Sum of power series (pdf)

38510987654321 222222222210

1

2 n

Summation of power series.

David Coulson, 2015 dtcoulson@gmail.com

38510987654321 222222222210

1

2 n

This total represents the number of blocks required to build a pyramid of ten layers.

38510987654321 222222222210

1

2 n

But doing so teaches you nothing about how power series behave in general. There is a theoretical approach that I like and which I have derived from first principles using a ton of algebra. I am going to share it over the next few pages. Be warned that there is a LOT of algebra involved, in fact some of the longest equations I have seen in years.

This could easily be worked out in a matter of minutes using a spreadsheet. A spreadsheet is definitely the fastest method for adding up a power series, if you take into consideration the time needed to go fetch a more theoretical method off the internet.

2222

1

2 ...321 NnN

Start by considering the graph of the totals as the number of terms increases. It looks as though the sum of a power series is another power series.

DCNBNANnN

23

1

2

2222

1

2 ...321 NnN

This is borne out by looking at the differences between terms, and then the differences between differences. If this eventually reaches a line of unchanging numbers, then the series is a polynomial function.

In this case, the constant line appears at the third difference, which means that the polynomial is a cubic.

DCNBNANnN

23

1

2

The other terms will be harder to work out, however.

Common sense tells us that D in this function has to be zero. The sum of no bricks has to be none!

14333

5222

1111

23

23

23

CBA

CBA

CBA

CNBNANnN

23

1

2One way to do it would be to look at any three (x,y) pairs and set up a 3x4 matrix.

This could definitely be done, but it would be a horrible process.

14333

5222

1111

23

23

23

CBA

CBA

CBA

CNBNANnN

23

1

2

A better way is to compare the (N+1)th term with the Nth.

21

21

1

2 1

NnnNN

111 1 23232CNBNANNCNBNAN

1

1

1 1

22

332

NNC

NNB

NNAN

1

12

133 12 22

C

NB

NNANN

Compare coefficients:

1

12

133 12 22

C

NB

NNANN

1

223

13

CBA

BA

A

This might look messy but it is set up so that it unzips nicely. Each coefficient can be identified directly, one line at a time.

61

21

31

C

B

A

NNNnN

612

213

31

1

2

This unzipping approach can be used to reveal the coefficients on higher power series. For example, the sum of cubic terms:

31

31

1

3 1

NnnNN

1 1 1 1 1 2233443NNDNNCNNBNNAN

DNCNBNANnN

234

1

3

0.1 1.1 0.0.1 .2.11 0.0.0.1 .3.3.11 1.0.0.0.0 - .4.6.4.11 1.3.3.1 DCBA

In this notation, only the coefficients are expressed. The rest is implied by the context.

1 1 1 1 1 2233443NNDNNCNNBNNAN

Expanding the bracketed terms in the normal way will turn this into an equation so big it won’t even fit on the page! But I can avoid this by introducing a better notation.

You can see that the end result is the combinatorial coefficients less the first term, obtained (for example) from Pascal’s triangle.

1 12. 3.3.1 4.6.4.1 1.3.3.1 DCBA

0.1 1.1 0.0.1 .2.11 0.0.0.1 .3.3.11 1.0.0.0.0 - .4.6.4.11 1.3.3.1 DCBA

If structured in a matrix, things get even clearer.

DCBA

CBAN

BAN

AN

1 1 1 1 1 : 1

2 3 4 3 :

3 6 3 :

4 1 :

2

3

The coefficients unzip from the top line downwards.

2

413

214

41

1

3 NNNnN

1111

234

36

4

1

3

3

1

D

C

B

A

1 12. 3.3.1 4.6.4.1 1.3.3.1 DCBA

NNNNNnN

n

30123

314

215

51

1

4 0

EDCBA

DCBAN

CBAN

BAN

AN

111111:1

23454:

36106:

4104:

51:

2

3

4

The same idea should apply to the next series up, namely the sum of all quartics:

... and the sum of quintics.

No, I don't want to solve that set of equations by hand either! But a spreadsheet makes quick work of it.

FEDCBA

EDCBAN

DCBAN

CBAN

BAN

AN

1111111:1

234565:

36101510:

4102010:

5155:

61:

2

3

4

5

NNNNNNnN

n

00 2

12134

1255

216

61

1

5

... and so on.

Some things to notice about all of this.

Fact 1: The leading coefficient is always the integral coefficient.

1

1

2

2

1

11

1

... NANANANn P

P

P

P

NP

This doesn’t surprise me in the least. As N gets very large, the summation starts to resemble an integration. The term with the highest exponent pulls away from the lesser terms. Therefore, in the extreme case where N tends towards infinity, the leading coefficient has to equal the integral coefficient.

Fact 2: The second coefficient is always ½. This really IS a surprise.

Some things to notice about all of this.

1

1

2

2211

11

1

... NANANNn PP

P

NP

Fact 3: The coefficients always add to 1.

If you can remember those three facts, then it will be easy to recall the coefficients for the first two power sums.

1

212

21

1

1 NNnN

1

612

213

31

1

2 NNNnN

And the good news is that these are the summations you are most likely to come across. We rarely if ever see the sums of cubics or quartics or higher powers.

There is a relationship between the sum for cubics and the sum for linear terms.

2

1

2

2

2

413

214

41

1

3 1

N

N

N

nNNNNn

This shortcut gives you one more summation that’s easy to recall.

It’s kind of hard to imagine what the sum of cubic terms could be applied to. The sum of linear terms is the number of logs stacked in a triangle. The sum of square terms is the number of bricks stacked in a pyramid (a three-dimensional triangle?). Therefore the sum of cubic terms is the number of hyperbricks stacked in a four-dimensional triangle, a hyper-pyramid.

That’s all.

- David C, 2015