- #1

nassboy

- 39

- 0

-1

1 -1

-3 7 -4

15 -63 82 -34

-105 729 -1638 1510 -496

945 -10545 36684 -56568 40540 -11056

Good Luck!

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In summary: In[10]:= (*Check the Series Repeating*)In[11]:= (*Check the Power Series*)In[12]:= (*Check the Series for convergence*)All three of those checks passed.Can somebody help me to find the pattern to generate the series?

- #1

nassboy

- 39

- 0

-1

1 -1

-3 7 -4

15 -63 82 -34

-105 729 -1638 1510 -496

945 -10545 36684 -56568 40540 -11056

Good Luck!

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- #2

Gokul43201

Staff Emeritus

Science Advisor

Gold Member

- 7,220

- 24

Just for my curiosity, where is this question from?

- #3

rasselas21

- 2

- 0

-10,395 : : : : : : : : sum of previous elements in row * -1

I'm likewise curious to as to what this sequence is to. It might help to figure out the sequence.

It would look roughly something like:

-10,395 161,051 -1,054,500 1,907,568 –1,207,882 1,641,870 -278,300

That is, each element should roughly be around the value given (within 25% error maybe?), except that the first value is correct.

- #4

nassboy

- 39

- 0

The sequence comes from the series solution to a differential equation,,,,

and the next line is...

-10395 186075 -946392 2174820 -2552292 1497688 -349504

- #5

rasselas21

- 2

- 0

What's the formula for producing some of these elements (like row 5 for example)?

- #6

nassboy

- 39

- 0

I've been working on finding a closed form solution or an approximate solution for a variant of the Poisson-Boltzmann equation.

y''[x]=A*(1-e^(-y[x])

I'm unable to solve it using methods such as separation of variables, but I have been able to generate a power series using recursive derivatives. I've pasted the mathematica code below...I've been unable to find the pattern to generate the series...can anybody help?

In[2]:= (*Define Recursive Derivative Function *)

In[3]:= Derivative[N_][Y][0]:=D[Y''[x],{x,N-2}] /. x->0

In[4]:= (*Define the Problem*)

In[5]:= Y''[x]=A*(1-E^(-Y[x]))

Out[5]= A (1-E^-Y[x])

In[6]:= Y''[0]=Y''[x] /. x->0

Out[6]= A (1-E^-Y[0])

In[7]:= Y'[0]=0

Out[7]= 0

In[8]:= (*Check the Series*)

In[9]:= Y[0]+Integrate[Integrate[Series[Y''[x],{x,0,10}],x],x]

Out[9]= Y[0]+1/2 (A-A E^-Y[0]) x^2+1/24 A^2 E^(-2 Y[0]) (-1+E^Y[0]) x^4-1/720 (A^3 E^(-3 Y[0]) (4-7 E^Y[0]+3 E^(2 Y[0]))) x^6+(A^4 E^(-4 Y[0]) (-34+82 E^Y[0]-63 E^(2 Y[0])+15 E^(3 Y[0])) x^8)/40320-((A^5 E^(-5 Y[0]) (496-1510 E^Y[0]+1638 E^(2 Y[0])-729 E^(3 Y[0])+105 E^(4 Y[0]))) x^10)/3628800+(A^6 E^(-6 Y[0]) (-11056+40540 E^Y[0]-56568 E^(2 Y[0])+36684 E^(3 Y[0])-10545 E^(4 Y[0])+945 E^(5 Y[0])) x^12)/479001600+O[x]^13

The purpose of finding the next line in a sequence is to identify patterns and relationships between numbers or objects in a sequence. This can help in predicting future values and understanding the underlying logic behind the sequence.

The next line in a numerical sequence can be determined by identifying any patterns or relationships between the given numbers. This can involve looking for common differences, ratios, or geometric patterns. Once a pattern is recognized, it can be used to predict the next number in the sequence.

There is no one specific method or formula for finding the next line in a sequence, as it depends on the type of sequence and the patterns present. Some common methods include using arithmetic or geometric sequences, recursive formulas, or visual patterns.

Yes, finding the next line in a sequence can have practical applications in various fields such as mathematics, computer science, and data analysis. It can be used to predict future trends, make informed decisions, and solve problems in different industries.

While finding the next line in a sequence can be a useful tool, it is not always applicable or accurate. Some sequences may have complex or unpredictable patterns, making it difficult to determine the next line. Additionally, it may not account for external factors that can affect the sequence.

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