Cauchy product

Cauchy product   (Redirected from Cesaro's theorem) Jump to navigation Jump to search In mathematics, more specifically in mathematical analysis, the Cauchy product is the discrete convolution of two infinite series. It is named after the French mathematician Augustin-Louis Cauchy.

Contents 1 Definitions 1.1 Cauchy product of two infinite series 1.2 Cauchy product of two power series 2 Convergence and Mertens' theorem 2.1 Example 2.2 Proof of Mertens' theorem 3 Cesàro's theorem 3.1 Theorem 4 Examples 5 Generalizations 5.1 Products of finitely many infinite series 5.2 Proof 6 Relation to convolution of functions 7 Notes 8 References Definitions The Cauchy product may apply to infinite series[1][2][3][4][5][6][7][8][9][10][11] or power series.[12][13] When people apply it to finite sequences[14] or finite series, that can be seen merely as a particular case of a product of series with a finite number of non-zero coefficients (see discrete convolution).

Convergence issues are discussed in the next section.

Cauchy product of two infinite series Let {textstyle sum _{i=0}^{infty }a_{i}} and {textstyle sum _{j=0}^{infty }b_{j}} be two infinite series with complex terms. The Cauchy product of these two infinite series is defined by a discrete convolution as follows: {displaystyle left(sum _{i=0}^{infty }a_{i}right)cdot left(sum _{j=0}^{infty }b_{j}right)=sum _{k=0}^{infty }c_{k}}     where     {displaystyle c_{k}=sum _{l=0}^{k}a_{l}b_{k-l}} . Cauchy product of two power series Consider the following two power series {displaystyle sum _{i=0}^{infty }a_{i}x^{i}}     and     {displaystyle sum _{j=0}^{infty }b_{j}x^{j}} with complex coefficients {displaystyle {a_{i}}} and {displaystyle {b_{j}}} . The Cauchy product of these two power series is defined by a discrete convolution as follows: {displaystyle left(sum _{i=0}^{infty }a_{i}x^{i}right)cdot left(sum _{j=0}^{infty }b_{j}x^{j}right)=sum _{k=0}^{infty }c_{k}x^{k}}     where     {displaystyle c_{k}=sum _{l=0}^{k}a_{l}b_{k-l}} . Convergence and Mertens' theorem Not to be confused with Mertens' theorems concerning distribution of prime numbers.

Let (an)n≥0 and (bn)n≥0 be real or complex sequences. It was proved by Franz Mertens that, if the series {textstyle sum _{n=0}^{infty }a_{n}} converges to A and {textstyle sum _{n=0}^{infty }b_{n}} converges to B, and at least one of them converges absolutely, then their Cauchy product converges to AB.[15] The theorem is still valid in a Banach algebra (see first line of the following proof).

It is not sufficient for both series to be convergent; if both sequences are conditionally convergent, the Cauchy product does not have to converge towards the product of the two series, as the following example shows: Example Consider the two alternating series with {displaystyle a_{n}=b_{n}={frac {(-1)^{n}}{sqrt {n+1}}},,} which are only conditionally convergent (the divergence of the series of the absolute values follows from the direct comparison test and the divergence of the harmonic series). The terms of their Cauchy product are given by {displaystyle c_{n}=sum _{k=0}^{n}{frac {(-1)^{k}}{sqrt {k+1}}}cdot {frac {(-1)^{n-k}}{sqrt {n-k+1}}}=(-1)^{n}sum _{k=0}^{n}{frac {1}{sqrt {(k+1)(n-k+1)}}}} for every integer n ≥ 0. Since for every k ∈ {0, 1, ..., n} we have the inequalities k + 1 ≤ n + 1 and n – k + 1 ≤ n + 1, it follows for the square root in the denominator that √(k + 1)(n − k + 1) ≤ n +1, hence, because there are n + 1 summands, {displaystyle |c_{n}|geq sum _{k=0}^{n}{frac {1}{n+1}}=1} for every integer n ≥ 0. Therefore, cn does not converge to zero as n → ∞, hence the series of the (cn)n≥0 diverges by the term test.

Proof of Mertens' theorem For simplicity, we will prove it for complex numbers. However, the proof we are about to give is formally identical for an arbitrary Banach algebra (not even commutativity or associativity is required).

