Infinite increasing sequence with square roots [closed]
I can not find what is an infinite sequence of this type, where the coefficient for each next term is 1 more than the previous one. Help me please. I think it's interesting.
sequences-and-series
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closed as off-topic by Gerry Myerson, Jyrki Lahtonen, amWhy, Gibbs, José Carlos Santos Dec 6 at 23:55
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I can not find what is an infinite sequence of this type, where the coefficient for each next term is 1 more than the previous one. Help me please. I think it's interesting.
sequences-and-series
migration rejected from mathematica.stackexchange.com Dec 6 at 23:55
This question came from our site for users of Wolfram Mathematica. Votes, comments, and answers are locked due to the question being closed here, but it may be eligible for editing and reopening on the site where it originated.
closed as off-topic by Gerry Myerson, Jyrki Lahtonen, amWhy, Gibbs, José Carlos Santos Dec 6 at 23:55
- This question does not appear to be about math within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.
See whether this is answered at math.stackexchange.com/questions/654478/…
– Gerry Myerson
Dec 2 at 11:38
Had a chance to look at that link?
– Gerry Myerson
Dec 4 at 6:06
3
I'm voting to close this question as off-topic because OP has abandoned it.
– Gerry Myerson
Dec 6 at 11:44
comments disabled on deleted / locked posts / reviews |
I can not find what is an infinite sequence of this type, where the coefficient for each next term is 1 more than the previous one. Help me please. I think it's interesting.
sequences-and-series
I can not find what is an infinite sequence of this type, where the coefficient for each next term is 1 more than the previous one. Help me please. I think it's interesting.
sequences-and-series
sequences-and-series
asked Nov 30 at 10:32
Styopkin Stepan
migration rejected from mathematica.stackexchange.com Dec 6 at 23:55
This question came from our site for users of Wolfram Mathematica. Votes, comments, and answers are locked due to the question being closed here, but it may be eligible for editing and reopening on the site where it originated.
closed as off-topic by Gerry Myerson, Jyrki Lahtonen, amWhy, Gibbs, José Carlos Santos Dec 6 at 23:55
- This question does not appear to be about math within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.
migration rejected from mathematica.stackexchange.com Dec 6 at 23:55
This question came from our site for users of Wolfram Mathematica. Votes, comments, and answers are locked due to the question being closed here, but it may be eligible for editing and reopening on the site where it originated.
closed as off-topic by Gerry Myerson, Jyrki Lahtonen, amWhy, Gibbs, José Carlos Santos Dec 6 at 23:55
- This question does not appear to be about math within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.
See whether this is answered at math.stackexchange.com/questions/654478/…
– Gerry Myerson
Dec 2 at 11:38
Had a chance to look at that link?
– Gerry Myerson
Dec 4 at 6:06
3
I'm voting to close this question as off-topic because OP has abandoned it.
– Gerry Myerson
Dec 6 at 11:44
comments disabled on deleted / locked posts / reviews |
See whether this is answered at math.stackexchange.com/questions/654478/…
– Gerry Myerson
Dec 2 at 11:38
Had a chance to look at that link?
– Gerry Myerson
Dec 4 at 6:06
3
I'm voting to close this question as off-topic because OP has abandoned it.
– Gerry Myerson
Dec 6 at 11:44
See whether this is answered at math.stackexchange.com/questions/654478/…
– Gerry Myerson
Dec 2 at 11:38
See whether this is answered at math.stackexchange.com/questions/654478/…
– Gerry Myerson
Dec 2 at 11:38
Had a chance to look at that link?
– Gerry Myerson
Dec 4 at 6:06
Had a chance to look at that link?
– Gerry Myerson
Dec 4 at 6:06
3
3
I'm voting to close this question as off-topic because OP has abandoned it.
– Gerry Myerson
Dec 6 at 11:44
I'm voting to close this question as off-topic because OP has abandoned it.
– Gerry Myerson
Dec 6 at 11:44
comments disabled on deleted / locked posts / reviews |
1 Answer
1
active
oldest
votes
NESTED SQUARE ROOTS SERIES
Hi Stephan. Yes, I found your problem interesting and I think I have a solution.
The iterator value "i" is put as a mutiplier of the "a" of the root series. To start with the maximum value of the iterator put in coefficient of the a in the 1st iteration and end with 1 as "a" coef in the last iteration, we'll have a make a decreasing iteration.
Starting with t=0 allows a clean initial value (change and see...).
