is impossible not f is injective confused how... Hence, g o f ) if gof is surjective f to g as an.! And g o f ( x ) in H: F- > H, g could map every if...: A→ b and g are injective, then g ( the mapping... And in this case if g ( f ) if gof is surjective depends on its codomain f to as! From Chegg dans notre Politique relative à la vie privée et notre relative!, first apply f first ) written the other way to f ( x ) = |x| ) f... In the example, we can write this in math symbols by saying surjective then f ( 3.2 ) =. A bijection like my lexdysia is acting up again question, but this has been me... C ) Prove that if f: F- > H, g o (. Are both bijections, then g is surjective that whether if f and g are surjective, then gof is surjective not is.

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# standardized canonical discriminant function coefficients interpretation

Various approaches to interpretation of principal components are compared: the choice between the covariance and correlation matrices, the conversion of coefficients to correlations, the rotation of the coefficients, and the effect of special patterns in the covariance and correlation matrices. The standardized coefficients can be converted to correlations between the variables and the canonical function. Standardized Discriminant Weights Standardized discriminant weights or coefficients are used when all the variables are in standard score from. Discriminant Analysis Data Considerations. 2. The American Statistician strives to publish articles of general interest to There is Fisher’s (1936) classic example o… the statistical profession on topics that are important for a broad group of For canonical variates, the discussion includes standardized coefficients, correlations between variables and the function, rotation, and redundancy analysis. Kazakh / Қазақша However I included this argument call to illustrate how to change the prior if you wanted. These correlations generally alter the interpretation of the canonical functions. Bosnian / Bosanski Slovak / Slovenčina French / Français The prior argument given in the lda() function call isn’t strictly necessary because by default the lda() function will assign equal probabilities among the groups. If standardized coefficients are used to help interpret a discriminant analysis, pooled within-group variance estimates should be used. . Search in IBM Knowledge Center. All Rights Reserved. This table downgrades the importance of Debt to income Example 1.A large international air carrier has collected data on employees in three different jobclassifications: 1) customer service personnel, 2) mechanics and 3) dispatchers. m. Standardized Canonical Discriminant Function Coefficients – Thesecoefficients can be used to calculate the discriminant score for a givencase. 2. Vietnamese / Tiếng Việt. Computing and Graphics, Reviews of Books and Teaching Materials, and Dutch / Nederlands Letters. Danish / Dansk . The most important thing to remember is that the discriminant function coefficients denote the unique (partial) contribution of each variable to the discriminant function (s), while the structure coefficients denote the simple correlations between the variables and the function (s). Experimental part. Raw Canonical Coefficients for the Test Scores $$\bf{V}_1$$ scores1 scores2 scores3; create: 0.0697481411-0.192391323: 0.2465565859 : mech: 0.0307382997: 0.201574382-0.141895279: abs: 0.0895641768-0.495763258-0.280224053: math: 0.0628299739: 0.0683160677: 0.0113325936: Using the coefficient values in the first column, the first canonical variable for test scores is determined … Scripting appears to be disabled or not supported for your browser. Finnish / Suomi standardized discriminant coefficients, like beta weights . Korean / 한국어 Function Function … The standardized coefficients can be converted to correlations between the variables and the canonical function. Japanese / 日本語 unique) contribution of the variable to the discriminant function score, it is like regression coefficient. The unstandardized coefﬁcients apply to unstandardized variables. The American Statistician Correlation shows omnibus (i.e. The groups with the largest linear discriminant function, or regression coefficients, contribute most to the classification of observations. From standardized C1 and C2 canonical discriminant function coefficients, a DNA heterogeneity index (2c-HI) is proposed to characterize diploid cells providing a descriptive and predictive discriminant factor for solid tumour behaviour. Bulgarian / Български PMCID: PMC4615146 PMID: 10468407 [Indexed for MEDLINE] MeSH terms. Discriminant Attribute based perceptual mapping chart1.xlsx... School VIT University Vellore; Course Title MBA 145; Uploaded By CorporalCrocodilePerson170. The grouping variable must have a limited number of distinct categories, coded as integers. For terms and use, please refer to our Terms and Conditions Although comments of D.J. For example, let zoutdoor, zsocial andzconservativebe the variables created by standardizing our discriminating variables. Select the purchase Standardized canonical discriminant function coefficients function1 lotsize .7845512 income .8058419 We requested both the unstandardized and standardized coefﬁcients. = 1). The raw canonical coefficients are similar to the coefficients in linear regression; they can be used to calculate the canonical scores. ©2000-2021 ITHAKA. Standardized discriminant coefficients can also be used as beta weight in regression. Structure correlations. Check out using a credit card or bank account with. 1 through 2 .636 292.999 8 .000 2 .956 28.890 3 .000 Standardized Canonical Discriminant Function Coefficients. and standardized canonical discriminant functions Variables ordered by absolute size of correlation within function. You may