# A very difficult problem

1. Feb 4, 2010

### DarkNess_wtc

i have thought about that for serval days= =

File size:
5.9 KB
Views:
150
2. Feb 4, 2010

### magwas

It is 35.565955205923352

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

3. Feb 4, 2010

4. Feb 4, 2010

### magwas

Just entered into the calculator.

5. Feb 4, 2010

### DarkNess_wtc

...

i want solving it without cal

6. Feb 4, 2010

### 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

7. Feb 4, 2010

### DarkNess_wtc

>that is 2(19-1) = 36

HOW come to this step?

8. Feb 4, 2010

### 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

Last edited: Feb 4, 2010
9. Feb 4, 2010

### g_edgar

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

10. Feb 4, 2010

### 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:

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

(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:

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

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.

11. Feb 6, 2010

### aracharya

You can also use the first 5 terms of the http://planetmath.org/encyclopedia/EulerMaclaurinSummationFormula.html" [Broken] (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%.

Last edited by a moderator: May 4, 2017
12. Feb 6, 2010

### aracharya

As stated in http://en.wikipedia.org/wiki/Bernoulli_numbers" [Broken] 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). \$$

Last edited by a moderator: May 4, 2017
13. Feb 6, 2010

### DarkNess_wtc

i can't understand
$$S_n = \sum_{i=1}^n \frac{1}{\sqrt{i}}\approx 2\sqrt{n} +\frac{1}{2\sqrt{n}} - \frac{1}{24n\sqrt{n}} - \frac{35}{24}$$

14. Feb 7, 2010

### 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 http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html" [Broken] 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 http://en.wikipedia.org/wiki/Bernoulli_number" [Broken].

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.

Last edited by a moderator: May 4, 2017
15. Feb 7, 2010

### aracharya

Last edited by a moderator: Apr 24, 2017