Matematické Fórum

vlado_bb · 02. 07. 2020 18:11

↑↑ Dacu: I do not use WolframAlpha.

And I work at a university, department of mathematics.

nejsem_tonda · 02. 07. 2020 18:38

What do you think of the answer given by "WolframAlpha"?

I don't understand what kind of answer you would be satisfied with.
Wolframalpha clearly interprets the symbol as
$\sum_{k=n}^1 k^2=1^2-1^2-2^2-3^2-4^2-\ldots-(n-1)^2$
I have no idea why WA chooses this interpretation.

Ferdish

What was your original idea anyway? Did you want to calculate the sum of squares of all positive integers from 1 to some chosen value of $n\in \mathbb{N}$ ?

Because if so, then no wonder Wolfram gives you such a senseless output. Because your notation is wrong. Sumation is defined as

$\sum_{i=m}^{n}a_i=a_{m}+a_{m+1}+a_{m+2}+\ldots +a_{n-1}+a_{n}$

where $i$ is the index of summation, $a_i$ is an indexed variable representing each term of the sum, $m$ is the LOWER bound of summation and $n$ is the UPPER bound of summation. The $i = m$ under the summation symbol means that the index $i$ starts out equal to $m$ . The index is incremented by one for each successive term, stopping when $i = n$ . I am convinced this mathematical convention is used in all of Europe, including Romania.

According to this definition, in your case the index of summation is $k$ , the lower bound is $n$ and the upper bound is 1. Wolfram has no knowledge about the value of $n$ so he does not know from which value to start and whether this value is indeed an integer or not.

So tell me, does this REALLY correspond with your desired calculation?

auditor

Hello,

It seems that for the formula $\sum_{k=1}^{n} (n+1-k)^{2}$ , Wolfram Alpha gives expected answers.

On the other hand, the general formula $\sum_{k=n}^{1} k^{2}$ equals to $-1/6 (n - 2) (2 n^2 + n + 3 )$ according to Wolfram Alpha.

Moreover Wolfram Alpha gives the zero sum for the same formula with a specific integer $n>1$ which does not correspond to the results based on the general formula for integers $n>2$ .

Similar behaviour applies to the formula $\sum_{k=n}^{1}k$ in Wolfram Alpha. It seems that Wolfram Alpha interprets such formulas differently than we expect.

My understanding is that the author just wanted to inform us about this.

Dacu · 07. 07. 2020 08:39

Hello all,

Given the graph $y=k^2$ and solving the recurrence equation $S_n=S_{n-1}-(n-1)^2)$ resulting from graph and with $S_1=1$ , we obtain $S_n=1-\frac{(n-1)n(2n-1)}{6}=\sum_{k=n}^1 k^2$ .

All the best,

Dacu

Dacu · 07. 07. 2020 16:32

vlado_bb napsal(a):
↑↑ Dacu: I do not use WolframAlpha.

And I work at a university, department of mathematics.

Hello,

I think that the "WolframAlpha" program can help a lot if the initial data is entered correctly.Do you think the "WolframAlpha" program is wrong?

All the best,

Dacu

Dacu · 07. 07. 2020 16:58 — Editoval Dacu (07. 07. 2020 17:02)

auditor napsal(a):
Hello,

It seems that for the formula $\sum_{k=1}^{n} (n+1-k)^{2}$ , Wolfram Alpha gives expected answers.

On the other hand, the general formula $\sum_{k=n}^{1} k^{2}$ equals to $-1/6 (n - 2) (2 n^2 + n + 3 )$ according to Wolfram Alpha.

Moreover Wolfram Alpha gives the zero sum for the same formula with a specific integer $n>1$ which does not correspond to the results based on the general formula for integers $n>2$ .

Similar behaviour applies to the formula $\sum_{k=n}^{1}k$ in Wolfram Alpha. It seems that Wolfram Alpha interprets such formulas differently than we expect.

My understanding is that the author just wanted to inform us about this.

Hello,

From the "WolframAlpha" read:

1) Odkaz

2) Odkaz

3) Odkaz

4) Odkaz

5) Odkaz

6) Odkaz

and so on.....so resulting , depends much on how you enter the initial data in the "WolframAlpha" program ....

All the best,

Dacu

Dacu · 07. 07. 2020 17:07

Ferdish napsal(a):
What was your original idea anyway? Did you want to calculate the sum of squares of all positive integers from 1 to some chosen value of $n\in \mathbb{N}$ ?

Hello,

$\sum_{k=n}^1 k^2$ where $n\in \mathbb{N}$ .

All the best,

Dacu

Ferdish · 07. 07. 2020 17:25

↑ Dacu:
I've seen this formula already, since the very beginning of this topic. But it is not an answer to my question. Try to describe your problem with your own words, don't use any formulas, if possible.

For instance, I still don't know what shall I choose as the initial value of $n$ to start my sumation from and what type of number it is (natural, integer, rational, real)? Don't recall to Wolfram this time, I wanna hear an answer from you directly.