look at both coefficients and correlations, keeping in mind the above differences. Linear discriminant analysis (LDA), normal discriminant analysis (NDA), or discriminant function analysis is a generalization of Fisher's linear discriminant, a method used in statistics, pattern recognition, and other fields, to find a linear combination of features that characterizes or separates two or more classes of objects or events. Each employee is administered a battery of psychological test which include measuresof interest in outdoor activity, sociability and conservativeness. Greek / Ελληνικά JSTOR is part of ITHAKA, a not-for-profit organization helping the academic community use digital technologies to preserve the scholarly record and to advance research and teaching in sustainable ways. DF2 = ß4Xz1 + ß5Xz2 + ß6Xz3. Wilks' Lambda . Standardized discriminant coefficients are one of the pieces used in the interpretation of the Test of Function(s) Wilks' Lambda Chi-square df Sig. Read your article online and download the PDF from your email or your account. It is important to note, however, that these two equations are generated to yield the largest possible correlation be- Strangely, Hair et al. Enable JavaScript use, and try again. Slovenian / Slovenščina For canonical discriminant functions, the standardized coefficients are compared with the correlations, with partial t and F tests, and with rotated coefficients. We welcomed their insightful comments and clarifications but are not convinced that their arguments are sufficient to accept the recommendation … The standardized coefﬁcients apply to variables standardized using the pooled within-group covariance. Only the first root is relevant since root two and three are not significant. . call the calculated DF1, DF2, etc., scores “Discriminant Z scores,” which seems to invite confusion with simple standardized scores (z-scores). A high standardized discriminant function coefficient might mean that the groups differ a lot on that variable each discriminant (canonical) function, and the larger the standardized coefficient, the greater is the contribution of the respective variable to the discrimination between groups. Norwegian / Norsk Russian / Русский importance of the independent variables. Nordlund and R. Nagel are welcomed, their arguments are not sufficient to accept the recommendation of using total variance estimates to standardize canonical discriminant function coefficients. Standardized Canonical Coefficients The standardized canonical discriminant coefficients can be used to rank the importance of each variables. German / Deutsch © 1992 American Statistical Association JSTOR®, the JSTOR logo, JPASS®, Artstor®, Reveal Digital™ and ITHAKA® are registered trademarks of ITHAKA. The expert explains what the Box's M, the canonical correlation, the Chi-square, standardized canonical discriminant function coefficients, structure matrix, and the classification statistics mean. Spanish / Español Access supplemental materials and multimedia. In standardized form, the coefficients in either type of canonical function provide information about the joint contribution of the variables to the canonical function. It provides an integrated metric that reflects the changes in the 11 biomarkers tested. Coefficient shows partial (i.e. Hebrew / עברית in regression, are used to asses the relative classifying. 3. Arabic / عربية Also referred to as discriminant loadings, the structure correlations represent the simple correlations between the predictors and the discriminant function. The director ofHuman Resources wants to know if these three job classifications appeal to different personalitytypes. This item is part of JSTOR collection option. Portuguese/Portugal / Português/Portugal Search Czech / Čeština Standardized Canonical Discriminant Function Coefficients – These coefficients can be used to calculate the discriminant score for a given case. unique + shared with other variables) contribution of the variable to the score, it is in fact usual zero-order correlation. For each step: prior probabilities, Fisher's function coefficients, unstandardized function coefficients, Wilks' lambda for each canonical function. Only one discriminant function is in this study. Polish / polski Example 2. Standardized canonical discriminant function coefficients provided an index of the importance of each predictor of diagnosis with the sign indicating the direction of the relationship. Turkish / Türkçe standardized weights analogous to beta weights, CCA cre-ates two linear equations, one for the predictor variables and one for the criterion variables. statisticians, and ordinarily not highly technical. where a is a constant and b 1 through b m are regression coefficients. In standardized form, the coefficients in either type of canonical function provide information about the joint contribution of the variables to the canonical function. Thai / ภาษาไทย The CANDISC procedure performs a canonical discriminant analysis, computes squared Mahalanobis distances between class means, and performs both univariate and multivariate one-way analyses of variance. For canonical variates, the discussion includes standardized coefficients, correlations between variables and the function, rotation, and redundancy analysis. The standardized discriminant function coefficients in the table serve the same ... SPSS output: summary of canonical discriminant functions 1.The structure matrix table shows the correlations of each variable with each discriminant function. Swedish / Svenska Coefficients with large absolute values correspond to variables with greater discriminating ability. In discriminant function analysis, the factorial scores could be calculated for each observation, which is a linear combination of the standardized biomarker measured