Use Euler's summation formula to prove the following....

In summary: Sorry, I made many mistakes earlier.## \int_{1}^{\infty}(t-[t])(\frac{1-3\log {t}}{t^4})dt=\int_{1}^{\infty}\frac{1-3\log {t}}{t^4}dt=\int_{1}^{\infty}\frac{1}{t^4}dt-\int_{1}^{\infty}\frac{3\log {t}}{t^4}dt=\frac{1}{3}-\frac{1}{3}=0 ##Also, the "[t]" from my textbook has no upper part when writing this, do you know the correct Latex command
  • #1
Math100
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Homework Statement
Use Euler's summation formula to prove that, for ## x\geq 2 ##,
## \sum_{n\leq x}\frac{ \log {n}}{n^3}=A-\frac{ \log {x}}{2x^2}-\frac{1}{4x^2}+O\frac{ \log {x}}{x^3} ##, where ## A ## is a constant.
Relevant Equations
None.
Proof:

Let ## x\geq 2 ##.
Then ## \frac{d}{dt}(\frac{ \log {t}}{t^3})=\frac{1-3\log {t}}{t^4} ##.
By Euler's summation formula, we have that
## \sum_{n\leq x}\frac{ \log {n}}{n^3}=\int_{1}^{x} \frac{\log {t}}{t^3}dt+\int_{1}^{x} (t-[t])(\frac{1-3\log {t}}{t^4})dt+(x-[x])\frac{log {x}}{x^3} ##
## =\frac{-\log {x}}{2x^2}-\frac{1}{4x^2}+\frac{1}{4}+(\int_{1}^{\infty}-\int_{x}^{\infty})(t-[t])\frac{1-3\log {t}}{t^4}dt+O(\frac{log {x}}{x^3}) ##.
Thus
\begin{align*}
&\left | \int_{x}^{\infty}(t-[t])\frac{1-3\log {t}}{t^4}dt \right |\leq 2\int_{x}^{\infty}\frac{\log {t}}{t^4}dt\\
&=2\cdot (\frac{1+3\log {x}}{9t^3})\\
&=O(\frac{\log {x}}{x^3}).\\
\end{align*}
Therefore, ## \sum_{n\leq x}\frac{\log {n}}{n^3}=A-\frac{\log {x}}{2x^2}-\frac{1}{4x^2}+O(\frac{\log {x}}{x^3}) ##, where ## A ## is a constant.
 
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  • #2
This needs a lot more explanation. Which formula did you use? I see

1664224917253.png


What I do not see is how this formula is used to get your step.
 
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  • #3
fresh_42 said:
This needs a lot more explanation. Which formula did you use? I see

View attachment 314701

What I do not see is how this formula is used to get your step.
Euler's summation formula:
If ## f ## has a continuous derivative ## f' ## on the interval ## [y, x] ##, where ## 0<y<x ##, then
## \sum_{y<n\leq x}f(n)=\int_{y}^{x}f(t)dt+\int_{y}^{x}(t-[t])f'(t)dt+f(x)([x]-x)-f(y)([y]-y) ##.

The formula which you see is something I've never seen before.
 
  • #4
Math100 said:
Euler's summation formula:
If ## f ## has a continuous derivative ## f' ## on the interval ## [y, x] ##, where ## 0<y<x ##, then
## \sum_{y<n\leq x}f(n)=\int_{y}^{x}f(t)dt+\int_{y}^{x}(t-[t])f'(t)dt+f(x)([x]-x)-f(y)([y]-y) ##.

The formula which you see is something I've never seen before.
O.k., let us assume this formula is correct, then you need to show

a) ##\displaystyle{ \int_1^\infty (t-[t])\left(\dfrac{1-3\log {t}}{t^4}\right)dt = O\left(\dfrac{\log x}{x^3}\right)}\\[30pt]##
b) ## \displaystyle{\left| \int_x^\infty (t-[t])\left(\dfrac{1-3\log {t}}{t^4}\right)dt\right|\leq 2 \int_x^\infty \dfrac{\log t}{t^4}\,dt}##
 
  • #5
fresh_42 said:
O.k., let us assume this formula is correct, then you need to show

a) ##\displaystyle{ \int_1^\infty (t-[t])\left(\dfrac{1-3\log {t}}{t^4}\right)dt = O\left(\dfrac{\log x}{x^3}\right)}\\[30pt]##
b) ## \displaystyle{\left| \int_x^\infty (t-[t])\left(\dfrac{1-3\log {t}}{t^4}\right)dt\right|\leq 2 \int_x^\infty \dfrac{\log t}{t^4}\,dt}##
## \left | \int_{x}^{\infty}(t-[t])(\frac{1-3\log {t}}{t^4})dt \right |\leq \int_{x}^{\infty}\frac{1-3\log {t}}{t^4}dt=-\frac{\log {x}}{x^3}=O(\frac{\log {x}}{x^3}) ##
 