Dacu · 07. 07. 2020 18:02 — Editoval Dacu (08. 07. 2020 12:56)

Ferdish napsal(a):
↑ Dacu:
I've seen this formula already, since the very beginning of this topic. But it is not an answer to my question. Try to describe your problem with your own words, don't use any formulas, if possible.

For instance, I still don't know what shall I choose as the initial value of $n$ to start my sumation from and what type of number it is (natural, integer, rational, real)? Don't recall to Wolfram this time, I wanna hear an answer from you directly.

I answered today at 8:39 AM ... I repeat:
"Given the graph $y=k^2$ and solving the recurrence equation $S_n=S_{n-1}-(n-1)^2)$ resulting from graph and with $S_1=1$ , we obtain $S_n=1-\frac{(n-1)n(2n-1)}{6}=\sum_{k=n}^1 k^2$ ."....where $n\in \mathbb N$ .Thank you very much!

Yes , I also tried to calculate the sums of the form $\sum_{k=r_1}^{r_2} f(k)$ where $r_1 , r_2 \in \mathbb R$ with $r_1\leq r_2$ or $r_1\geq r_2$ , as well as for $\sum_{k=c_1}^{c_2} f(k)$ or $\sum_{k=c_2}^{c_1} f(k)$ where $c_1 , c_2 \in \mathbb C$ .

All the best,

Dacu

Ferdish

↑ Dacu:
Okay, it is clear that we do not understand each other. The language barrier between us two is just too big.

I hope someone else will help ya. Good luck and see you around!

Dacu · 07. 07. 2020 18:20 — Editoval Dacu (08. 07. 2020 09:38)

Ferdish napsal(a):
↑ Dacu:
Okay, it is clear that we do not understand each other. The language barrier between us two is just too big.

I hope someone else will help ya. Good luck and see you around!

I'm very sorry, but the mathematical language is universally valid and I don't think the computer program "WolframAlpha" is wrong ...I hope I didn't upset you with something!?!

Thank you very much!

All the best,

Dacu

misaH · 07. 07. 2020 19:19

↑ Ferdish:

No - proste robí sumu od k=1 do k=1 z k^2 a (aj podľa grafu (?) ) mu to vyjde 1, presne ako podľa formuly z WA, hehe...

Všetko ostatné je nepodstatné.

🌞

vlado_bb · 07. 07. 2020 20:08

↑ Dacu:Can you please formulate your problem without using the $\sum$ symbol?

Dacu · 08. 07. 2020 12:35 — Editoval Dacu (08. 07. 2020 12:49)

vlado_bb napsal(a):
↑ Dacu:Can you please formulate your problem without using the $\sum$ symbol?

Hello,

Mr. Moderator , the problem is not for Media school (Střední škola)!Initially the problem was posted to forum "Zajímavé příklady a úlohy-Zajímavé úlohy z matematické analýzy"....I considered that it is an interesting mathematical analysis problem...I do understand the fact that you do not like the "WolframAlpha" program, but then I would ask you to explain to me why this program gives an answer that seems surprising to you....Thank you very much!

All the best,

Dacu

vlado_bb · 08. 07. 2020 13:46

↑ Dacu: It is not the case that I do not like WoLphramAlpha - I simply do not use it.

To your problem - what do you understand by $\sum_{n=2}^1 a_n$ ? Is it $a_2+a_1$ ?

Dacu · 10. 07. 2020 06:53

vlado_bb napsal(a):
↑ Dacu: It is not the case that I do not like WoLphramAlpha - I simply do not use it.

To your problem - what do you understand by $\sum_{n=2}^1 a_n$ ? Is it $a_2+a_1$ ?

Hello,

Is it wrong to use the calculation program "WolframAlpha"?
Initially I thought that $\sum_{k=n}^{1} k^2=-\sum _{k=1}^{n} k^2$ , but checking with "WolframAlpha" I was very surprised by the answer of this calculation program ...In this regard, please explain the answer given by the calculation program "WolframAlpha"!Thank you very much!

All the best,

Dacu

vlado_bb · 10. 07. 2020 07:23

↑ Dacu: $\sum_{k=n}^{1} k^2=-\sum _{k=1}^{n} k^2$ - this is not a generally used convention and it is not possible to expect that some software will interpret it this way. A software could give either an error message or some different answer.

There is nothing wrong in using WolframAlpha, only it is recommended to formulate problems in line with usual conventions.

jarrro · 10. 07. 2020 10:42 — Editoval jarrro (10. 07. 2020 10:50)

I think for $\mathbb{Z}\ni m\leq n\in\mathbb{Z}$ and function $a:\mathbb{C}\to\mathbb{C}$ we can define $\sum_{k=m}^{n}{a{$k$}}$ by natural way by formula $\sum_{k=m}^{n}{a{$k$}}:=a{$m$}+a{$m+1$}\cdots+a{$m+k-1$}+\cdots +a{$n$}=:S{$m,n$}$
I think Dacu is finding analytical generalization of $S$ to real or complex arguments. It is obvious that there are infinitely many analytic functions for which the values in whole numbers are the same as in $S$ (I hope. If not, correct me please). So one can give some additional conditions for such generalization $f$ . The natural conditions are $f{$z-1,w$}=a{$z-1$}+f{$z,w$}\\ f{$z,w+1$}=f{$z,w$}+a{$w+1$}$ for arbitrary complex numbers $z,w$ .
Good question is if these conditions are sufficient for uniqueness.
But if one even solve this generalization problem, it $\color{red}{\text{C A N N O T}}$ be written $f{$z,w$}=\sum_{n=z}^{w}{a{$n$}}$ , because the right hand side symbol make no sense for complex numbers which don't satisfy the condition $\mathbb{Z}\ni z\leq w\in\mathbb{Z}$ .