In[1]:= For[i = 5; t = 0, i >= 1, i -= 1, t = (t + i a)^(1/2)]; t
Out[1]= Sqrt[a + Sqrt[2 a + Sqrt[3 a + Sqrt[Sqrt[5] Sqrt[a] + 4 a]]]]
Of course at the end of the above displayed result Sqrt[Sqrt[5] Sqrt[a]+4 a] should read Sqrt[4 a +Sqrt[5 a] ], which is in the desired form.
Let' s see what this gives with a=4 and 5 iterations:
In[2]:= For[i = 5; t = 0, i >= 1, i -= 1, t = (t + i 4)^(1/2)]; t
N[t]
Out[2]= Sqrt[4 + Sqrt[8 + Sqrt[12 + Sqrt[16 + 2 Sqrt[5]]]]]
Out[3]= 2.73377
With the same a=4 and 15 iterations the output remains very close. The series is rapidly convergent.
In[4]:= For[i = 15; t = 0, i >= 1, i -= 1, t = (t + i 4)^(1/2)]; N[t]
Out[4]= 2.73397
Let's build a summarized functions[a_,tini_,imax] where we enter the "a" value, the start value tini of the function, and the maximum number of iterations imax:
In[5]:= s[a_, tini_, imax_] :=
For[i = imax; t = tini, i >= 1, i -= 1, t = (t + i a)^(1/2)]
Let's see the data for a=5, 0 at start, and 10 iterations, so s[6,0,10]
To show the result we have, optionally for check to ask the nested final value "t", and its approx value N[t]
In[6]:= s[6, 0, 10]
t
N[t]
Out[7]= Sqrt[6 + Sqrt[
12 + Sqrt[18 + Sqrt[
24 + Sqrt[30 + Sqrt[36 + Sqrt[42 + Sqrt[48 + Sqrt[54 + 2 Sqrt[15]]]]]]]]]]
Out[8]= 3.17874
We see that 10 more iterations do not change the first 5 decimal digits.
In[9]:= s[6, 0, 20]
N[t]
Out[10]= 3.17874
Voila!
Olivier
PS: I don`t know how to past with MTM output format, neither how to attach a notebook (it's my 1st post...). But you Will easily copy and past the formulas in a new notebook.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
NESTED SQUARE ROOTS SERIES
Hi Stephan. Yes, I found your problem interesting and I think I have a solution.
The iterator value "i" is put as a mutiplier of the "a" of the root series. To start with the maximum value of the iterator put in coefficient of the a in the 1st iteration and end with 1 as "a" coef in the last iteration, we'll have a make a decreasing iteration.
Starting with t=0 allows a clean initial value (change and see...).
In[1]:= For[i = 5; t = 0, i >= 1, i -= 1, t = (t + i a)^(1/2)]; t
Out[1]= Sqrt[a + Sqrt[2 a + Sqrt[3 a + Sqrt[Sqrt[5] Sqrt[a] + 4 a]]]]
Of course at the end of the above displayed result Sqrt[Sqrt[5] Sqrt[a]+4 a] should read Sqrt[4 a +Sqrt[5 a] ], which is in the desired form.
Let' s see what this gives with a=4 and 5 iterations:
In[2]:= For[i = 5; t = 0, i >= 1, i -= 1, t = (t + i 4)^(1/2)]; t
N[t]
Out[2]= Sqrt[4 + Sqrt[8 + Sqrt[12 + Sqrt[16 + 2 Sqrt[5]]]]]
Out[3]= 2.73377
With the same a=4 and 15 iterations the output remains very close. The series is rapidly convergent.
In[4]:= For[i = 15; t = 0, i >= 1, i -= 1, t = (t + i 4)^(1/2)]; N[t]
Out[4]= 2.73397
Let's build a summarized functions[a_,tini_,imax] where we enter the "a" value, the start value tini of the function, and the maximum number of iterations imax:
In[5]:= s[a_, tini_, imax_] :=
For[i = imax; t = tini, i >= 1, i -= 1, t = (t + i a)^(1/2)]
Let's see the data for a=5, 0 at start, and 10 iterations, so s[6,0,10]
To show the result we have, optionally for check to ask the nested final value "t", and its approx value N[t]
In[6]:= s[6, 0, 10]
t
N[t]
Out[7]= Sqrt[6 + Sqrt[
12 + Sqrt[18 + Sqrt[
24 + Sqrt[30 + Sqrt[36 + Sqrt[42 + Sqrt[48 + Sqrt[54 + 2 Sqrt[15]]]]]]]]]]
Out[8]= 3.17874
We see that 10 more iterations do not change the first 5 decimal digits.