  • #6
Math100 said:
## \left | \int_{x}^{\infty}(t-[t])(\frac{1-3\log {t}}{t^4})dt \right |\leq \int_{x}^{\infty}\frac{1-3\log {t}}{t^4}dt=-\frac{\log {x}}{x^3}=O(\frac{\log {x}}{x^3}) ##
Why? This cannot be since we have a positive number on the left and a negative on the right!
\begin{align*}
\left | \int_{x}^{\infty}(t-[t])\dfrac{1-3\log {t}}{t^4}dt \right | &\leq \int_{x}^\infty \left|t-[t]\right|\cdot \left|\dfrac{1-3\log {t}}{t^4} \right|\,dt \\
& \leq \int_{x}^\infty \left|\dfrac{1-3\log {t}}{t^4}\right|\,dt \\
&\leq \int_{x}^\infty \dfrac{1}{t^4}\,dt + 3 \int_{x}^\infty \dfrac{\log t}{t^4} \,dt\\
&\leq \dfrac{1}{3x^3} + \infty
\end{align*}
This means that we must take more care. We only have
\begin{align*}
\left | \int_{x}^{\infty}(t-[t])\dfrac{1-3\log {t}}{t^4}dt \right | &\leq \int_{x}^\infty \left|t-[t]\right|\cdot \left|\dfrac{1-3\log {t}}{t^4} \right|\,dt \\
& \leq \int_{x}^\infty \left|\dfrac{1-3\log {t}}{t^4}\right|\,dt
\end{align*}
And now? We only know that ##x>1##. So why have you split the integrals at all? What about the first integral ##\int_1^\infty## and how to deal with the zero of ##1-3\log t##?
 
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  • #7
fresh_42 said:
Why? This cannot be since we have a positive number on the left and a negative on the right!
\begin{align*}
\left | \int_{x}^{\infty}(t-[t])\dfrac{1-3\log {t}}{t^4}dt \right | &\leq \int_{x}^\infty \left|t-[t]\right|\cdot \left|\dfrac{1-3\log {t}}{t^4} \right|\,dt \\
& \leq \int_{x}^\infty \left|\dfrac{1-3\log {t}}{t^4}\right|\,dt \\
&\leq \int_{x}^\infty \dfrac{1}{t^4}\,dt + 3 \int_{x}^\infty \dfrac{\log t}{t^4} \,dt\\
&\leq \dfrac{1}{3x^3} + \infty
\end{align*}
This means that we must take more care. We only have
\begin{align*}
\left | \int_{x}^{\infty}(t-[t])\dfrac{1-3\log {t}}{t^4}dt \right | &\leq \int_{x}^\infty \left|t-[t]\right|\cdot \left|\dfrac{1-3\log {t}}{t^4} \right|\,dt \\
& \leq \int_{x}^\infty \left|\dfrac{1-3\log {t}}{t^4}\right|\,dt
\end{align*}
And now? We only know that ##x>1##. So why have you split the integrals at all? What about the first integral ##\int_1^\infty## and how to deal with the zero of ##1-3\log t##?
Sorry, I made many mistakes earlier.
## \int_{1}^{\infty}(t-[t])(\frac{1-3\log {t}}{t^4})dt=\int_{1}^{\infty}\frac{1-3\log {t}}{t^4}dt=\int_{1}^{\infty}\frac{1}{t^4}dt-\int_{1}^{\infty}\frac{3\log {t}}{t^4}dt=\frac{1}{3}-\frac{1}{3}=0 ##
Also, the "[t]" from my textbook has no upper part when writing this, do you know the correct Latex command for this? And what does this symbolize/indicate?
 
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  • #8
Math100 said:
Sorry, I made many mistakes earlier.
## \int_{1}^{\infty}(t-[t])(\frac{1-3\log {t}}{t^4})dt=\int_{1}^{\infty}\frac{1-3\log {t}}{t^4}dt=\int_{1}^{\infty}\frac{1}{t^4}dt-\int_{1}^{\infty}\frac{3\log {t}}{t^4}dt=\frac{1}{3}-\frac{1}{3}=0 ##
Also, the "[t]" from my textbook has no upper part when writing this, do you know the correct Latex command for this? And what does this symbolize/indicate?
And how did you get rid of ##(t-[t])## in the integrand?