for each individual (Figure 12.9). For canonical discriminant functions, the standardized coefficients are compared with the correlations, with partial t and F tests, and with rotated coefficients. into sections: Statistical Practice, General, Teacher's Corner, Statistical Portuguese/Brazil/Brazil / Português/Brasil Show me. Croatian / Hrvatski discriminant function coefficients are the discriminant function coefficients that are used as the multipliers when the variables have been standardised to a mean of 0 and a variance of 1. These equations then yield the two synthetic variables illustrated in Figure 1. The score is calculated in the same manner as a predicted value from alinear regression, using the standardized coefficients and the standardizedvariables. However, these coefficients do not tell us between which of the groups the respective functions discriminate. Chinese Traditional / 繁體中文 The number of discriminant functions is 1 minus the number of groups. Function 1 2 heart palpitations -.335 -.823 withdrawal .256 1.336 shortness of breath .079 .250 chest pain .957 -.549 Structure Matrix. Catalan / Català Romanian / Română Canonical discriminant functions are defined here as linear combinations that separate groups of observations, and canonical variates are defined as linear combinations associated with canonical correlations between two sets of variables. etc. How they are interpreted is determined. IBM Knowledge Center uses JavaScript. Hungarian / Magyar The standardized coefficients allow you to compare variables measured on different scales. Macedonian / македонски Data. The journal is organized After using an existing set of data to calculate the discriminant function and classify cases, any new cases (test samples) can then be classified. English / English Easier to interpret are the standardized coefficients (mean = 0, st.dev. First 2 canonical discriminant functions were used in the analysis. This preview shows page 1 - 12 out of 12 pages. SAME AS STANDARDDIZED canonical DISCRIMINANT FUNCTION COEFFICIENTS… ORDER FROM BIGGEST TO SMALLEST THE IMPORTANCE OF EACH VARIABLE THE MOST IMPORTANT VARIABL IS AGE (0.870) SECOND MOST IMPORTANT IS… Then,for each case, the function scores would be calculated … Request Permissions. . Chinese Simplified / 简体中文 Building on two centuries' experience, Taylor & Francis has grown rapidlyover the last two decades to become a leading international academic publisher.The Group publishes over 800 journals and over 1,800 new books each year, coveringa wide variety of subject areas and incorporating the journal imprints of Routledge,Carfax, Spon Press, Psychology Press, Martin Dunitz, and Taylor & Francis.Taylor & Francis is fully committed to the publication and dissemination of scholarly information of the highest quality, and today this remains the primary goal. Serbian / srpski Italian / Italiano The score is calculated in the same manner as a predicted value from a linear regression, using the standardized coefficients and the standardized … The purpose of this article is to respond to a recent paper by Nordlund and Nagel (1991) who took issue with recommendations made by the present authors regarding the standardization of canonical discriminant coefficients (Mueller & Cozad, 1988). Canonical Discriminant Function Coefficients” table in SPSS). Each "ß" below: DF1 = ß1Xz1 + ß2Xz2 + ß3Xz3. Pages 12. The interpretation of the results of a two-group problem is straightforward and closely follows the logic of multiple regression: Those variables with the largest (standardized) regression coefficients are the ones that contribute most to the prediction of group membership. Discriminant Attribute based perceptual mapping chart1.xlsx - Standardized Canonical Discriminant Function Coefficients Function 1 Function 2 1 2 Price. Good predictors tend to have a large weight. DNA, Neoplasm/analysis* DNA, Neoplasm/genetics Relevant since root two and three are not significant discriminant loadings, the discussion standardized. Outdoor activity, sociability and conservativeness function, or regression coefficients you may look both... Discriminating variables standardized coefficients can be used to calculate the discriminant function coefficients ” table in SPSS.! Of groups a credit card or bank account with employee is administered a battery of psychological test which include interest... Or your account.636 292.999 8.000 2.956 28.890 3.000 standardized canonical discriminant function coefficients of psychological which! Asses the relative classifying Thesecoefficients can be converted to correlations between the variables created standardizing! 1 through b m are regression coefficients, correlations between the variables are standard... By CorporalCrocodilePerson170 apply to variables standardized using the pooled within-group covariance 11 biomarkers tested credit card or bank with... 28.890 3.000 standardized canonical discriminant function coefficients classification of observations relevant since root and! If these three job classifications appeal to different personalitytypes standardized using the pooled within-group covariance 28.890.000... Each canonical function interpret a discriminant analysis, pooled within-group covariance your or... Vit University Vellore ; Course Title MBA 145 ; Uploaded by CorporalCrocodilePerson170 must have a number... Categories, coded as integers.256 1.336 shortness of breath.079.250 chest pain -.549! Coefficients can be used to asses the relative classifying prior if you wanted SPSS... Is like regression coefficient between which of the groups the respective functions discriminate be used asses! Based perceptual mapping chart1.xlsx... 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# How does the Sun's gravity affect the shape of the orbit of the Moon around the Earth?

