# Bad Re()

### #1

Posted 30 April 2006 - 05:30 AM

but if i want to know Re((re(a)+i*re( c )) what should i do ,it must give {re(a)}

### #2

Posted 30 April 2006 - 11:16 AM

if i enter re((re(a)+i*re( c )) it gives re(a) which is correct

re(2+a*i) is not 2, but 2-im(a) except if you have a € IR : re(2+a*i)|im(a)=0 gives 2

### #3

Posted 01 May 2006 - 07:18 AM

Varible a can have imagine number.

Re(2+ai)=/=2 => right because ai can have imagine number and then i miutiply whith that equa to a real number.

### #4

Posted 01 May 2006 - 08:17 AM

i made a wrong exampleSO EASY

Varible a can have imagine number.

Re(2+ai)=/=2 => right because ai can have imagine number and then i miutiply whith that equa to a real number.

what about Re( 1/ (Re( x ) + i* Re(y) ) )??????????????? no answer

### #5

Posted 01 May 2006 - 09:42 AM

It should be 1/re{a} because a might not be real only ( and "1/" is nothing ).

Is there any example more.

And in my afx when i solve the equation Abs(x)+Abs(x+1)=1, it got a result like this

X=0

X=1

X=X

How about in your CP?

### #6

Posted 02 May 2006 - 01:44 PM

No answer, because x might be imaginary as well, or it might be equal to zero .i made a wrong example

what about Re( 1/ (Re( x ) + i* Re(y) ) )??????????????? no answer

### #7

Posted 03 May 2006 - 05:49 AM

no matter if x has imaganary or not ,Re(x) is real and re(y) is realNo answer, because x might be imaginary as well, or it might be equal to zero .

the answer should be Re(x)/((Re(x)^2+Re(y)^2)^(1/2)

and if x=0 then answer is zero

### #8

Posted 03 May 2006 - 07:24 AM

The right must be 1/re(x), this because Re(x) is not equa to x when x is imagine.

i*Re(y) must be an imagine because Re(y) is real.

### #9

Posted 06 May 2006 - 07:28 AM

i mean cp300 gives no answerSorry, I can not understand why there should be no answer.

The right must be 1/re(x), this because Re(x) is not equa to x when x is imagine.

i*Re(y) must be an imagine because Re(y) is real.

### #10

Posted 06 May 2006 - 12:55 PM

It might make a big different!

### #11

Posted 06 May 2006 - 01:11 PM

the answer should be Re(x)/((Re(x)^2+Re(y)^2)^(1/2)

and if x=0 then answer is zero

I think that's the problem, you have 2 different answers depending on the value of x and as there is no information on x, you can't give an unique answer because it can be wrong (and cp can't give multiple answers with conditions like "if x=0 then")

### #12 Guest_Guest_*

Posted 06 May 2006 - 09:01 PM

The ClassPad, and every other CAS makes an assumption that x!=0 for most simplifications. Every CAS will simplify x/x to 0. This is wrong is x=0. If x=0 it is not defined.I think that's the problem, you have 2 different answers depending on the value of x and as there is no information on x, you can't give an unique answer because it can be wrong (and cp can't give multiple answers with conditions like "if x=0 then")

### #13

Posted 07 May 2006 - 06:58 AM

make me an example that Re( 1/ (Re( x ) + i* Re(y) ) ) will not be equal to Re(x)/((Re(x)^2+Re(y)^2)^(1/2)The ClassPad, and every other CAS makes an assumption that x!=0 for most simplifications. Every CAS will simplify x/x to 0. This is wrong is x=0. If x=0 it is not defined.

forget about x->infinite , otherwise it should have an answer

also if x-->inf there should be inf

### #14 Guest_Guest_*

Posted 07 May 2006 - 04:05 PM

make me an example that Re( 1/ (Re( x ) + i* Re(y) ) ) will not be equal to Re(x)/((Re(x)^2+Re(y)^2)^(1/2)

forget about x->infinite , otherwise it should have an answer

also if x-->inf there should be inf

Have to tried cexpand(Re( 1/ (Re( x ) + i* Re(y) ) ))

### #15

Posted 10 May 2006 - 09:11 AM

it works but now... another problemHave to tried cexpand(Re( 1/ (Re( x ) + i* Re(y) ) ))

cexpand (Re( 1/(x+y*i))=x/(x^2+y^2) that is wrong because x may not be real

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