Question Number 213398 by issac last updated on 04/Nov/24
$$\mathrm{One}\:\mathrm{simple}\:\mathrm{Equation} \\ $$$$\mathrm{pls}\:\mathrm{prove}\:\mathrm{this}\:\mathrm{property} \\ $$$$\underset{{j}=\mathrm{1}} {\overset{{N}} {\sum}}\:{a}_{{j}} \centerdot\underset{{k}=\mathrm{1}} {\overset{{M}} {\sum}}{b}_{{k}} =\underset{{j}=\mathrm{1}} {\overset{{N}} {\sum}}\centerdot\underset{{k}=\mathrm{1}} {\overset{{M}} {\sum}}\:{a}_{{j}} {b}_{{k}} .. \\ $$$$\:\:\mathrm{and} \\ $$$$\underset{{j}=\mathrm{0}} {\overset{{N}} {\sum}}\:{f}\left({a}+\frac{{b}−{a}}{{N}}{j}\right)\centerdot\frac{{b}−{a}}{{N}}\centerdot\underset{{k}=\mathrm{0}} {\overset{{M}} {\sum}}\:\mathrm{g}\left({a}+\frac{{b}−{a}}{{M}}{k}\right)\centerdot\frac{{b}−{a}}{{M}} \\ $$$$=\underset{{j}=\mathrm{0}} {\overset{{N}} {\sum}}\:\underset{{k}=\mathrm{0}} {\overset{{M}} {\sum}}\:{f}\left({a}+\frac{{b}−{a}}{{N}}{j}\right)\mathrm{g}\left({a}+\frac{{b}−{a}}{{M}}{k}\right)\frac{\left({b}−{a}\right)^{\mathrm{2}} }{{MN}} \\ $$$$\mathrm{But}….. \\ $$$$\mathrm{that}\:\mathrm{Sum}\:\mathrm{not}\:\mathrm{euqal}\:\mathrm{to}\:\int_{{a}} ^{\:{b}} \:{f}\left({z}\right)\mathrm{g}\left({z}\right)\mathrm{d}{z}… \\ $$$$\mathrm{why}\:\mathrm{integral}\:\mathrm{form}\:\mathrm{dosen}'\mathrm{t}\:\mathrm{work} \\ $$$$\:\mathrm{like}\:\mathrm{Summation} \\ $$$$\underset{{j}=\mathrm{1}} {\overset{{N}} {\sum}}\:{f}\left({j}\right)\:\centerdot\underset{{k}=\mathrm{1}} {\overset{{M}} {\sum}}\:\mathrm{g}\left({k}\right)=\underset{{j}=\mathrm{1}} {\overset{{N}} {\sum}}\underset{{k}=\mathrm{1}} {\overset{{M}} {\sum}}\:{f}\left({j}\right)\mathrm{g}\left({k}\right) \\ $$$$\:\mathrm{is}\:\mathrm{True}\:\mathrm{But}….. \\ $$$$\int_{{a}} ^{{b}} \:{f}\left({u}\right)\mathrm{d}{u}\centerdot\:\int_{{a}} ^{{b}} \:\mathrm{g}\left({v}\right)\mathrm{d}{v}\:\:\mathrm{isn}'\mathrm{t}\:\mathrm{equal}\:\mathrm{to}\: \\ $$$$\int_{{a}} ^{{b}} \:{f}\left({w}\right)\mathrm{g}\left({w}\right)\mathrm{d}{w} \\ $$
Answered by MrGaster last updated on 04/Nov/24
$$\underset{{j}=\mathrm{1}} {\overset{{N}} {\sum}}{a}_{{j}} \centerdot\underset{{k}=\mathrm{1}} {\overset{{M}} {\sum}}{b}_{{k}} =\underset{{j}=\mathrm{1}} {\overset{{N}} {\sum}}\centerdot\underset{{k}=\mathrm{1}} {\overset{{M}} {\sum}}{a}_{{j}} {b}_{{k}} \\ $$$$\underset{{j}=\mathrm{0}} {\overset{{N}} {\sum}}{f}\left({a}+\frac{{b}−{a}}{{N}}{j}\right)\centerdot\frac{{b}−{a}}{{N}}\centerdot\underset{{k}=\mathrm{0}} {\overset{{M}} {\sum}}\mathrm{g}\left({a}+\frac{{b}−{a}}{{M}}{k}\right)\centerdot\frac{{b}−{a}}{{M}}\neq\int_{{a}} ^{{b}} {f}\left({z}\right)\mathrm{g}\left({z}\right){dz} \\ $$$$\int_{{a}} ^{{b}} {f}\left({u}\right){du}\centerdot\int_{{a}} ^{{b}} \mathrm{g}\left({v}\right){dv}\neq\int_{{a}} ^{{b}} {f}\left({w}\right)\mathrm{g}\left({w}\right){dw} \\ $$$$\int_{{a}} ^{{b}} {f}\left({u}\right){du}\centerdot\int_{{a}} ^{{b}} \mathrm{g}\left({v}\right)=\left(\int_{{a}} ^{{b}} {f}\left({v}\right){dv}\right)\left(\int_{{a}} ^{{b}} \mathrm{g}\left({v}\right){dv}\right) \\ $$$$=\int_{{a}} ^{{b}} \int_{{a}} ^{{b}} {f}\left({u}\right)\mathrm{g}\left({v}\right){dudv} \\ $$$$\neq\int_{{a}} ^{{b}} {f}\left({w}\right)\mathrm{g}\left({w}\right){dw} \\ $$