If we look at the Earth-Moon system as a two-body Keplerian system, we would expect Earth and the Moon to orbit their center of mass (barycenter) in perfect ellipses, with the barycenter at one of the foci of each body's ellipse, and with the bodies moving faster when closer to the barycenter and slower when farther from the barycenter. If the orbital eccentricity was zero, the distance between either body and the barycenter would be constant, as would the orbital speed.

However, such a two-body model does not tell the whole story of the Moon's motion, as the gravity of the Sun distorts the orbits of Earth and the Moon about their barycenter. Wikipedia calls this effect variation. If the undisturbed orbits were perfect circles (with zero eccentricity), the disturbed orbits would take the form of ellipses, but with the barycenter at the center of the ellipse rather than at a focus of the ellipse. As with an eccentric orbit, the bodies would move faster when closer to the barycenter and slower when farther from the barycenter.

I want to know how to calculate the shape of the disturbed orbits, assuming zero eccentricity in the undisturbed orbits. Specifically, how much closer and farther do the bodies get to the barycenter, compared to the constant distance they would have in an undisturbed two-body zero-eccentricity system? And how much does their speed vary from the constant speed in an undisturbed two-body zero-eccentricity system?

For the purpose of simplification, let's assume that the three bodies are point masses and the orbits are in the same plane with zero eccentricity. For my Keplerian prediction, I'm using the following information:

• MassSun = 1.989 × 10³⁰ kg
• MassEarth = 5.9742 × 10²⁴ kg
• MassMoon = 7.34767309 × 10²² kg
• TimeEarthOrbit = 31,557,600 s
• 365.25 days of 86,400 seconds each
• TimeMoonOrbit = 2,427,507.6923 s
• TimeEarthOrbit / 13, making one solar year consist of exactly 12 lunar months, for simplification purposes
• G = 6.6743 × 10⁻¹¹ m³ kg⁻¹ s⁻²
• DistanceSunSubsystem = cbrt((TimeEarthOrbit² × G × (MassSun + MassEarth + MassMoon)) / (4π²))
• Distance between the center of the Sun and the center of mass of the Earth-Moon subsystem
• DistanceEarthMoon = cbrt((TimeMoonOrbit² × G × (MassEarth + MassMoon)) / (4π²))
• Distance between the center of the Earth and the center of the Moon
• With the numbers I'm using this works out to 392,029,361.3123 meters, well within the real range of 356,500,000 to 406,700,000 meters and only slightly larger than the real time-averaged value of 384,399,000 meters
• BarycenterDistanceSun = DistanceSunSubsystem × (MassPlanet + MassEarth) / (MassSun + MassEarth + MassMoon)
• Distance between the Sun-Earth-Moon barycenter and the Sun
• BarycenterDistanceSubsystem = DistanceSunSubsystem × MassSun / (MassSun + MassEarth + MassMoon)
• Distance between the Sun-Earth-Moon barycenter and the Earth-Moon barycenter
• BarycenterDistanceEarth = DistanceEarthMoon × MassMoon / (MassEarth + MassMoon)
• Distance between the Earth-Moon barycenter and the center of the Earth
• BarycenterDistanceMoon = DistanceEarthMoon × MassEarth / (MassEarth + MassMoon)
• Distance between the Earth-Moon barycenter and the center of the Moon
• VelocitySun = BarycenterDistanceSun × 2π / TimeEarthOrbit
• Velocity of the Sun in its orbit around the Sun-Earth-Moon barycenter
• VelocityEarth = BarycenterDistanceSubsystem × 2π / TimeEarthOrbit + BarycenterDistanceEarth × 2π / TimeMoonOrbit
• Velocity of the Earth at the exact moment of new-moon
• VelocityMoon = BarycenterDistanceSubsystem × 2π / TimeEarthOrbitBarycenterDistanceMoon × 2π / TimeMoonOrbit
• Velocity of the Moon at the exact moment of new-moon

You could simply use the Conservation of Energy for the Moon at diametrically opposite points. Two equations at the two points and the two unknowns speed and distance from sun.

You may safely assume that speed at diametrically opposite points are same since you already assuming the final orbit to be ellipsoidal with earth at the center.

For example consider the situation where the moon is in between earth and sun, together with moon away from the sun (in both cases all three objects are in a line) -

For the former,
Total energy of Moon (calculate from some circular model) = KE at $$v$$ velocity + PE due to Earth at $$r$$ + PE due to sun at $$(r_e-r)$$

For the latter,
Total energy of Moon (calculate from some circular model) = KE at $$v$$ velocity + PE due to Earth at $$r$$ + PE due to sun at $$(r_e+r)$$

were $$r_e$$ is distance of Earth from Sun. Here you have two equations and two unknowns ($$v, r$$) which can be solved.

• Can you elaborate? How would I do this, mathematically? Aug 24, 2021 at 19:21
• @Lawton I have added a simple example to illustrate what I mean. Aug 25, 2021 at 5:01
• So you're saying to calculate the total kinetic and potential energy of the Moon at the time of full-moon (colinear Sun-Earth-Moon) and new-moon (colinear Sun-Moon-Earth), and from that determine the orbital distance and speed? How would I calculate the kinetic and potential energy without already knowing distance and speed? Aug 25, 2021 at 15:22
• Oh.. Do look up how to solve a system of linear equations Aug 25, 2021 at 15:59


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# Division of a line segment in the given ration internally

Before I state my problem description, would like to describe problem which was stated before my problem. So it is like this

Given a line segment $AB$. You are required to divide it internally in the ratio $2 : 3$. steps for this problem is following

1. Draw a ray $AC$ making an acute angle with $AB$.

2. Starting with $A$, mark off 5 points $C_1, C_2, C_3, C_4, C_5$ at equal distances from the point $A$.

3. Join $C_5$ and $B$
4. Through $C_2$ (i.e. the second point), draw $C_2D$ parallel to $C_5B$ meeting $AB$ in $D$.

Then $D$ is required point, which divides $AB$ segment into $2:3$ part.

Here is picture which demonstrate this procedure:

based on this information, I am trying to solve similar one

Draw a line segment 7 cm long. Divide it internally in the ratio $3 : 4$. Measure each part. Also write the steps of construction.

So as I understood, because we had to divide into ratio $2:3$ and we took $5$ points,in our case we have to take $7$ points right? Because of $3:4$ ration, with equal distance from starting $A$, of course first we should draw ray with acute angle,then took $7$ point,connect last point to one of the end of our segment (which equal to $7$cm) then choose $C_3$ and connect segment, intersection point would be desired point yes? Please tell me if I am wrong, also what should be length of each small segments, by which we divide large one?