I hope I don't write stupidities. If yes, be free to write me it.

Dacu · 11. 07. 2020 07:23

vlado_bb napsal(a):
↑ Dacu: $\sum_{k=n}^{1} k^2=-\sum _{k=1}^{n} k^2$ - this is not a generally used convention and it is not possible to expect that some software will interpret it this way. A software could give either an error message or some different answer.

There is nothing wrong in using WolframAlpha, only it is recommended to formulate problems in line with usual conventions.

Hello,

Apparently you're right, but I repeat:
"Given the graph $y=k^2$ and solving the recurrence equation $S_n=S_{n-1}-(n-1)^2)$ resulting from graph and with $S_1=1$ , we obtain $S_n=1-\frac{(n-1)n(2n-1)}{6}=\sum_{k=n}^1 k^2$ .".
In this sense, in soon I will post a drawing that shows exactly and what the "WolframAlpha" computer program says ....

All the best,

Dacu

jarrro · 11. 07. 2020 10:24 — Editoval jarrro (11. 07. 2020 13:16)

↑ Dacu:you only found (maybe only one from many) generalizations of such sum, so correct is only first equality in $S_n=1-\frac{(n-1)n(2n-1)}{6}=\sum_{k=n}^1 k^2$
The second one (by me, and obviuosly by other discutters) make no sense.
It's the same as writing that $\zeta{$-1$}=\sum_{k=1}^{\infty}{k}$

Dacu · 16. 07. 2020 19:55 — Editoval Dacu (18. 07. 2020 05:59)

Hello all,

The graph below supports my reasoning. Correct? Thank you very much!

Odkaz

All the best,

Dacu

Matematické Fórum

#26 02. 07. 2020 18:11

Re: Součet některých přirozených čísel

#27 02. 07. 2020 18:38

Re: Součet některých přirozených čísel

#28 02. 07. 2020 20:33 — Editoval Ferdish (02. 07. 2020 20:40)

Re: Součet některých přirozených čísel

#29 03. 07. 2020 09:54 — Editoval auditor (03. 07. 2020 10:49)

Re: Součet některých přirozených čísel

#30 07. 07. 2020 08:39

Re: Součet některých přirozených čísel

#31 07. 07. 2020 16:32

Re: Součet některých přirozených čísel

vlado_bb napsal(a):

#32 07. 07. 2020 16:58 — Editoval Dacu (07. 07. 2020 17:02)

Re: Součet některých přirozených čísel

auditor napsal(a):

#33 07. 07. 2020 17:07

Re: Součet některých přirozených čísel

Ferdish napsal(a):

#34 07. 07. 2020 17:25

Re: Součet některých přirozených čísel

#35 07. 07. 2020 18:02 — Editoval Dacu (08. 07. 2020 12:56)

Re: Součet některých přirozených čísel

Ferdish napsal(a):

#36 07. 07. 2020 18:11 — Editoval Ferdish (07. 07. 2020 18:11)

Re: Součet některých přirozených čísel

#37 07. 07. 2020 18:20 — Editoval Dacu (08. 07. 2020 09:38)

Re: Součet některých přirozených čísel

Ferdish napsal(a):

#38 07. 07. 2020 19:19

Re: Součet některých přirozených čísel

#39 07. 07. 2020 20:08

Re: Součet některých přirozených čísel

#40 08. 07. 2020 12:35 — Editoval Dacu (08. 07. 2020 12:49)

Re: Součet některých přirozených čísel

vlado_bb napsal(a):

#41 08. 07. 2020 13:46

Re: Součet některých přirozených čísel

#42 10. 07. 2020 06:53

Re: Součet některých přirozených čísel

vlado_bb napsal(a):

#43 10. 07. 2020 07:23

Re: Součet některých přirozených čísel

#44 10. 07. 2020 10:42 — Editoval jarrro (10. 07. 2020 10:50)

Re: Součet některých přirozených čísel

#45 11. 07. 2020 07:23

Re: Součet některých přirozených čísel

vlado_bb napsal(a):

#46 11. 07. 2020 10:24 — Editoval jarrro (11. 07. 2020 13:16)

Re: Součet některých přirozených čísel

#47 16. 07. 2020 19:55 — Editoval Dacu (18. 07. 2020 05:59)

Re: Součet některých přirozených čísel

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