Skip to main content

series expansion - List manipulation - various




OK, Please forgive the messiness of this, but I am working with something like:


d = (q = 8;
f = (lop =
Transpose[{Flatten[
Reverse[Table[
Gamma[y], {y, 1, q + 1}]] /. {ComplexInfinity -> 0}],
Flatten[Table[Binomial[n, k], {n, q, q}, {k, 0, n}]],
Flatten[Table[x^n, {n, 0, q}]]}];
{#1*#2*#3} & @@@ lop);

Flatten[f];
Total[f]);
d

which gives:


{40320 + 40320 x + 20160 x^2 + 6720 x^3 + 1680 x^4 + 336 x^5 + 56 x^6 + 8 x^7 + x^8}

I am then comparing the roots with another series, eg:


Normal[Series[E^x, {x, 0, 8}]]


which, in this case are the same.


However, I have lots of different series & was wondering if I could manipulate them as stated below.


(NB - I realise there is no 'ComplexInfinity' output in the above example, but have left it in to demonstrate that I have tried several replace techniques for the subsequent questions.)



I have several questions on list manipulation & at the risk of asking the same question multiple times, I have compiled them into the following single question:


(a) Changing alternate (or some other pattern) signs / operators:


From this:


1 + x + (x^2)/2! +  (x^3)/3! + (x^4)/4! +  (x^5)/5! ...

I would like to do get this:



1 - x + (x^2)/2! -  (x^3)/3! + (x^4)/4! -  (x^5)/5! ...

(b) Multiplying alternate (or some other pattern) elements by a given value (eg - I):


From this:


1 + x + (x^2)/2! +  (x^3)/3! + (x^4)/4! +  (x^5)/5! ...

I would like to do get this:


1 - I x + (x^2)/2! -  I(x^3)/3! + (x^4)/4! -  I(x^5)/5! ...

(c) Convert all Imaginary values in a list to real values (or vice versa), noting that there may be no pattern in the distribution here:



From this:


1 + I x + (x^2)/2! +  (x^3)/3! + I(x^4)/4! +  I(x^5)/5! ...

I would like to do get this:


1 + x + (x^2)/2! +  (x^3)/3! + (x^4)/4! +  (x^5)/5! ...

or this:


I + I x + I(x^2)/2! +  I(x^3)/3! + I (x^4)/4! +  I(x^5)/5! ...

I have tried various select & replace methods, but have had no luck so far.




Answer



Well you can begin by wrapping your expression in Hold or HoldForm then using ReleaseHold when you want to Evaluate them. For your first example:


expr = HoldForm[1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + (x^5)/5!]

Now you can do:


Replace[expr /. {Times[c : Power[x, n_?OddQ], d_] :> Times[-1, c, d]}, x -> - x, 2]

To get:


1 - x + x^2/2! - x^3/3! + x^4/4! - x^5/5!


For the next one:


Replace[expr /. {Times[c : Power[x, n_?OddQ], d_] :> Times[-I, c, d]}, x -> - I x, 2]

Which gives:


1 - I x + x^2/2! - (I x^3)/3! + x^4/4! - (I x^5)/5!

For the third:


expr2 = 1 + I x + (x^2)/2! + (x^3)/3! + I (x^4)/4! + I (x^5)/5!;

To change the imaginary numbers do:



expr2 /. Complex[0, c_] :> c

To change negatives to positives in e.g.


expr3 = -1 - x + (x^2)/2! - (x^3)/3! + (x^4)/4! - (x^5)/5!

Simply do:


expr3 /. {Times[-1, c_] :> Times[1, c], Rational[-1, d_] :> Rational[1, d], n_?Negative :> -1 n}

To get:


1 + x + x^2/2 + x^3/6 + x^4/24 + x^5/120

Comments

Popular posts from this blog

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...