How to solve this recurrence relations? - recurrence

I am trying solve this recurrence relations. I read a similar question in this site but it wasn't my answer.
T(n)=T(sqrt(n)) if n>4
T(n)=1 if n=4
thanks in advance.

According to your definition:
T(x) = 1 for all numbers of type 4^(2^n) for n being a natural number
T(x) not defined for all others

Related

Is it possible to run a mixed MANOVA in SPSS?

Is it possible to run a 3 (condition) x 2 (time) MANOVA in SPSS? I think I found a way but here it is said that it is not possible to run a MANOVA in the mixed procedure (https://www.ibm.com/support/pages/manova-or-mancova-analysis-available-mixed-procedure-spss).
Thank you very much for your help!
Kind regards,
K.

Unexpected result with apoc.text.sorensenDiceSimilarity?

A bit confused in regards to string similarity using Sorensen-Dice.
Apparently it makes a difference in what order parameters are being passed.
WITH
apoc.text.sorensenDiceSimilarity("+46xxxxx2260", "+46xxxxx2226") as score1,
apoc.text.sorensenDiceSimilarity("+46xxxxx2226", "+46xxxxx2260") as score2
RETURN
score1, score2
One of these scores (i.e. similarity coefficients) will say 1.0, the other 0.909090...
Does not make sense to me, but perhaps there's something with the algorithm I'm not aware of?
Any insight is appreciated.
P.S. "Neo4j Kernel", "3.5.9", "community"
This is definitely a bug and a good catch!
As alternative you can do below query which uses apoc functions as toSet and intersection and text function, split. There is hack on the query that uses ROUND(10^4/10^4) to use 4-decimal places. If you like my answer, please vote and accept it. Thanks.
WITH apoc.coll.toSet(split("+46xxxxx2260","")) as set1, apoc.coll.toSet(split("+46xxxxx2226","")) as set2
WITH set1, set2, apoc.coll.intersection(set1, set2) as common
RETURN ROUND(2*size(common)*10^4/(size(set1)+size(set2)))/10^4 as sorensenDiceSimilarity
Result:
0.9091

F#'s DescriptiveStatistics unknown?

I am currently converting a project in F# to Python, I am a relative amateur in F#, please...what on earth does this line do...
let stats = new DescriptiveStatistics(list)
I cant find a good definition of DescriptiveStatistics anywhere! Please help!
Many thanks
That is very likely from MathNet.
Check the beginning of your F# for the line:
using MathNet.Numerics.Statistics;
https://numerics.mathdotnet.com/DescriptiveStatistics.html
"...in case you need to gather a whole set of statistical characteristics in one pass, is provided by the DescriptiveStatistics class:..." Thanks for your help guys

SPSS string field to numeric conversion

I have a field in my data with low, middle and high. I want to replace low with 1, middle with 2 and high with 3.
Is there a good way to do this?
(I'm very Beginner for SPSS software.)
Else you could simply write this as Syntax:
string new_variable (a1).
RECODE
old_variable
("low"="1") ("middle"="2") ("high"="3") INTO new_variable.
EXECUTE.
Got a link which solve my problem and sharing for others, who came accross for same problem.
http://www.unige.ch/ses/sococ/cl/spss/tasks/createdummies.html
Depending on what you're measuring you could consider to type the field as a ordinal value type. It is not technically an interval value.
if oldString="low" newNum=1.
if oldString="middle" newNum=2.
if oldString="high" newNum=3.

Can I create an HDF5 link to a hyperslab?

Is it possible to create a link to just a hyperslab of a dataset in HDF5?
For example, I have one dataset of size 1000 x 3, representing (a,b,c) as a function of time, let's say. And now I want a link that points just to the 'a' data (1000 x 1). Is this possible?
[Having googled this extensively, I learned the valuable lesson that "link" is essentially useless in a google query. And I can't tell from the HDF5 documentation, so I'm sorry if this is stupid.]
Having asked at the (very helpful) HDF5 helpdesk, I find that the answer is no. For anyone else looking for this functionality: redesign your code/data structure.
Unfortunately for me, the code is not mine, and the data structure is set by other, stubborn people.

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