Commented by MrGaster last updated on 04/Nov/24
Ps:First, explain the first equation:
The distributive property of summation can be expressed as:
The sum over all j from 1 to N, multiplied by the sum over all k from 1 to M, equals the sum over all j from 1 to N and k from 1 to M of the product. Specifically:
(Σ from j=1 to N of a_j) * (Σ from k=1 to M of b_k) = Σ from j=1 to N and k=1 to M of (a_j * b_k)
This equation is correct because it demonstrates the distributive property of summation. The expression on the left is the product of the results of two summations, while the expression on the right is the sum of the products of each pair of a_j and b_k, and these two methods yield the same result.
Next, explain why the following expression does not equal the integral form:
(Σ from j=0 to N of f(a + (b-a)/N * j) * (b-a)/N) * (Σ from k=0 to M of g(a + (b-a)/M * k) * (b-a)/M) does not equal ∫ from a to b of f(z) * g(z) dz
This expression is the product of two Riemann sums, and it does not equal the product of the integrals of f(z) and g(z). This is because the product of integrals involves a double integral, whereas the product of Riemann sums is not equivalent to a double integral. Specifically, the product of Riemann sums is the product of the sums of two separate integrals, not a double integral.
Finally, explain why the following integral expression does not hold:
∫ from a to b of f(u) du * ∫ from a to b of g(v) dv does not equal ∫ from a to b of f(w) * g(w) dw
This inequality holds because the expression on the left is the product of two separate integrals, which equals a double integral:
∫ from a to b ∫ from a to b of f(u) * g(v) du dv
The expression on the right is a single integral of the product of f(w) and g(w). These two expressions are not equivalent because the double integral involves integrating over two variables, while the single integral involves integrating over only one variable. In general, these two expressions will not be equal unless f(u) and g(v) satisfy specific conditions, such as f(u) = g(u) or they are independent over the interval of integration.
Commented by Frix last updated on 04/Nov/24
$$\mathrm{chatGPT}? \\ $$
Answered by Frix last updated on 07/Nov/24
$$\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}\underset{{a}} {\overset{{b}} {\int}}{g}\left({x}\right){dx}=\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){g}\left({x}\right){dx} \\ $$$$\Rightarrow \\ $$$$\left(\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}\right)^{\mathrm{2}} =\underset{{a}} {\overset{{b}} {\int}}\left({f}\left({x}\right)\right)^{\mathrm{2}} {dx} \\ $$$$\mathrm{Simply}\:\mathrm{let}\:{f}\left({x}\right)={x}^{\alpha} \\ $$$$\:\:\:\:\:\left(\underset{{a}} {\overset{{b}} {\int}}{x}^{\alpha} {dx}\right)^{\mathrm{2}} =\frac{\left({b}^{\alpha+\mathrm{1}} −{a}^{\alpha+\mathrm{1}} \right)^{\mathrm{2}} }{\left(\alpha+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\underset{{a}} {\overset{{b}} {\int}}{x}^{\mathrm{2}\alpha} {dx}=\frac{{b}^{\mathrm{2}\alpha+\mathrm{1}} −{a}^{\mathrm{2}\alpha+\mathrm{1}} }{\mathrm{2}\alpha+\mathrm{1}} \\ $$$$\mathrm{How}\:\mathrm{could}\:\mathrm{they}\:\mathrm{be}\:\mathrm{equal}? \\ $$