Assume without loss of generality that the series {textstyle sum _{n=0}^{infty }a_{n}} converges absolutely. Define the partial sums {displaystyle A_{n}=sum _{i=0}^{n}a_{i},quad B_{n}=sum _{i=0}^{n}b_{i}quad {text{and}}quad C_{n}=sum _{i=0}^{n}c_{i}} with {displaystyle c_{i}=sum _{k=0}^{i}a_{k}b_{i-k},.} Then {displaystyle C_{n}=sum _{i=0}^{n}a_{n-i}B_{i}} by rearrangement, hence {displaystyle C_{n}=sum _{i=0}^{n}a_{n-i}(B_{i}-B)+A_{n}B,.}         (1) Fix ε > 0. Since {textstyle sum _{kin mathbb {N} }|a_{k}|-1} and {textstyle s>-1} , suppose the sequence {textstyle (a_{n})_{ngeq 0}} is {textstyle (C,;r)} summable with sum A and {textstyle (b_{n})_{ngeq 0}} is {textstyle (C,;s)} summable with sum B. Then their Cauchy product is {textstyle (C,;r+s+1)} summable with sum AB.

Examples For some {textstyle x,yin mathbb {R} } , let {textstyle a_{n}=x^{n}/n!} and {textstyle b_{n}=y^{n}/n!} . Then {displaystyle c_{n}=sum _{i=0}^{n}{frac {x^{i}}{i!}}{frac {y^{n-i}}{(n-i)!}}={frac {1}{n!}}sum _{i=0}^{n}{binom {n}{i}}x^{i}y^{n-i}={frac {(x+y)^{n}}{n!}}} by definition and the binomial formula. Since, formally, {textstyle exp(x)=sum a_{n}} and {textstyle exp(y)=sum b_{n}} , we have shown that {textstyle exp(x+y)=sum c_{n}} . Since the limit of the Cauchy product of two absolutely convergent series is equal to the product of the limits of those series, we have proven the formula {textstyle exp(x+y)=exp(x)exp(y)} for all {textstyle x,yin mathbb {R} } . As a second example, let {textstyle a_{n}=b_{n}=1} for all {textstyle nin mathbb {N} } . Then {textstyle c_{n}=n+1} for all {displaystyle nin mathbb {N} } so the Cauchy product {displaystyle sum c_{n}=(1,1+2,1+2+3,1+2+3+4,dots )} does not converge. Generalizations All of the foregoing applies to sequences in {textstyle mathbb {C} } (complex numbers). The Cauchy product can be defined for series in the {textstyle mathbb {R} ^{n}} spaces (Euclidean spaces) where multiplication is the inner product. In this case, we have the result that if two series converge absolutely then their Cauchy product converges absolutely to the inner product of the limits.

Products of finitely many infinite series Let {displaystyle nin mathbb {N} } such that {displaystyle ngeq 2} (actually the following is also true for {displaystyle n=1} but the statement becomes trivial in that case) and let {textstyle sum _{k_{1}=0}^{infty }a_{1,k_{1}},ldots ,sum _{k_{n}=0}^{infty }a_{n,k_{n}}} be infinite series with complex coefficients, from which all except the {displaystyle n} th one converge absolutely, and the {displaystyle n} th one converges. Then the limit {displaystyle lim _{Nto infty }sum _{k_{1}+ldots +k_{n}leq N}a_{1,k_{1}}cdots a_{n,k_{n}}} exists and we have: {displaystyle prod _{j=1}^{n}left(sum _{k_{j}=0}^{infty }a_{j,k_{j}}right)=lim _{Nto infty }sum _{k_{1}+ldots +k_{n}leq N}a_{1,k_{1}}cdots a_{n,k_{n}}} Proof Because {displaystyle forall Nin mathbb {N} :sum _{k_{1}+ldots +k_{n}leq N}a_{1,k_{1}}cdots a_{n,k_{n}}=sum _{k_{1}=0}^{N}sum _{k_{2}=0}^{k_{1}}cdots sum _{k_{n}=0}^{k_{n-1}}a_{1,k_{n}}a_{2,k_{n-1}-k_{n}}cdots a_{n,k_{1}-k_{2}}} the statement can be proven by induction over {displaystyle n} : The case for {displaystyle n=2} is identical to the claim about the Cauchy product. This is our induction base.