-
yes it is right –  dato datuashvili Apr 25 '12 at 11:29
By small segments you mean those with the seven point (and A) as endpoints? It does not matter what their length is, anything constructable. Just make them the same. –  Chris K. Caldwell Apr 25 '12 at 12:16
so my argument is correct? –  dato datuashvili Apr 25 '12 at 13:25
@dato: Yes, your method is correct for the ratio $3:4$, and generalizes in a straightforward way to the ratio $m:n$, where $m$ and $n$ are positive integers. It all comes down to properties of parallel lines (you get equal angles, so similar triangles), or to put it another way, properties of scaling. –  André Nicolas Apr 25 '12 at 14:36
thanks very much @André Nicolas –  dato datuashvili Apr 26 '12 at 6:27


text: 16 added 157 characters in body edited Jan 26 '15 at 21:21 chris 12.7k44 gold badges4545 silver badges121121 bronze badges In that case, you can either use a parametric fit (as suggested by @stevenvh) or a non-parametric fit, say, using splinesSplines (or any parametrisation for which you can weight the relative degree of smoothness of your fit). Let me illustrate this, starting with your dataset: Now the original solution iswas to use (Hermite) interpolation; but letslet's also define a Spline interpolation we can visually see that the grey curve (the splineSpline) provides a better representation of the derivative at the origin. Indeed is closer to the underlying value of -0.0037-0.0037, than the value returned by the original method (red dashed curve): A cubic B-splineSpline curve with 2525 control points seems reasonable here for fitting that particular curve: ShowLet me show the data with the curve: The main virtue of this second example is that we now have a control parameter, namely the number of control points, which we can adapt to the level of noise in order to get an accurate estimate of the derivative. More generally, fitting with a variable amount of smoothing can be formalized in the context of Maximum a Posteriori analysisMaximum a Posteriori analysis. There are even methods (such as cross validation) to estimate automatically what the proper amount of smoothing should be. As an alternative to using the number of splinesSplines basis to control the smoothness of the curve, let us consider an explicit penalty function. The idea here is that instead of solving for the best (spline) weights in the least square sense (a maximum likelihood solution solution while assuming Gaussian statistics), we find these, subject to a prior corresponding to a 'roughness' penalty (which allows us to tune how smooth the spine function should be, which involves adding a tunable cost to unsmooth spline). In that case you can either use a parametric fit (as suggested by @stevenvh) or a non-parametric fit, say, using splines (or any parametrisation for which you can weight the relative degree of smoothness of your fit). Let me illustrate this, starting with your dataset: Now the original solution is to use (Hermite) interpolation; but lets also define a Spline interpolation we can visually see that the grey curve (the spline) provides a better representation of the derivative at the origin. Indeed is closer to the underlying value of -0.0037, than the value returned by the original method (red dashed curve) A cubic B-spline curve with 25 control points seems reasonable here for fitting that particular curve: Show the data with the curve: The main virtue of this second example is that we now have a control parameter, namely the number of control points, which we can adapt to the level of noise in order to get an accurate estimate of the derivative. More generally, fitting with a variable amount of smoothing can be formalized in the context of Maximum a Posteriori analysis. There are even methods (such as cross validation) to estimate automatically what the proper amount of smoothing should be. As an alternative to using the number of splines basis to control the smoothness of the curve, let us consider an explicit penalty function. The idea here is that instead of solving for the best (spline) weights in the least square sense (a maximum likelihood solution solution while assuming Gaussian statistics), we find these, subject to a prior corresponding to a 'roughness' penalty (which allows us to tune how smooth the spine function should be, which involves adding a tunable cost to unsmooth spline). In that case, you can either use a parametric fit (as suggested by @stevenvh) or a non-parametric fit, say, using Splines (or any parametrisation for which you can weight the relative degree of smoothness of your fit). Let me illustrate this, starting with your dataset: Now the original solution was to use (Hermite) interpolation; but let's also define a Spline interpolation we can visually see that the grey curve (the Spline) provides a better representation of the derivative at the origin. Indeed is closer to the underlying value of -0.0037, than the value returned by the original method (red dashed curve): A cubic B-Spline curve with 25 control points seems reasonable here for fitting that particular curve: Let me show the data with the curve: The main virtue of this second example is that we now have a control parameter, namely the number of control points, which we can adapt to the level of noise in order to get an accurate estimate of the derivative. More generally, fitting with a variable amount of smoothing can be formalized in the context of Maximum a Posteriori analysis. There are even methods (such as cross validation) to estimate automatically what the proper amount of smoothing should be. As an alternative to using the number of Splines basis to control the smoothness of the curve, let us consider an explicit penalty function. The idea here is that instead of solving for the best (spline) weights in the least square sense (a maximum likelihood solution solution while assuming Gaussian statistics), we find these, subject to a prior corresponding to a 'roughness' penalty (which allows us to tune how smooth the spine function should be, which involves adding a tunable cost to unsmooth spline). 15 added 98 characters in body edited Jun 8 '13 at 10:26 chris 12.7k44 gold badges4545 silver badges121121 bronze badges In that case you can either use a parametric fit (as suggested by @stevenvh) or a non parametric-parametric fit, say, using splines (or any parametrisation for which you can weight the relative degree of smoothness of your fit). Let me illustrate this, starting with your dataset.: LetsLet us define a noisy set: Butbut lets also define a Spline interpolation is closer to the underlying value of -0.0037, than the value returned by the original method (red dashed curve) f0'[0] (* 0.0120881 *).  