You could use
\begin{align*}
\left|\int_1^\infty (t-[t])\left(\dfrac{1-3\log t}{t^4}\right)\right|\,dt &\leq \int_1^\infty |(t-[t])|\left|\dfrac{1-3\log t}{t^4}\right|\,dt\\
&\leq \int_1^\infty \left|\dfrac{1-3\log t}{t^4}\right|\,dt\\
&= \dfrac{2}{3e}\\
&= O\left(\dfrac{\log x}{x^3}\right)
\end{align*}
and explain how the integral is achieved, and you need the triangle inequality in your original formula since the absolute values have to come in from somewhere. But one step after the other. We still need an upper bound for the ##\int_x^\infty ## part.
\lfoor 5.3 \rfloor = [5.3] =5\, , \,\lceil 5.3 \rceil = 6 results in ##\lfloor 5.3 \rfloor = [5.3] =5\, , \,\lceil 5.3 \rceil = 6##
This means the use of ##[\ldots]## is o.k.
 
  • #9
fresh_42 said:
And how did you get rid of ##(t-[t])## in the integrand?

You could use
\begin{align*}
\left|\int_1^\infty (t-[t])\left(\dfrac{1-3\log t}{t^4}\right)\right|\,dt &\leq \int_1^\infty |(t-[t])|\left|\dfrac{1-3\log t}{t^4}\right|\,dt\\
&\leq \int_1^\infty \left|\dfrac{1-3\log t}{t^4}\right|\,dt\\
&= \dfrac{2}{3e}\\
&= O\left(\dfrac{\log x}{x^3}\right)
\end{align*}
and explain how the integral is achieved, and you need the triangle inequality in your original formula since the absolute values have to come in from somewhere. But one step after the other. We still need an upper bound for the ##\int_x^\infty ## part.
\lfoor 5.3 \rfloor = [5.3] =5\, , \,\lceil 5.3 \rceil = 6 results in ##\lfloor 5.3 \rfloor = [5.3] =5\, , \,\lceil 5.3 \rceil = 6##
This means the use of ##[\ldots]## is o.k.
I do not know either. I just ended up getting rid of it to evaluate the integral without it, and to see if it matches the answer. But do you know how to? And without the upper part in bracket, the floor function, how do these values work in here?
 
  • #10
From the beginning!

You have as relevant equations not none, but
$$
\sum_{y<n\leq x}f(n)=\int_{y}^{x}f(t)dt+\int_{y}^{x}(t-[t])f'(t)dt+f(x)([x]-x)-f(y)([y]-y)
$$

Applied to our sum, we get
\begin{align*}
\sum_{n\leq x}\dfrac{ \log {n}}{n^3}&=\int_{1}^{x}\dfrac{ \log {t}}{t^3}\,dt+\int_{1}^{x}(t-[t])\left(\dfrac{ \log {t}}{t^3}\right)'\,dt+\dfrac{ \log {x}}{x^3}([x]-x)-\dfrac{ \log {1}}{1^3}([1]-1)\\
&=\left(\dfrac{1}{4}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}\right)+\int_1^x(t-[t])\dfrac{1-3\log t}{t^4}\,dt +O\left(\dfrac{\log x}{x^3}\right)\\
&= \left(\dfrac{1}{4}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}\right)+\int_1^2(t-[t])\dfrac{1-3\log t}{t^4}\,dt+\underbrace{\int_2^x(t-|t|)\dfrac{1-3\log t}{t^4}\,dt}_{=O(x^{-3}\log x)}\\
&\phantom{=}+O\left(\dfrac{\log x}{x^3}\right)\\
&= \left(\dfrac{1}{4}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}\right) +\dfrac{\log(16)-3}{16}+O\left(\dfrac{\log x}{x^3}\right)\\
&=\underbrace{\dfrac{1}{4}+\dfrac{\log(16)-3}{16}}_{=:A}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}+O\left(\dfrac{\log x}{x^3}\right)
\end{align*}

I used WolframAlpha for the integrals. You should calculate them explicitly!
 
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  • #11
fresh_42 said:
From the beginning!