In[9]:= s[6, 0, 20]
N[t]
Out[10]= 3.17874
Voila!
Olivier
PS: I don`t know how to past with MTM output format, neither how to attach a notebook (it's my 1st post...). But you Will easily copy and past the formulas in a new notebook.
add a comment |
NESTED SQUARE ROOTS SERIES
Hi Stephan. Yes, I found your problem interesting and I think I have a solution.
The iterator value "i" is put as a mutiplier of the "a" of the root series. To start with the maximum value of the iterator put in coefficient of the a in the 1st iteration and end with 1 as "a" coef in the last iteration, we'll have a make a decreasing iteration.
Starting with t=0 allows a clean initial value (change and see...).
In[1]:= For[i = 5; t = 0, i >= 1, i -= 1, t = (t + i a)^(1/2)]; t
Out[1]= Sqrt[a + Sqrt[2 a + Sqrt[3 a + Sqrt[Sqrt[5] Sqrt[a] + 4 a]]]]
Of course at the end of the above displayed result Sqrt[Sqrt[5] Sqrt[a]+4 a] should read Sqrt[4 a +Sqrt[5 a] ], which is in the desired form.
Let' s see what this gives with a=4 and 5 iterations:
In[2]:= For[i = 5; t = 0, i >= 1, i -= 1, t = (t + i 4)^(1/2)]; t
N[t]
Out[2]= Sqrt[4 + Sqrt[8 + Sqrt[12 + Sqrt[16 + 2 Sqrt[5]]]]]
Out[3]= 2.73377
With the same a=4 and 15 iterations the output remains very close. The series is rapidly convergent.
In[4]:= For[i = 15; t = 0, i >= 1, i -= 1, t = (t + i 4)^(1/2)]; N[t]
Out[4]= 2.73397
Let's build a summarized functions[a_,tini_,imax] where we enter the "a" value, the start value tini of the function, and the maximum number of iterations imax:
In[5]:= s[a_, tini_, imax_] :=
For[i = imax; t = tini, i >= 1, i -= 1, t = (t + i a)^(1/2)]
Let's see the data for a=5, 0 at start, and 10 iterations, so s[6,0,10]
To show the result we have, optionally for check to ask the nested final value "t", and its approx value N[t]
In[6]:= s[6, 0, 10]
t
N[t]
Out[7]= Sqrt[6 + Sqrt[
12 + Sqrt[18 + Sqrt[
24 + Sqrt[30 + Sqrt[36 + Sqrt[42 + Sqrt[48 + Sqrt[54 + 2 Sqrt[15]]]]]]]]]]
Out[8]= 3.17874
We see that 10 more iterations do not change the first 5 decimal digits.
In[9]:= s[6, 0, 20]
N[t]
Out[10]= 3.17874
Voila!
Olivier
PS: I don`t know how to past with MTM output format, neither how to attach a notebook (it's my 1st post...). But you Will easily copy and past the formulas in a new notebook.
add a comment |
NESTED SQUARE ROOTS SERIES
Hi Stephan. Yes, I found your problem interesting and I think I have a solution.
The iterator value "i" is put as a mutiplier of the "a" of the root series. To start with the maximum value of the iterator put in coefficient of the a in the 1st iteration and end with 1 as "a" coef in the last iteration, we'll have a make a decreasing iteration.
Starting with t=0 allows a clean initial value (change and see...).
In[1]:= For[i = 5; t = 0, i >= 1, i -= 1, t = (t + i a)^(1/2)]; t
Out[1]= Sqrt[a + Sqrt[2 a + Sqrt[3 a + Sqrt[Sqrt[5] Sqrt[a] + 4 a]]]]
Of course at the end of the above displayed result Sqrt[Sqrt[5] Sqrt[a]+4 a] should read Sqrt[4 a +Sqrt[5 a] ], which is in the desired form.
Let' s see what this gives with a=4 and 5 iterations:
In[2]:= For[i = 5; t = 0, i >= 1, i -= 1, t = (t + i 4)^(1/2)]; t
N[t]
Out[2]= Sqrt[4 + Sqrt[8 + Sqrt[12 + Sqrt[16 + 2 Sqrt[5]]]]]
Out[3]= 2.73377
With the same a=4 and 15 iterations the output remains very close. The series is rapidly convergent.