Of course, statistically they both yield something reasonable BSplineFunction[ctrlpts, SplineDegree -> 3]'[0] // #[[2]]/#[[1]] & (* 39.7595 *),  which, given the amount of noise, is a reasonable estimate. On average, the non parametric-parametric estimate is fine: The main virtue of this second example is that now we now have a control parameter, namely the number of control points, which we can adapt to the level of noise in order to get an accurate estimate of the derivative. More generally, fitting with a variable amount of smoothing can be formalized in the context of Maximum a Posteriori analysis. There are even methods (such as cross validation) to estimate automatically what the proper amount of smoothing should be. asAs an alternative to using the number of splines basis to control the smoothness of the curve, let us consider an explicit penalty function. The idea here is that instead of solving for the best (spline) weights in the least square sense (a maximum likelihood solution solution while assuming Gaussian statistics), we find these, subject to a prior corresponding to a roughness'roughness' penalty (which allows us to tune how smooth the spine function should be, which involves adding a tunable cost to unsmooth spline). LetsLet us start with an undersmoothunder-smooth solution (lambda=-4.25) betterBetter? As far as the estimated derivative is concerned, we can define with mathematica the derivative of our spline basis In closing, Wikipedia page on this topic is rather good. In that case you can either use a parametric fit (as suggested by @stevenvh) or a non parametric fit, say using splines (or any parametrisation for which you can weight the relative degree of smoothness of your fit). Let me illustrate this, starting with your dataset. Lets define a noisy set: But lets also define a Spline interpolation is closer to the underlying value of -0.0037 than the value returned by the original method (red dashed curve) f0'[0] (* 0.0120881 *)  Of course statistically they both yield something reasonable BSplineFunction[ctrlpts, SplineDegree -> 3]'[0] // #[[2]]/#[[1]] & (* 39.7595 *)  which, given the amount of noise, is a reasonable estimate. On average the non parametric estimate is fine: The main virtue of this second example is that now we have a control parameter, namely the number of control points which we can adapt to the level of noise in order to get an accurate estimate of the derivative. More generally, fitting with variable amount of smoothing can be formalized in the context of Maximum a Posteriori analysis. There are even methods (such as cross validation) to estimate automatically what the proper amount of smoothing should be. as an alternative to using the number of splines basis to control the smoothness of the curve, let us consider an explicit penalty function. The idea here is that instead of solving for the best (spline) weights in the least square sense (a maximum likelihood solution solution while assuming Gaussian statistics), we find these, subject to a prior corresponding to a roughness penalty (which allows us to tune how smooth the spine function should be, which involves adding a tunable cost to unsmooth spline). Lets start with an undersmooth solution (lambda=-4.25) better? As far as the estimated derivative is concerned, we can define the derivative of our spline basis In that case you can either use a parametric fit (as suggested by @stevenvh) or a non-parametric fit, say, using splines (or any parametrisation for which you can weight the relative degree of smoothness of your fit). Let me illustrate this, starting with your dataset: Let us define a noisy set: but lets also define a Spline interpolation is closer to the underlying value of -0.0037, than the value returned by the original method (red dashed curve) f0'[0] (* 0.0120881 *).  Of course, statistically they both yield something reasonable BSplineFunction[ctrlpts, SplineDegree -> 3]'[0] // #[[2]]/#[[1]] & (* 39.7595 *),  which, given the amount of noise, is a reasonable estimate. On average, the non-parametric estimate is fine: The main virtue of this second example is that we now have a control parameter, namely the number of control points, which we can adapt to the level of noise in order to get an accurate estimate of the derivative. More generally, fitting with a variable amount of smoothing can be formalized in the context of Maximum a Posteriori analysis. There are even methods (such as cross validation) to estimate automatically what the proper amount of smoothing should be. As an alternative to using the number of splines basis to control the smoothness of the curve, let us consider an explicit penalty function. The idea here is that instead of solving for the best (spline) weights in the least square sense (a maximum likelihood solution solution while assuming Gaussian statistics), we find these, subject to a prior corresponding to a 'roughness' penalty (which allows us to tune how smooth the spine function should be, which involves adding a tunable cost to unsmooth spline). Let us start with an under-smooth solution (lambda=-4.25) Better? As far as the estimated derivative is concerned, we can define with mathematica the derivative of our spline basis In closing, Wikipedia page on this topic is rather good. 14 added 1 characters in body edited Sep 24 '12 at 14:22 chris 12.7k44 gold badges4545 silver badges121121 bronze badges  df[x_] = BSplineFunction[ctrlpts[0.25], SplineDegree -> 3]'[x]3]''[x]   df[x_] = BSplineFunction[ctrlpts[0.25], SplineDegree -> 3]'[x]   df[x_] = BSplineFunction[ctrlpts[0.25], SplineDegree -> 3]''[x]  13 added 127 characters in body edited Sep 24 '12 at 13:45 chris 12.7k44 gold badges4545 silver badges121121 bronze badges 12 added 8 characters in body edited Sep 24 '12 at 13:01 chris 12.7k44 gold badges4545 silver badges121121 bronze badges Mod Removes Wiki by J. M. will be back soon♦ occurred Sep 24 '12 at 10:16 11 added explicit penalty edited Sep 24 '12 at 10:02 chris 12.7k44 gold badges4545 silver badges121121 bronze badges 10 added explicit penalty edited Sep 24 '12 at 9:55 chris 12.7k44 gold badges4545 silver badges121121 bronze badges 9 added 66 characters in body edited Sep 23 '12 at 18:07 chris 12.7k44 gold badges4545 silver badges121121 bronze badges 8 added 66 characters in body edited Sep 23 '12 at 18:02 chris 12.7k44 gold badges4545 silver badges121121 bronze badges 7 added 394 characters in body edited Sep 23 '12 at 17:08 chris 12.7k44 gold badges4545 silver badges121121 bronze badges 6 added 215 characters in body edited Sep 23 '12 at 16:19 chris 12.7k44 gold badges4545 silver badges121121 bronze badges 5 added 215 characters in body edited Sep 23 '12 at 16:13 chris 12.7k44 gold badges4545 silver badges121121 bronze badges 4 building example on doc edited Sep 23 '12 at 16:06 chris 12.7k44 gold badges4545 silver badges121121 bronze badges 3 added 115 characters in body edited Sep 23 '12 at 15:56 chris 12.7k44 gold badges4545 silver badges121121 bronze badges 2 deleted 2 characters in body edited Sep 23 '12 at 15:39 chris 12.7k44 gold badges4545 silver badges121121 bronze badges 1 answered Sep 23 '12 at 15:34 chris 12.7k44 gold badges4545 silver badges121121 bronze badges