You have as relevant equations not none, but
$$
\sum_{y<n\leq x}f(n)=\int_{y}^{x}f(t)dt+\int_{y}^{x}(t-[t])f'(t)dt+f(x)([x]-x)-f(y)([y]-y)
$$

Applied to our sum, we get
\begin{align*}
\sum_{n\leq x}\dfrac{ \log {n}}{n^3}&=\int_{1}^{x}\dfrac{ \log {t}}{t^3}\,dt+\int_{1}^{x}(t-[t])\left(\dfrac{ \log {t}}{t^3}\right)'\,dt+\dfrac{ \log {x}}{x^3}([x]-x)-\dfrac{ \log {1}}{1^3}([1]-1)\\
&=\left(\dfrac{1}{4}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}\right)+\int_1^x(t-[t])\dfrac{1-3\log t}{t^4}\,dt +O\left(\dfrac{\log x}{x^3}\right)\\
&= \left(\dfrac{1}{4}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}\right)+\int_1^2(t-[t])\dfrac{1-3\log t}{t^4}\,dt+\underbrace{\int_2^x(t-|t|)\dfrac{1-3\log t}{t^4}\,dt}_{=O(x^{-3}\log x)}\\
&\phantom{=}+O\left(\dfrac{\log x}{x^3}\right)\\
&= \left(\dfrac{1}{4}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}\right) +\dfrac{\log(16)-3}{16}+O\left(\dfrac{\log x}{x^3}\right)\\
&=\underbrace{\dfrac{1}{4}+\dfrac{\log(16)-3}{16}}_{=:A}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}+O\left(\dfrac{\log x}{x^3}\right)
\end{align*}

I used WolframAlpha for the integrals. You should calculate them explicitly!
But how does ## \frac{\log {x}}{x^3}([x]-x)=O(\frac{\log {x}}{x^3}) ##? And how to find ## \int_{1}^{2}(t-[t])\frac{1-3\log {t}}{t^4}dt, \int_{2}^{x}(t-[t])\frac{1-3\log {t}}{t^4}dt ##?
 
  • #12
Math100 said:
But how does ## \frac{\log {x}}{x^3}([x]-x)=O(\frac{\log {x}}{x^3}) ##?
##-\dfrac{\log {x}}{x^3} \leq \dfrac{\log {x}}{x^3}([x]-x) \leq 0##
It gets as negative as ##\dfrac{\log {x}}{x^3}## gets.

The entire expression is positive. We must show that
$$
\sum_{n\leq x}\dfrac{ \log {n}}{n^3}=A-\dfrac{ \log {x}}{2x^2}-\dfrac{1}{4x^2}+O\left(\dfrac{ \log {x}}{x^3}\right)
$$
which means
$$
\sum_{n\leq x}\dfrac{ \log {n}}{n^3}=A-\dfrac{ \log {x}}{2x^2}-\dfrac{1}{4x^2}+C\cdot \left(\dfrac{ \log {x}}{x^3}\right)
$$
with some constant ##C##. That means that ##C## swallows all coefficients in front of terms ##\dfrac{\log x}{x^3}## whether they are negative, positive, or zero. The result, in the end, will be positive, but even this is irrelevant.

Math100 said:
And how to find ## \int_{1}^{2}(t-[t])\frac{1-3\log {t}}{t^4}dt, \int_{2}^{x}(t-[t])\frac{1-3\log {t}}{t^4}dt ##?

As I said, I used WolframAlpha, I haven't tried to integrate them. The split at ##2## was because the first integral ##\int_1^2## has the zero of the function, so I needed an exact value. The right integrand is always zero or negative, so
$$
0\geq (t-[t])\dfrac{1-3\log {t}}{t^4} \geq \dfrac{1-3\log {t}}{t^4}
$$
and
$$
\int_{2}^{x} 0\,dt = 0 \geq \int_{2}^{x}(t-[t])\dfrac{1-3\log {t}}{t^4}\,dt \geq \int_{2}^{x}\dfrac{1-3\log {t}}{t^4}\,dt =\dfrac{\log x}{x^3}-\dfrac{\log 2}{8}
$$
which is again ##O\left(\dfrac{ \log {x}}{x^3}\right)## and a constant that ends up in ##A.## I should have better defined ##A=\dfrac{1}{4}+\dfrac{\log (16) -3}{16}-\dfrac{\log(2)}{8}=\dfrac{1}{16}+\dfrac{\log(2)}{8}.##
 
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  • #13
fresh_42 said:
##-\dfrac{\log {x}}{x^3} \leq \dfrac{\log {x}}{x^3}([x]-x) \leq 0##
It gets as negative as ##\dfrac{\log {x}}{x^3}## gets.