The induction step goes as follows: Let the claim be true for an {displaystyle nin mathbb {N} } such that {displaystyle ngeq 2} , and let {textstyle sum _{k_{1}=0}^{infty }a_{1,k_{1}},ldots ,sum _{k_{n+1}=0}^{infty }a_{n+1,k_{n+1}}} be infinite series with complex coefficients, from which all except the {displaystyle n+1} th one converge absolutely, and the {displaystyle n+1} -th one converges. We first apply the induction hypothesis to the series {textstyle sum _{k_{1}=0}^{infty }|a_{1,k_{1}}|,ldots ,sum _{k_{n}=0}^{infty }|a_{n,k_{n}}|} . We obtain that the series {displaystyle sum _{k_{1}=0}^{infty }sum _{k_{2}=0}^{k_{1}}cdots sum _{k_{n}=0}^{k_{n-1}}|a_{1,k_{n}}a_{2,k_{n-1}-k_{n}}cdots a_{n,k_{1}-k_{2}}|} converges, and hence, by the triangle inequality and the sandwich criterion, the series {displaystyle sum _{k_{1}=0}^{infty }left|sum _{k_{2}=0}^{k_{1}}cdots sum _{k_{n}=0}^{k_{n-1}}a_{1,k_{n}}a_{2,k_{n-1}-k_{n}}cdots a_{n,k_{1}-k_{2}}right|} converges, and hence the series {displaystyle sum _{k_{1}=0}^{infty }sum _{k_{2}=0}^{k_{1}}cdots sum _{k_{n}=0}^{k_{n-1}}a_{1,k_{n}}a_{2,k_{n-1}-k_{n}}cdots a_{n,k_{1}-k_{2}}} converges absolutely. Therefore, by the induction hypothesis, by what Mertens proved, and by renaming of variables, we have: {displaystyle {begin{aligned}prod _{j=1}^{n+1}left(sum _{k_{j}=0}^{infty }a_{j,k_{j}}right)&=left(sum _{k_{n+1}=0}^{infty }overbrace {a_{n+1,k_{n+1}}} ^{=:a_{k_{n+1}}}right)left(sum _{k_{1}=0}^{infty }overbrace {sum _{k_{2}=0}^{k_{1}}cdots sum _{k_{n}=0}^{k_{n-1}}a_{1,k_{n}}a_{2,k_{n-1}-k_{n}}cdots a_{n,k_{1}-k_{2}}} ^{=:b_{k_{1}}}right)\&=left(sum _{k_{1}=0}^{infty }overbrace {sum _{k_{2}=0}^{k_{1}}sum _{k_{3}=0}^{k_{2}}cdots sum _{k_{n}=0}^{k_{n-1}}a_{1,k_{n}}a_{2,k_{n-1}-k_{n}}cdots a_{n,k_{1}-k_{2}}} ^{=:a_{k_{1}}}right)left(sum _{k_{n+1}=0}^{infty }overbrace {a_{n+1,k_{n+1}}} ^{=:b_{k_{n+1}}}right)\&=left(sum _{k_{1}=0}^{infty }overbrace {sum _{k_{3}=0}^{k_{1}}sum _{k_{4}=0}^{k_{3}}cdots sum _{k_{n}+1=0}^{k_{n}}a_{1,k_{n+1}}a_{2,k_{n}-k_{n+1}}cdots a_{n,k_{1}-k_{3}}} ^{=:a_{k_{1}}}right)left(sum _{k_{2}=0}^{infty }overbrace {a_{n+1,k_{2}}} ^{=:b_{n+1,k_{2}}=:b_{k_{2}}}right)\&=left(sum _{k_{1}=0}^{infty }a_{k_{1}}right)left(sum _{k_{2}=0}^{infty }b_{k_{2}}right)\&=left(sum _{k_{1}=0}^{infty }sum _{k_{2}=0}^{k_{1}}a_{k_{2}}b_{k_{1}-k_{2}}right)\&=left(sum _{k_{1}=0}^{infty }sum _{k_{2}=0}^{k_{1}}left(overbrace {sum _{k_{3}=0}^{k_{2}}cdots sum _{k_{n}+1=0}^{k_{n}}a_{1,k_{n+1}}a_{2,k_{n}-k_{n+1}}cdots a_{n,k_{2}-k_{3}}} ^{=:a_{k_{2}}}right)left(overbrace {a_{n+1,k_{1}-k_{2}}} ^{=:b_{k_{1}-k_{2}}}right)right)\&=left(sum _{k_{1}=0}^{infty }sum _{k_{2}=0}^{k_{1}}overbrace {sum _{k_{3}=0}^{k_{2}}cdots sum _{k_{n}+1=0}^{k_{n}}a_{1,k_{n+1}}a_{2,k_{n}-k_{n+1}}cdots a_{n,k_{2}-k_{3}}} ^{=:a_{k_{2}}}overbrace {a_{n+1,k_{1}-k_{2}}} ^{=:b_{k_{1}-k_{2}}}right)\&=sum _{k_{1}=0}^{infty }sum _{k_{2}=0}^{k_{1}}a_{n+1,k_{1}-k_{2}}sum _{k_{3}=0}^{k_{2}}cdots sum _{k_{n+1}=0}^{k_{n}}a_{1,k_{n+1}}a_{2,k_{n}-k_{n+1}}cdots a_{n,k_{2}-k_{3}}end{aligned}}} Therefore, the formula also holds for {displaystyle n+1} .