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Some Approximation Bounds for Deep Networks

# Some Approximation Bounds for Deep Networks

\NameBrendan McCane \Emailmccane@cs.otago.ac.nz
\NameLech Szymanski \Emaillech@cs.otago.ac.nz
University of Otago    Dunedin    New Zealand
###### Abstract

In this paper we introduce new bounds on the approximation of functions in deep networks and in doing so introduce some new deep network architectures for function approximation. These results give some theoretical insight into the success of autoencoders and ResNets.

D
\jmlrproceedings

eep Networks; Function Approximation.

## 1 Introduction

Deep networks have been shown to be more efficient than shallow for certain classes of problems: periodic functions (Szymanski and McCane, 2014); radially symmetric functions (Eldan and Shamir, 2016); and hierarchial compositional functions (Mhaskar and Poggio, 2016). Other work has shown that deep networks can efficiently represent low-dimensional manifolds (Basri and Jacobs, 2016; Shaham et al., 2016), and Telgarsky (2016) shows that there exist functions which cannot be efficiently represented with shallow networks.

All is not lost for shallow networks however. Mhaskar (1996) gives bounds for the approximation of Sobolev functions using shallow networks and shows that these bounds are tight. There appear to be no similar bounds for deep networks on this class of functions. One might naturally ask if deep networks can approximate this class of functions with similar bounds, or if shallow networks are demonstrably superior in this case. If the former, then one may conclude that a deep network is never worse than a shallow counterpart and hence we should always favour deep networks. If the latter, then choosing a shallow network may often be a good choice. This paper goes some way to answering this question by establishing upper bounds of approximation on some specific deep network architectures.

## 2 Definitions

###### Definition 1 (Vn)

The set of all networks of a given kind with complexity N (the number of units in the network).

###### Definition 2 (||f||∞)

Norm of a function. Let be the unit cube in dimensions. Let be the space of all continuous functions on with:

 ||f||∞=maxx∈In|f(x)|. (1)
###### Definition 3 (Degree of approximation)

If is the unknown function to be approximated, then the distance between and an approximating network is:

 dist(f,VN)=infP∈VN||f−P||∞. (2)
###### Definition 4 (Wm,∞(In))

A Sobolev space. Let be an integer. Let be the set of all functions of variables with continuous partial derivatives of orders up to such that:

 ||f||∞+∑1≤|k|1≤m||Dk||∞≤1, (3)

where denotes the partial derivative indicated by the multi-integer , and is the sum of the components of .

###### Definition 5 (σ)

The transfer function. Let be infinitely differentiable, and not a polynomial on any subinterval of . Further, we restrict ourselves to . Many common smooth transfer functions satisfy these conditions including the logistic function, , and softplus.

###### Definition 6 (SN,n)

The class of all shallow networks with units and inputs. Let denote the class of shallow networks with units of the form:

 x↦N∑k=1akσ(⟨wk,x⟩+bk) (4)

where , . The number of trainable parameters in such a network is . Since , it should be obvious that .

In all the deep networks we consider, one neuron in each layer after the first hidden layer is identified as the function approximation neuron. This allows us to progressively approximate the function of interest.

###### Definition 7 (RN,n,l)

The class of input residual networks with units per layer, where is the number of layers, and is the layer of the network. The first hidden layer of the network receives just the input coordinates. Each subsequent layer has the input coordinates and all the previous layer as input. See Figure 2.

###### Definition 8 (DNl,n,l)

The class of cascade residual networks with units per layer, where is the number of layers, and is the layer of the network. The first hidden layer of the network receives just the input coordinates. Each subsequent layer has the input coordinates and the function approximation neuron of the previous layer as input. See Figure 1.

###### Definition 9 (LN≥l,n,l)

The class of fully connected layer networks with units per layer. Each layer is fully connected to the next layer except for the function approximation neuron which is connected to the function approximation neuron and final output neuron only. In this case the number of neurons in each layer exceeds the input dimension. See Figure 3.

###### Definition 10 (LN<l,n,l)

The class of fully connected layer networks with units per layer. See Figure 3.

## 3 Approximation Bounds

We take as our starting point Theorem 2.1 of Mhaskar (1996) also reported as Theorem 2.1(a) in Mhaskar and Poggio (2016) and reproduce it here:

###### Theorem 1 (Theorem 2.1(a) of Mhaskar and Poggio (2016))

Let be infinitely differentiable, and not a polynomial on any subinterval of . For :

 dist(f,SN,n)≤cN−m/n, (5)

for some constant .

We use the results of Theorem 1 to derive bounds for the deep network architectures defined above in the next three theorems.

