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floor sum
https://forumdematematica.org/viewtopic.php?f=22&t=171		 Página 1 de 1
Autor:  kinu [ 28 jan 2012, 05:45 ]
Título da Pergunta:  floor sum
Calculate Sum of \(\left[\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{5}}+\frac{1}{\sqrt[3]{6}}+................+\frac{1}{\sqrt[3]{1000000}}\right]\)

where \(\left[x\right] =\) Greatest Integer function
Autor:  Rui Carpentier [ 30 jan 2012, 15:19 ]
Título da Pergunta:  Re: floor sum
Hint:
Use the partition \(\{3, 4, \dots ,1000000\}\) for the interval \([3,1000000]\) and show (by using superior and inferior sums) that:

\(\int_{3}^{1000000}\frac{1}{\sqrt[3]{x}}dx-(\frac{1}{\sqrt[3]{3}}-\frac{1}{\sqrt[3]{1000000}}) <\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{5}}+\cdots+\frac{1}{\sqrt[3]{1000000}} < \int_{3}^{1000000}\frac{1}{\sqrt[3]{x}}dx\)
Autor:  kinu [ 02 fev 2012, 17:38 ]
Título da Pergunta:  Re: floor sum
Thanks Rui Carpentier

would you like to explain in detail
Autor:  Rui Carpentier [ 02 fev 2012, 20:33 ]
Título da Pergunta:  Re: floor sum
Given a partition \(\{x_0,\dots ,x_n\}\) of an interval \([a,b]\) we know that:

\(\sum_{i=0}^{n-1} m_i(x_{i+1}-x_i)\leq\int_a^b f(x)dx\leq\sum_{i=0}^{n-1} M_i(x_{i+1}-x_i)\) where \(m_i=\inf \{f(x):x\in [x_i,x_{i+1}]\}\) and \(M_i=\sup \{f(x):x\in [x_i,x_{i+1}]\}\)

In our case, where \(f(x)=\frac{1}{\sqrt[3]{x}}\), \([a,b]=[3,1000000]\) and the partition is \(\{3,4,5,\dots ,1000000\}\) (i.e. \(x_{i+1}=x_i+1\)), we have

\(\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{5}}+\cdots +\frac{1}{\sqrt[3]{1000000}}\leq \int_3^{1000000}\frac{1}{\sqrt[3]{x}}dx\leq \frac{1}{\sqrt[3]{3}}+\frac{1}{\sqrt[3]{4}}+\cdots +\frac{1}{\sqrt[3]{999999}}\)

(note that \(\inf \{\frac{1}{\sqrt[3]{x}}:x\in [x_i,x_{i+1}]\}=\frac{1}{\sqrt[3]{x_{i+1}}}\) and \(\sup \{\frac{1}{\sqrt[3]{x}}:x\in [x_i,x_{i+1}]\}=\frac{1}{\sqrt[3]{x_i}}\))

Since \(\frac{1}{\sqrt[3]{3}}+\frac{1}{\sqrt[3]{4}}+\cdots +\frac{1}{\sqrt[3]{999999}}=\frac{1}{\sqrt[3]{4}}+\cdots +\frac{1}{\sqrt[3]{1000000}}+\frac{1}{\sqrt[3]{3}}-\frac{1}{\sqrt[3]{1000000}}\), we have that

\(\int_3^{1000000}\frac{1}{\sqrt[3]{x}}dx -\frac{1}{\sqrt[3]{3}}+\frac{1}{\sqrt[3]{1000000}}\leq \frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{5}}+\cdots +\frac{1}{\sqrt[3]{1000000}}\).

Thus, \(\int_3^{1000000}\frac{1}{\sqrt[3]{x}}dx -\frac{1}{\sqrt[3]{3}}+\frac{1}{\sqrt[3]{1000000}}\leq \frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{5}}+\cdots +\frac{1}{\sqrt[3]{1000000}}\leq \int_3^{1000000}\frac{1}{\sqrt[3]{x}}dx\).

Since \(\int_3^{1000000}\frac{1}{\sqrt[3]{x}}dx =\frac{3\sqrt[3]{1000000}^2}{2}-\frac{3\sqrt[3]{3}^2}{2}=1500-\frac{3\sqrt[3]{3}^2}{2}=1496.87987...\) and \(\int_3^{1000000}\frac{1}{\sqrt[3]{x}}dx -\frac{1}{\sqrt[3]{3}}+\frac{1}{\sqrt[3]{1000000}}=1496.19651...\), the answer is 1496.
Autor:  kinu [ 06 fev 2012, 17:53 ]
Título da Pergunta:  Re: floor sum
Thanks Rui Carpenter
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