Relation to convolution of functions A finite sequence can be viewed as an infinite sequence with only finitely many nonzero terms, or in other words as a function {displaystyle f:mathbb {N} to mathbb {C} } with finite support. For any complex-valued functions f, g on {displaystyle mathbb {N} } with finite support, one can take their convolution: {displaystyle (f*g)(n)=sum _{i+j=n}f(i)g(j).} Then {textstyle sum (f*g)(n)} is the same thing as the Cauchy product of {textstyle sum f(n)} and {textstyle sum g(n)} .

More generally, given a unital semigroup S, one can form the semigroup algebra {displaystyle mathbb {C} [S]} of S, with the multiplication given by convolution. If one takes, for example, {displaystyle S=mathbb {N} ^{d}} , then the multiplication on {displaystyle mathbb {C} [S]} is a generalization of the Cauchy product to higher dimension.

Notes ^ Canuto & Tabacco 2015, p. 20. ^ Bloch 2011, p. 463. ^ Friedman & Kandel 2011, p. 204. ^ Ghorpade & Limaye 2006, p. 416. ^ Hijab 2011, p. 43. ^ Montesinos, Zizler & Zizler 2015, p. 98. ^ Oberguggenberger & Ostermann 2011, p. 322. ^ Pedersen 2015, p. 210. ^ Ponnusamy 2012, p. 200. ^ Pugh 2015, p. 210. ^ Sohrab 2014, p. 73. ^ Canuto & Tabacco 2015, p. 53. ^ Mathonline, Cauchy Product of Power Series. ^ Weisstein, Cauchy Product. ^ Rudin, Walter (1976). Principles of Mathematical Analysis. McGraw-Hill. p. 74. References Apostol, Tom M. (1974), Mathematical Analysis (2nd ed.), Addison Wesley, p. 204, ISBN 978-0-201-00288-1. Bloch, Ethan D. (2011), The Real Numbers and Real Analysis, Springer, ISBN 9780387721767. Canuto, Claudio; Tabacco, Anita (2015), Mathematical Analysis II (2nd ed.), Springer. Friedman, Menahem; Kandel, Abraham (2011), Calculus Light, Springer, ISBN 9783642178481. Ghorpade, Sudhir R.; Limaye, Balmohan V. (2006), A Course in Calculus and Real Analysis, Springer. Hardy, G. H. (1949), Divergent Series, Oxford University Press, p. 227–229. Hijab, Omar (2011), Introduction to Calculus and Classical Analysis (3rd ed.), Springer. Mathonline, Cauchy Product of Power Series. Montesinos, Vicente; Zizler, Peter; Zizler, Václav (2015), An Introduction to Modern Analysis, Springer. Oberguggenberger, Michael; Ostermann, Alexander (2011), Analysis for Computer Scientists, Springer. Pedersen, Steen (2015), From Calculus to Analysis, Springer. Ponnusamy, S. (2012), Foundations of Mathematical Analysis, Birkhäuser, ISBN 9780817682927. Pugh, Charles C. (2015), Real Mathematical Analysis (2nd ed.), Springer. Sohrab, Houshang H. (2014), Basic Real Analysis (2nd ed.), Birkhäuser. Weisstein, Eric W., "Cauchy Product", From MathWorld – A Wolfram Web Resource. Categories: Augustin-Louis CauchyReal analysisComplex analysisSequences and series

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