This question is incomplete, the complete question is;
We know that the remainder R[tex]_n[/tex] will satisfy | R[tex]_n[/tex] | ≤ b[tex]_{ n + 1[/tex] = 1 / ( n + 1 )9[tex]^{ n + 1[/tex].
We must make n large enough so that this is less than 0.0001.
Rounding to five decimal places,
we have b₂ = _________ , b₃ =_________and b₄ =__________
Answer:
b₂ = 0.00617, b = 0.00046 and b₄ = 0.00004
Step-by-step explanation:
Given the data in the question;
| R[tex]_n[/tex] | ≤ b[tex]_{ n + 1[/tex] = 1 / ( n + 1 )9[tex]^{ n + 1[/tex]
Now,
b[tex]_{ n + 1[/tex] = 1 / ( n + 1 )9[tex]^{ n + 1[/tex]
b₂ = b[tex]_{ 1 + 1[/tex] = 1 / ( 1 + 1 )9[tex]^{ 1 + 1[/tex] = 1 / (2)9² = 1 / 162 = 0.00617 { 5 decimal places }
b₃ = b[tex]_{ 2 + 1[/tex] = 1 / ( 2 + 1 )9[tex]^{ 2 + 1[/tex] = 1 / (3)9³ = 1 / 2187 = 0.00046 { 5 decimal places }
b₄ = b[tex]_{ 3 + 1[/tex] = 1 / ( 3 + 1 )9[tex]^{ 3 + 1[/tex] = 1 / (4)9⁴ = 1 / 19683 = 0.00004 { 5 decimal places }
Therefore, b₂ = 0.00062, b = 0.00046 and b₄ = 0.00004
