A very difficult problem

[DarkNess_wtc]

i have thought about that for serval days= =

please help



http://www.physicsforums.com/attachment.php?attachmentid=23523&stc=1&d=1265280816

[magwas]

It is 35.565955205923352

But I can't tell you about \sum_{2}^{n} \frac{1}{\sqrt{n}}.

[DarkNess_wtc]

how about steps?

[magwas]

Just entered into the calculator.

[DarkNess_wtc]

...

i want solving it without cal

[Prathyush]

In what form do u want the answer.
you can approximate the sum by integral 1/sqrt(n) between 1 and 19.
notice that 19^2 is 361
that is 2(19-1) = 36
The answer is close enough

[DarkNess_wtc]

>that is 2(19-1) = 36

HOW come to this step?

[Prathyush]

\sum_{1}^{361} \frac{1}{\sqrt{n}}
is approximately equal to
\int_{1}^{361} \frac{1}{\sqrt{n}}. =2(\sqrt{a}-\sqrt{b})
=2(\sqrt{361}-\sqrt{1})
=36

[g_edgar]

Why do you think there is any answer better than 35.565955205923352 ??

[elibj123]

I don't think there's an analytical solution, but you can approximate it.
First approximation is integral,

Next approximations would involve some play with operators.

Let's define D\equiv \frac{d}{dx} as the derivative operator, and observe the operator \sigma \equiv e^{D}

Functions of operators are defined via their Taylor Series, so:

e^{D} f=\sum^{\infty}_{k=0}\frac{D^{k}f(x)}{k!}=\sum^{\i nfty}_{k=0}\frac{1}{k!}\frac{d^{k}f(x)}{dx^{k}}(x+ 1-x)^k=f(x+1)

(Notice that the last sum is the Taylor Series of f(z) around the point x)

So \sigma is some sort of a delay\shift operator.

Let's now observe the operator S defined as

F(n)=Sf=\sum^{n}_{k=1}f(k)

You can notice that \sigma Sf=\sum^{n+1}_{k=1}f(k)

So \sigma S f-Sf=f(n+1)=\sigma f

And conclude S(\sigma -1)=\sigma

And if you recall the definition of \sigma then:

S=(1-\sigma^{-1})^{-1}=\frac{1}{1-e^{-D}}

S(D-\frac{D^{2}}{2}+\frac{D^{3}}{6}-...)=1

Knowing the expansion of the exponential function, you can approximate the Taylor expansion of S. The interesting fact is that S has a simple pole at "D=0", which means that in the expansion of S you will have a negative power of D:

S=D^{-1}+...

Which means that integration (the inverse of D) is the first approximation to discrete summation.

A better approximation would be

S=D^{-1}+\frac{1}{2}+\frac{1}{4}D

Operate S on \frac{1}{\sqrt{n}}
And get:

\sum^{n}_{k=2}\frac{1}{\sqrt{k}}=\int^{n}_{k=2}\fr ac{dk}{\sqrt{k}}+\frac{1}{2\sqrt{n}}-\frac{1}{8n\sqrt{n}}

And after some tiresome calculation for n=361

=35.1977

Which I think is a pretty good approximation, if only to show that the operator thing is correct.
Taking further derivatives would make a better approximation.

[aracharya]

You can also use the first 5 terms of the Euler-Maclaurin formula (http://planetmath.org/encyclopedia/EulerMaclaurinSummationFormula.html) (i.e. the integral, the terms involving the first derivative, and the terms involving the second derivative) to approximate the sum as follows :


S_n = \sum_{i=1}^n \frac{1}{\sqrt{i}}\approx 2\sqrt{n} +\frac{1}{2\sqrt{n}} - \frac{1}{24n\sqrt{n}} -
\frac{35}{24}


The result is then :


S_{361} \approx 36.56797638


The error relative to the true result is about 0.0055%.

[aracharya]

As stated in Bernoulli number (http://en.wikipedia.org/wiki/Bernoulli_numbers) if the first Bernoulli number is chosen according to the convention B_1 = -1/2 then the Euler–MacLaurin formula is :


\sum\limits_{a\leq k<b}f(k)=\int_a^b f(x)\,dx \ + \sum\limits_{k=1}^m \frac{B_k}{k!}\left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m).


whereas if we choose the convention B_1 = 1/2, then the Euler–MacLaurin formula is :


\sum\limits_{a<k\leq b} f(k)=\int_a^b f(x)\,dx + \sum\limits_{k=1}^m \frac{B_k}{k!} \left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m). \

[DarkNess_wtc]

i can't understand


S_n = \sum_{i=1}^n \frac{1}{\sqrt{i}}\approx 2\sqrt{n} +\frac{1}{2\sqrt{n}} - \frac{1}{24n\sqrt{n}} -
\frac{35}{24}



can you explain it please?

[aracharya]

All I am doing is substituting f(n) = 1/\sqrt{n} into the Euler-Maclaurin formula. This time from Mathworld (since the formula is expanded in full), the Euler-Maclaurin (http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html) formula can be written as :


\sum_{k=1}^{n-1}f(k)=\int_0^n f(k) dk-1/2[f(n)-f(0)]+1/(12)[f^{'}(n)-f^{'}(0)]-1/(720)[f^{'''}(n)-f^{'''}(0)]+1/(30240)[f^{(5)}(n)-f^{(5)}(0)]-1/(1209600)[f^{(7)}(n)-f^{(7)}(0)]+...


Note that i have written 1/2[f(n)-f(0)] instead of 1/2[f(n)+f(0)]. The former is consistent with wikipedia Bernouilli numbers (http://en.wikipedia.org/wiki/Bernoulli_number).

Then I make the approximation :


\sum_{k=1}^{n-1}f(k)\approx \int_0^n f(k)dk-1/2[f(n)-f(0)]+1/(12)[f^{'}(n)-f^{'}(0)]


Then I substitute f(k) = 1/\sqrt{k}, and then add f(n) to the result since the summation in the formula is up to (n-1) where as we want it up to n

All you have to do is to subsitute f(k) = 1/\sqrt{k} into the formula above, and verify my result.

[aracharya]

Note that the operator method of elibj123 is one way of deriving the Euler-Maclaurin formula, as is shown in Chapter 25 of Quantum Calculus by Victor Kac and Pokman Cheung (http://books.google.co.uk/books?id=mK8HXHSTbB0C&dq=Quantum+Calculus+by+Victor+Kac+and+Pokman+Cheun g&printsec=frontcover&source=bn&hl=en&ei=96luS9LML47KjAf54didCA&sa=X&oi=book_result&ct=result&resnum=4&ved=0CBUQ6AEwAw#v=onepage&q=&f=false). You can also find some basic material on it at "Art of approximation in science and engineering by Sanjoy Mahajan" (http://web.mit.edu/6.055/old/S2009/notes/operators.pdf)