The entire expression is positive. We must show that
$$
\sum_{n\leq x}\dfrac{ \log {n}}{n^3}=A-\dfrac{ \log {x}}{2x^2}-\dfrac{1}{4x^2}+O\left(\dfrac{ \log {x}}{x^3}\right)
$$
which means
$$
\sum_{n\leq x}\dfrac{ \log {n}}{n^3}=A-\dfrac{ \log {x}}{2x^2}-\dfrac{1}{4x^2}+C\cdot \left(\dfrac{ \log {x}}{x^3}\right)
$$
with some constant ##C##. That means that ##C## swallows all coefficients in front of terms ##\dfrac{\log x}{x^3}## whether they are negative, positive, or zero. The result, in the end, will be positive, but even this is irrelevant.
As I said, I used WolframAlpha, I haven't tried to integrate them. The split at ##2## was because the first integral ##\int_1^2## has the zero of the function, so I needed an exact value. The right integrand is always zero or negative, so
$$
0\geq (t-[t])\dfrac{1-3\log {t}}{t^4} \geq \dfrac{1-3\log {t}}{t^4}
$$
and
$$
\int_{2}^{x} 0\,dt = 0 \geq \int_{2}^{x}(t-[t])\dfrac{1-3\log {t}}{t^4}\,dt \geq \int_{2}^{x}\dfrac{1-3\log {t}}{t^4}\,dt =\dfrac{\log x}{x^3}-\dfrac{\log 2}{8}
$$
which is again ##O\left(\dfrac{ \log {x}}{x^3}\right)## and a constant that ends up in ##A.## I should have better defined ##A=\dfrac{1}{4}+\dfrac{\log (16) -3}{16}-\dfrac{\log(2)}{8}=\dfrac{1}{16}+\dfrac{\log(2)}{8}.##
Also, how did you get ## \frac{\log {16}-3}{16} ##? Did the ## -\frac{\log {2}}{8} ## get swallowed by A or C?
 
  • #14
Math100 said:
Also, how did you get ## \frac{\log {16}-3}{16} ##?
WA
Math100 said:
Did the ## -\frac{\log {2}}{8} ## get swallowed by A or C?
A is easier.
 
  • #15
fresh_42 said:
WA

A is easier.
WA?
 
  • #17
Math100 said:
WA?
It's not difficult. You already did it, too.
$$
\int_1^2 (x-[x])\cdot \dfrac{1-3\log x}{x^4}\,dt= \underbrace{\int_1^2\dfrac{1-3\log x}{x^3}\,dt}_{=3(\log(4)-1)/16}- \underbrace{\int_1^2\dfrac{1-3\log x}{x^4}\,dt}_{=\log(2)/8}
$$
 
  • #18
fresh_42 said:
It's not difficult. You already did it, too.
$$
\int_1^2 (x-[x])\cdot \dfrac{1-3\log x}{x^4}\,dt= \underbrace{\int_1^2\dfrac{1-3\log x}{x^3}\,dt}_{=3(\log(4)-1)/16}- \underbrace{\int_1^2\dfrac{1-3\log x}{x^4}\,dt}_{=\log(2)/8}
$$
## \int_{1}^{2}(t-[t])\frac{1-3\log {t}}{t^4}dt ##
## =\int_{1}^{2}\frac{1-3\log {t}}{t^3}dt-\int_{1}^{2}\frac{1-3\log {t}}{t^4}dt ##
## =[-\frac{3}{16}+\frac{3}{8}\log {2}]-\frac{1}{8}\log {2} ##
## =-\frac{3}{16}+\log {2^{\frac{1}{4}}}=-\frac{3-\log {16}}{16} ##
 
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Related to Use Euler's summation formula to prove the following....

1. What is Euler's summation formula?

Euler's summation formula is a mathematical formula used to approximate the summation of a function over a range of values. It is based on the concept of Riemann sums and can be used to calculate definite integrals.

2. How do you use Euler's summation formula?

To use Euler's summation formula, you need to have a function that you want to approximate the summation of and a range of values over which you want to calculate the summation. You then plug in the values into the formula and solve for the result.

3. What are the key components of Euler's summation formula?

The key components of Euler's summation formula are the function to be summed, the range of values over which the summation will be calculated, and the constants involved in the formula (such as the step size and the number of terms).

4. What are the applications of Euler's summation formula?

Euler's summation formula has various applications in mathematics and science. It can be used to approximate definite integrals, calculate areas under curves, and solve differential equations. It is also used in statistics to estimate the sum of a series of data points.

5. How is Euler's summation formula derived?

Euler's summation formula is derived using the concept of Riemann sums and the properties of definite integrals. It involves taking the limit of an infinite series of approximations and simplifying it to a closed-form formula. The derivation is quite complex and involves advanced mathematical concepts.

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