# How to neglect the terms containing second and higher powers of parameter in a function

I want to neglect the terms containing second and higher powers of a from some equations. Since the equations are lengthy expressions, I was using Mathematica to neglect these terms. I have tried but I was unable. Can anyone please help me in this regard?

For example, for the following equation, I want to remove the terms containing a^2 and higher powers than 2.

(2 r^4 (4 a^2 + (27 c13)/(16 (-1 + c13)) + r^2 (Q + (-2 + r) r))^3 ((
81 c13)/(-1 + c13) + 16 r^2 (2 Q + (-3 + r) r) +
24 a Sqrt[-((54 c13)/(-1 + c13)) + 16 r^2 (-Q + r)])), (-((
a^2 (-16 r^2 (Q + (-2 + r) r) +
c13 (27 + 16 Q r^2 - 32 r^3 + 16 r^4))^2)/(
4 (-1 + c13)^2 r^2)) +
16 a^3 Sqrt[1 + (27 c13)/(16 (-1 + c13) r^4) + (Q - 2 r)/r^2]
Sqrt[(27 c13)/(-1 + c13) +
16 r^2 (Q + (-2 + r) r)] (-12 a +
Sqrt[-((54 c13)/(-1 + c13)) + 16 r^2 (-Q + r)]) + (
4 a^2 ((27 c13)/(16 (-1 + c13)) + r^2 (Q + (-2 + r) r)) (-12 a +
Sqrt[-((54 c13)/(-1 + c13)) + 16 r^2 (-Q + r)])^2)/
r^2 + (((27 c13)/(16 (-1 + c13)) + r^2 (Q + (-2 + r) r)) (4 a^2 + (
27 c13)/(16 (-1 + c13)) + r^2 (Q + (-2 + r) r)) (-12 a +
Sqrt[-((54 c13)/(-1 + c13)) + 16 r^2 (-Q + r)])^2)/
r^2)/((4 a^2 + (27 c13)/(16 (-1 + c13)) +
r^2 (Q + (-2 + r) r))^2 ((81 c13)/(-1 + c13) +
16 r^2 (2 Q + (-3 + r) r) +
24 a Sqrt[-((54 c13)/(-1 + c13)) + 16 r^2 (-Q + r)]))

• You know that there is a ',' in the expression that you wrote, right? This must be a typo. It's in the third line of the expression – DiSp0sablE_H3r0 Sep 15 '20 at 10:55
• yes, this is typo. – MMS Sep 15 '20 at 10:59
• Thanks @Bill, yes it is "yourexpression /. a^_ -> 0" working , but it is not working when we have higher powers of a after multiplication, like, if we have a (x + a), then after multiplication, we have term a^2, so it is not neglecting using this command. – MMS Sep 15 '20 at 11:06
• Mathematica's pattern matching only applies when it finds exactly the form of the expression that you are looking for. To expose those powers of a try Expand[yourexpression] /. a^_ -> 0 which should find more cases where there are powers of a in yourexpression. But even this may not be enough if you are expecting more and more complicated examples to work. That was why I asked you to check VERY carefully. – Bill Sep 15 '20 at 15:27

Normal[Series[(a + x) (a - x) (a - 2 x), {a, 0, 1}]]

-a x^2 + 2 x^3

• This command is not working in my case because the equations are in square root, like (Sqrt[1 + (27 c13)/(16 (-1 + c13) r^4) + (Q - 2 r)/r^2] Sqrt[ 64 a^2 + (27 c13)/(-1 + c13) + 16 r^2 (Q + (-2 + r) r) + a^3 (x + a)])/Sqrt[(81 c13)/(-1 + c13) + 16 r^2 (2 Q + (-3 + r) r) + 24 a (-4 a + Sqrt Sqrt[8 a^2 - (27 c13)/(-1 + c13) + 8 r^2 (-Q + r)])] – MMS Sep 15 '20 at 11:38