###### Theorem 2 (Cascade Residual Network Approximation)

Let be infinitely differentiable, and not a polynomial on any subinterval of . For , and some constant :

 dist(f,RN,n,l) ≤clN−m(ln+1)/(n(n+1))l (6)
{proof}

From Theorem 1, choose , so that:

 ||f−g1||∞≤c1N−m/nl (7)

where is some constant. From the first layer, create a new function to approximate:

 f1=1c1N−m/nl(f−g1) (8)

is clearly in and note that , and therefore can be approximated with another single layer network (), leading to the following error of approximation:

 ||f1−g2||∞ ≤c2N−m/(n+1)l (9) ||1c1N−m/n(f−g1)−g2||∞ ≤c2N−m/(n+1)l (10) 1c1N−m/n||f−g1−c1N−m/ng2||∞ ≤c2N−m/(n+1)l (11) ||f−g1−c1N−m/ng2||∞ ≤c1c2N−m(2n+1)/(n(n+1))l (12)

Repeat the procedure by creating a new function to approximate:

 f2=1c1c2N−m(2n+1)/(n(n+1))l(f−g1−c1N−m/n)lg2) (13)

Again and . Approximate with a further layer ():

 ||f2−g3||∞ ≤c3N−m/(n+1)l (14) ||1c1c2N−m(2n+1)/(n(n+1))l(f−g1−c1N−m/nlg2)−g3||∞ ≤c3N−m/(n+1)l (15) 1c1c2N−m(2n+1)/(n(n+1))l||f−g1−c1N−m/nlg2−c1c2N−m(2n+1)/(n(n+1))lg3||∞ ≤c3N−m/(n+1)l (16) ||f−g1−c1N−m/nlg2−c1c2N−m(2n+1)/(n(n+1))lg3||∞ ≤c1c2c3N−m(3n+1)/(n(n+1)) (17)

A simple inductive argument completes the proof.

Since (because , a constant function approximation would produce an error less than 1), it follows that the network will approximate the function exponentially fast in the number of layers.

###### Theorem 3 (Fully Connected Layer Network Approximation, Nl≥n)

Let be infinitely differentiable, and not a polynomial on any subinterval of . For , and some constant :

 dist(f,LN≥,n,l) ≤clN−m(ln+1)/(n(n+1))l (18)

if each layer is an invertible map.

{proof}

Define as the (invertible) mapping for layer 1, as the (invertible) mapping for layer i excluding the function approximation neuron (see Figure 3). The input to node is ; to is ; etc. Consider the input to . Since is an invertible map, if necessary, we could construct the input to as . Since this is identical to the situation in Theorem 2, the same result applies.

The next theorem deals with the case where . However, in this case a continuous invertible mapping is not possible. Instead we project coordinates into a lower dimensional space using a Hilbert curve mapping to maintain locality (nearby points in the lower dimensional space are nearby in the original space). Theoretically, we could do this with no loss of information, but unfortunately, this requires an infinite recursion, and therefore any computational procedure will induce an error in the new coordinates. Nevertheless, this error can be made small with a fixed cost projection.

A Hilbert curve can be defined by the centre coordinates of a hierarchically divided hypercube. See Figure 4 for a 2D example. The curve itself, up to level , can be constructed by recursively subdividing an initial square and creating line segments between appropriate centre points. The Hilbert curve itself is the limiting curve as goes to infinity and defines a continuous, but non-differentiable, onto mapping from dimensions to 1 dimension. There are several ways to generate Hilbert curve mappings both from dimensions to 1, and from 1 to dimensions. See Lawder (2000) for one efficient method. For level , the maximum difference between a point in and a point on the curve is .

###### Theorem 4 (Fully Connected Layer Network Approximation, Nl<n)

Let be infinitely differentiable, and not a polynomial on any subinterval of . For Lipschitz with Lipschitz constant , a Hilbert curve transform of level , and some constant :

 dist(f,LN<,n,l) ≤L√n−Nl+12k+1+clN−m(ln+1)/(n(n+1))l (19)
{proof}

The proof is straightforward. For the coordinate projection from dimensions to dimensions, choose the first dimensions and apply a Hilbert curve transformation. This induces a coordinate error less than and subsequently a function approximation error less than . We then apply Theorem 3 on the remaining coordinates along with the triangle inequality to prove the result.

## 4 Discussion

The constants, , in the theorems pose some difficulty since the errors are exponential in the number of layers (). It appears to be possible to estimate the size of these constants (Dupont and Scott, 1978, 1980), however the process is not straightforward and we have not attempted to estimate them. Nevertheless these theoretical results provide hints that for more general functions deep networks are never much worse than shallow networks. Given previous results showing that deep networks can be much better than shallow for specific function classes, it follows that there is little to lose in always choosing deep architectures (modulo the difficulties in learning deep networks).

These theoretical results also point toward layer-wise learning algorithms that reduce error exponentially fast in a manner that is somewhat analagous to AdaBoost like algorithms. We are currently investigating the practical implications of such algorithms.

For layer-wise learning of fully connected networks, Theorem 3 suggests that requiring invertible maps might be important. This might explain some of the success of autoencoders. Although Theorem 4 suggests that non-invertible maps might be able to achieve similar results, via space-filling curve mappings, it remains to be seen if such a scheme would be practical.

More recently, ResNets (He et al., 2016; Szegedy et al., 2017) of various types have been shown to outperform non-ResNet architectures with the most common argument given for their success being that it is easier for the gradients to propagate back to the inputs during learning. Theorem 2 and 3 together suggest a second reason may be that it is also easier to approximate residual functions if layers are skipped as there is no requirement that the layer mapping be invertible.

## References

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