# [✓] Is there one term missing when using Table?

Posted 1 year ago
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 Hey guys:When I use "Table" to get a series calculation results for A from 0 to 1 with gap 0.1. So there should be 11 elements in the list at last while "Table" only gives me 10 results and miss the last one when A=1. Really appreciate someone can tell me what happened and how to improve this. You may also find the commands in the attachment. $PreRead = (# /. s_String /; StringMatchQ[s, NumberString] && Precision@ToExpression@s == MachinePrecision :> s <> "50." &); skew[A_, \[Alpha]_, \[Beta]_, \[Delta]_] = (2 Sqrt[\[Delta]] (((-1 + A) A (\[Alpha] - \[Beta])^2 (A^2 (\[Alpha] - \[Beta])^2 - A (\[Alpha] - \[Beta]) (\[Alpha] - \[Beta] - \[Delta]) + \ \[Beta] (\[Alpha] + \[Delta])) (-(-1 + A) \[Beta] \[Delta] + \[Alpha] (\[Beta] + A \[Delta])))/(((-1 + A) \[Alpha] - A \[Beta] - 3 \[Delta]) ((-1 + A) \[Alpha] - A \[Beta] - 2 \[Delta]) (\[Alpha] - A \[Alpha] + A \[Beta] + \[Delta])^3) + (-3 (-1 + A) \[Beta] \[Delta] + \[Alpha] (\[Beta] + 3 A \[Delta]))/(\[Alpha] - A \[Alpha] + A \[Beta] + 3 \[Delta]) - ((-1 + A) A (\[Alpha] - \[Beta])^2 (-3 (-1 + A) \[Beta] \[Delta] + \[Alpha] (2 \[Beta] + 3 A \[Delta])))/((\[Alpha] - A \[Alpha] + A \[Beta] + \[Delta]) (\[Alpha] - A \[Alpha] + A \[Beta] + 2 \[Delta]) (\[Alpha] - A \[Alpha] + A \[Beta] + 3 \[Delta]))))/(((-1 + A) A (\[Alpha] - \[Beta])^2 (-(-1 + A) \[Beta] \[Delta] + \[Alpha] (\[Beta] + A \[Delta])))/(((-1 + A) \[Alpha] - A \[Beta] - 2 \[Delta]) (\[Alpha] - A \[Alpha] + A \[Beta] + \[Delta])^2) + (-2 (-1 + A) \[Beta] \[Delta] + \[Alpha] (\[Beta] + 2 A \[Delta]))/(\[Alpha] - A \[Alpha] + A \[Beta] + 2 \[Delta]))^(3/2); Table[skew[A, 20.1, 2.98, 0.015], {A, 0, 1, 0.1}] // N Length[%] {0.141895, 0.139203, 0.135927, 0.131859, 0.126677, 0.11985, 0.110442, \ 0.0966346, 0.0744006, 0.033185} 10 Thanks in advance Attachments: Answer 6 Replies Sort By: Posted 1 year ago  Thank you guys. Your comments answered my question well and I have learned a lot. Answer Posted 1 year ago  Just to clarify, this is the normal behavior In[7]:= Length@Table[x, {x, 0, 1, .1}] Out[7]= 11 When you do something drastic like $PreRead = (# /. s_String /; StringMatchQ[s, NumberString] && Precision@ToExpression@s == MachinePrecision :> s <> "50." &); then you need to be ready for the consequences, among which is In[10]:= Length@Table[x, {x, 0, 1, .1}] Out[10]= 10 Since you want your answers to such a high number of digits, why not just keep everything as rational and use N at the end? skew[Amp_, αmp_, βmp_, δmp_] := Module[{A = Rationalize@Amp, α = Rationalize@αmp, β = Rationalize@βmp, δ = Rationalize@δmp}, (2 Sqrt[δ] (((-1 + A) A (α - β)^2 (A^2 (α - β)^2 - A (α - β) (α - β - δ) \ + β (α + δ)) (-(-1 + A) β δ + α (β + A δ)))/(((-1 + A) α - A β - 3 δ) ((-1 + A) α - A β - 2 δ) (α - A α + A β + δ)^3) + (-3 (-1 + A) β δ + α (β + 3 A δ))/(α - A α + A β + 3 δ) - ((-1 + A) A (α - β)^2 (-3 (-1 + A) β δ + α (2 β + 3 A δ)))/((α - A α + A β + δ) (α - A α + A β + 2 δ) (α - A α + A β + 3 δ))))/(((-1 + A) A (α - β)^2 (-(-1 + A) β δ + α (β + A δ)))/(((-1 + A) α - A β - 2 δ) (α - A α + A β + δ)^2) + (-2 (-1 + A) β δ + α (β + 2 A δ))/(α - A α + A β + 2 δ))^(3/2) ]; In[2]:= Table[skew[A, 20.1, 2.98, 0.015], {A, 0, 1, 0.1}] // N[#, 50] & Length[%] Out[2]= {0.14189513095212063367049531962987546141568326643434, \ 0.13920344101328578263969756404215809382766534147239, \ 0.13592716321563841983794812006874993948237693570027, \ 0.13185917327785972103954298320739470452252943280392, \ 0.12667686295955584799447506972227890015016781166517, \ 0.11984973711991386295060361312735642563028850992476, \ 0.11044152395284289944251345711595700385738865557709, \ 0.096634643366104288816678293713331415198513960340867, \ 0.074400586745371447280910852144971096789513221053062, \ 0.033185045336993739355804248832880796377132424223379, \ 0.054635836470815303531952862734950163493012524454082} Out[3]= 11 
Answer
Posted 1 year ago
 0.1 is never really 0.1 in binary representation: Table[A, {A, 0, 1, 0.150}] if you copy the last number you will see: 0.9000000000000000000000000000000000000000000000000000000000000000000000000000350. i.e. a little bit more than 0.9, use exact arithmetic to be sure it is 11, or use Subdivide: Table[A, {A, Subdivide[0, 1, 10]}] 
Answer
Posted 1 year ago
 Sorry, I still do not what is happening here. I set up a table for k from 1 to 2 with gap 0.1. It turns out that the results are the same no matter gap being 0.1 or 1/10. Please see more details in the attachment. Attachments:
Answer
Posted 1 year ago
 Perhaps it's because if you sum ten instances of .150. you get someting slightly more than 1: In[2]:= .150. + .150. + .150. + .150. + .150. + .150. + .150. \ + .150. + .150. + .150. // FullForm Out[2] 1.0000000000000000000000000000000000000000000000000000000000000000000000\ 000000350. 
Answer
Posted 1 year ago
 Hint In[1]:= ... Length[Table[skew[A, 20.1, 2.98, 0.015], {A, 0, 1, 0.1}]] Out[3]= 10 versus In[1]:= ... Length[Table[skew[A, 20.1, 2.98, 0.015], {A, 0, 1, 1/10}]] Out[3]= 11 
Answer
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