What to use in Smath studio to simplify irrational expression

What to use in Smath studio to simplify irrational expression - Сообщения

#1 Опубликовано: 03.11.2010 19:17:18
Deni

Deni

0 сообщений из 2 понравились пользователям.

Группа: User

Here is the expression that I've been trying to simplify



"a" and "b" are not suposed to be defined.
Every time I put that expression in and put equals on the end, it says "a is not defined" wherein there is way to simplify this expression with no "a" or "b" given (not on computer), but i don't know how to do it in SMath Studio.

Also if there is a way, could it show a step-by-step simplifying of the expression.

Thanks for the help in advance.
Deni
#2 Опубликовано: 04.11.2010 04:23:59
Radovan Omorjan

Radovan Omorjan

325 сообщений из 2052 понравились пользователям.

Группа: Moderator

Hello Deni,
Wrote

Here is the expression that I've been trying to simplify



"a" and "b" are not suposed to be defined.
Every time I put that expression in and put equals on the end, it says "a is not defined" wherein there is way to simplify this expression with no "a" or "b" given (not on computer), but i don't know how to do it in SMath Studio


When you use the numerical equal to "=" everything in the given expression must be defined (have their numerical values). Therefore, you can not calculate the value of the above expression if the variables "a" and "b" do not have their values.

Regarding the simplification of the given expression - you can do two things at the moment.
1. Select a part of an expression and choose Calculation=>Simplify. Below that expression you will have the simplified part of the expression (in the way SMath symbolic engine is doing this). For instance if you select the Denominator and choose Calculation=>Simplify you will get this:
[MATH=eng]{sqrt(a-b)/{sqrt(a+b)-sqrt(a-b)}-{a-b}/sqrt(a^2-b^2-a+b)}/sqrt({a^2}/{b^2}-1)[/MATH]
[MATH=eng]sqrt(a-b)*sqrt(a+b)*(1/{b^2})^{1/2}[/MATH]
2. If you select the entire expression or use symbolic equal to "->", you will get this:
[MATH=eng]{sqrt(a-b)/{sqrt(a+b)-sqrt(a-b)}-{a-b}/sqrt(a^2-b^2-a+b)}/sqrt({a^2}/{b^2}-1)—{sqrt(a*(-1+a)+b*(1-b))-(sqrt(a+b)-sqrt(a-b))*sqrt(a-b)}/{sqrt(a+b)*sqrt(a*(-1+a)+b*(1-b))*(sqrt(a+b)-sqrt(a-b))*(1/{b^2})^{1/2}}[/MATH]

Regards,
Radovan
When Sisyphus climbed to the top of a hill, they said: "Wrong boulder!"
#3 Опубликовано: 04.11.2010 14:53:09
Deni

Deni

0 сообщений из 2 понравились пользователям.

Группа: User

That was a very useful explanation, a big Thank you Radovan, I appreciate it very much.
#4 Опубликовано: 04.11.2010 15:11:12
Radovan Omorjan

Radovan Omorjan

325 сообщений из 2052 понравились пользователям.

Группа: Moderator

You are welcome
Enjoy using SMath.

Regards,
Radovan
When Sisyphus climbed to the top of a hill, they said: "Wrong boulder!"
  • Новые сообщения Новые сообщения
  • Нет новых сообщений Нет новых сообщений