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如果你也在线性代数linearalgebra这个学科遇到相关的难题,请随时添加vx号联系我们的代写客服。我们会为你提供专业的服务。 linearalgebra™长期致力于留学生网课服务,涵盖各个网络学科课程:金融学Finance,经济学Economics,数学Mathematics,会计Accounting,文学Literature,艺术Arts等等。除了网课全程托管外,linearalgebra™也可接受单独网课任务。无论遇到了什么网课困难,都能帮你完美解决! 0.27 Properties of Developed FunctionsThe practical utility of the process of development in the algebra of logic lies in the fact that developed functions possess the following property:The sum or the product of two functions developed with respect to the same letters is obtained simply by finding the sum or the product of their coefficients. The negative of a developed function is obtained simply by replacing the coefficients of its development by their negatives.We shall now demonstrate these propositions in the case of two variables; this demonstration will of course be of universal application.Let the developed functions be\[a_{1}xy+b_{1}xy^{\prime}+c_{1}x^{\prime}y+d_{1}x^{\prime}y^{ \prime},\] \[a_{2}xy+b_{2}xy^{\prime}+c_{2}x^{\prime}y+d_{2}x^{\prime}y^{ \prime}.\]1. I say that their sum is \[(a_{1}+a_{2})xy+(b_{1}+b_{2})xy^{\prime}+(c_{1}+c_{2})x^{\prime}y+(d_{1}+d_{2} )x^{\prime}y^{\prime}.\] This result is derived directly from the distributive law. 2. I say that their product is \[a_{1}a_{2}xy+b_{1}b_{2}xy^{\prime}+c_{1}c_{2}x^{\prime}y+d_{1}d_{2}x^{\prime} y^{\prime},\] for if we find their product according to the general rule (by applying the distributive law), the products of two terms of different constituents will be zero; therefore there will remain only the products of the terms of the same constituent, and, as (by the law of tautology) the product of this constituent multiplied by itself is equal to itself, it is only necessary to obtain the product of the coefficients.3. Finally, I say that the negative of\[axy+bxy^{\prime}+cx^{\prime}y+dx^{\prime}y^{\prime}\]is\[a^{\prime}xy+b^{\prime}xy^{\prime}+c^{\prime}x^{\prime}y+d^{\prime}x^{\prime}y^ {\prime}.\]In order to verify this statement, it is sufficient to prove that the product of these two functions is zero and that their sum is equal to 1. Thus\[(axy+bxy^{\prime}+cx^{\prime}y+dx^{\prime}y^{\prime})(a^{\prime} xy+b^{\prime}xy^{\prime}+c^{\prime}x^{\prime}y+d^{\prime}x^{\prime}y^{\prime})\] \[\qquad=(aa^{\prime}xy+bb^{\prime}xy^{\prime}+cc^{\prime}x^{ \prime}y+dd^{\prime}x^{\prime}y^{\prime})\] \[\qquad=(0\cdot xy+0\cdot xy^{\prime}+0\cdot x^{\prime}y+0\cdot x^ {\prime}y^{\prime})=0\] \[(axy+bxy^{\prime}+cx^{\prime}y+dx^{\prime}y^{\prime})+(a^{\prime }xy+b^{\prime}xy^{\prime}+c^{\prime}x^{\prime}y+d^{\prime}x^{\prime}y^{ \prime})\] \[\qquad=[(a+a^{\prime})xy+(b+b^{\prime})xy^{\prime}+(c+c^{\prime })x^{\prime}y+(d+d^{\prime})x^{\prime}y^{\prime}]\] \[\qquad=(1xy+1xy^{\prime}+1x^{\prime}y+1x^{\prime}y^{\prime})=1.\]_Special Case.–_We have the equalities:\[(ab+a^{\prime}b^{\prime})^{\prime} =ab^{\prime}+a^{\prime}b,\] \[(ab^{\prime}+a^{\prime}b^{\prime})^{\prime} =ab+a^{\prime}b^{\prime},\]which may easily be demonstrated in many ways; for instance, by observing that the two sums \((ab+a^{\prime}b^{\prime})\) and \((ab^{\prime}+a^{\prime}b)\) combined form the development of 1; or again by _performing_ the negation \((ab+a^{\prime}b^{\prime})^{\prime}\) by means of De Morgan’s formulas (SS0.25).From these equalities we can deduce the following equality:\[(ab^{\prime}+ab=0)=(ab+a^{\prime}b^{\prime}=1),\]which result might also have been obtained in another way by observing that (SS0.18)\[(a=b)=(ab^{\prime}+a^{\prime}b=0)=[(a+b^{\prime})(a^{\prime}+b)=1],\]and by performing the multiplication indicated in the last equality.Theorem.–_We have the following equivalences:32_W. Stanley Jevons, _Pure Logic_, 1864, p. 61.\[(a=bc^{\prime}+b^{\prime}c)=(b=ac^{\prime}+a^{\prime}c)=(c=ab^{\prime}+a^{ \prime}b).\]For, reducing the first of these equalities so that its second member will be 0,\[a(bc+b^{\prime}c^{\prime})+a^{\prime}(bc^{\prime}+b^{\prime}c) =0,\] \[abc+ab^{\prime}c^{\prime}+a^{\prime}bc^{\prime}+a^{\prime}b^{ \prime}c =0.\]Now it is clear that the first member of this equality is symmetrical with respect to the three terms \(a,b,c\). We may therefore conclude that, if the two other equalities which differ from the first only in the permutation of these three letters be similarly transformed, the same result will be obtained, which proves the proposed equivalence._Corollary.–_If we have at the same time the three inclusions:\[a

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