In[4]:= For[i = 15; t = 0, i >= 1, i -= 1, t = (t + i 4)^(1/2)]; N[t]
Out[4]= 2.73397
Let's build a summarized functions[a_,tini_,imax] where we enter the "a" value, the start value tini of the function, and the maximum number of iterations imax:
In[5]:= s[a_, tini_, imax_] :=
For[i = imax; t = tini, i >= 1, i -= 1, t = (t + i a)^(1/2)]
Let's see the data for a=5, 0 at start, and 10 iterations, so s[6,0,10]
To show the result we have, optionally for check to ask the nested final value "t", and its approx value N[t]
In[6]:= s[6, 0, 10]
t
N[t]
Out[7]= Sqrt[6 + Sqrt[
12 + Sqrt[18 + Sqrt[
24 + Sqrt[30 + Sqrt[36 + Sqrt[42 + Sqrt[48 + Sqrt[54 + 2 Sqrt[15]]]]]]]]]]
Out[8]= 3.17874
We see that 10 more iterations do not change the first 5 decimal digits.
In[9]:= s[6, 0, 20]
N[t]
Out[10]= 3.17874
Voila!
Olivier
PS: I don`t know how to past with MTM output format, neither how to attach a notebook (it's my 1st post...). But you Will easily copy and past the formulas in a new notebook.
NESTED SQUARE ROOTS SERIES
Hi Stephan. Yes, I found your problem interesting and I think I have a solution.
The iterator value "i" is put as a mutiplier of the "a" of the root series. To start with the maximum value of the iterator put in coefficient of the a in the 1st iteration and end with 1 as "a" coef in the last iteration, we'll have a make a decreasing iteration.
Starting with t=0 allows a clean initial value (change and see...).
In[1]:= For[i = 5; t = 0, i >= 1, i -= 1, t = (t + i a)^(1/2)]; t
Out[1]= Sqrt[a + Sqrt[2 a + Sqrt[3 a + Sqrt[Sqrt[5] Sqrt[a] + 4 a]]]]
Of course at the end of the above displayed result Sqrt[Sqrt[5] Sqrt[a]+4 a] should read Sqrt[4 a +Sqrt[5 a] ], which is in the desired form.
Let' s see what this gives with a=4 and 5 iterations:
In[2]:= For[i = 5; t = 0, i >= 1, i -= 1, t = (t + i 4)^(1/2)]; t
N[t]
Out[2]= Sqrt[4 + Sqrt[8 + Sqrt[12 + Sqrt[16 + 2 Sqrt[5]]]]]
Out[3]= 2.73377
With the same a=4 and 15 iterations the output remains very close. The series is rapidly convergent.
In[4]:= For[i = 15; t = 0, i >= 1, i -= 1, t = (t + i 4)^(1/2)]; N[t]
Out[4]= 2.73397
Let's build a summarized functions[a_,tini_,imax] where we enter the "a" value, the start value tini of the function, and the maximum number of iterations imax:
In[5]:= s[a_, tini_, imax_] :=
For[i = imax; t = tini, i >= 1, i -= 1, t = (t + i a)^(1/2)]
Let's see the data for a=5, 0 at start, and 10 iterations, so s[6,0,10]
To show the result we have, optionally for check to ask the nested final value "t", and its approx value N[t]
In[6]:= s[6, 0, 10]
t
N[t]
Out[7]= Sqrt[6 + Sqrt[
12 + Sqrt[18 + Sqrt[
24 + Sqrt[30 + Sqrt[36 + Sqrt[42 + Sqrt[48 + Sqrt[54 + 2 Sqrt[15]]]]]]]]]]
Out[8]= 3.17874
We see that 10 more iterations do not change the first 5 decimal digits.
In[9]:= s[6, 0, 20]
N[t]
Out[10]= 3.17874
Voila!
Olivier
PS: I don`t know how to past with MTM output format, neither how to attach a notebook (it's my 1st post...). But you Will easily copy and past the formulas in a new notebook.
answered Dec 4 at 16:54
Olivier33
1
1
add a comment |
add a comment |
See whether this is answered at math.stackexchange.com/questions/654478/…
– Gerry Myerson
Dec 2 at 11:38
Had a chance to look at that link?
– Gerry Myerson
Dec 4 at 6:06
3
I'm voting to close this question as off-topic because OP has abandoned it.
– Gerry Myerson
Dec 6 at 11:44