Is any suggestion how can I solve partial fractions on my Algebra FX?

Thank you.

# partial fractions

Started by
mikrogen
, Nov 26 2002 03:49 PM

8 replies to this topic

### #1

Posted 26 November 2002 - 03:49 PM

### #2

Posted 27 November 2002 - 04:58 AM

Nope!

### #3

Posted 27 November 2002 - 08:44 AM

what are you getting at with partial fonctions... ? Could you be more precise...!

### #4

Posted 27 November 2002 - 09:19 AM

I can be more precize but Im afraid it wont help you

for example: I need x/((x+1)(x-1)) transform to 1/(x+1) + 1/(x-1). I think it should be called partial fractions.

How to force Casio Algebra FX2.0 to do it? (any program?)

for example: I need x/((x+1)(x-1)) transform to 1/(x+1) + 1/(x-1). I think it should be called partial fractions.

How to force Casio Algebra FX2.0 to do it? (any program?)

### #5

Posted 27 November 2002 - 03:20 PM

that's wrong :

x is 2

x/((x+1)(x-1)) equal 2/3

x/(x+1) + x/(x-1) equal 2/3 + 2 !!!

x is 2

x/((x+1)(x-1)) equal 2/3

x/(x+1) + x/(x-1) equal 2/3 + 2 !!!

### #6

Posted 27 November 2002 - 04:07 PM

Firstly:

x/((x+1)(x-1) = 1/(2(x+1) )+ 1/(2(x-1)) ...

I don't think there's a built in funciton for this on the calc...

However, in cas mode you could integrate the expression (this will splitt the fracion up) then differentiate the answer, and you will have you partial fraciton (or whatever it was called.. I don't know the english name for it )

Anyway, there are easy mathematical approaches do this problem:

*find the zeros of each factor

*maintaine 1 and 1 factor, and set x = that factors zero (to evaluate the other factors in the expression)

*sum all the new fraction

... hope you understand (and please correct me if I'm wrong... been a while since I did this)

x/((x+1)(x-1) = 1/(2(x+1) )+ 1/(2(x-1)) ...

I don't think there's a built in funciton for this on the calc...

However, in cas mode you could integrate the expression (this will splitt the fracion up) then differentiate the answer, and you will have you partial fraciton (or whatever it was called.. I don't know the english name for it )

Anyway, there are easy mathematical approaches do this problem:

*find the zeros of each factor

*maintaine 1 and 1 factor, and set x = that factors zero (to evaluate the other factors in the expression)

*sum all the new fraction

... hope you understand (and please correct me if I'm wrong... been a while since I did this)

### #7

Posted 27 November 2002 - 07:01 PM

2x/((x+1)(x-1)) equals to 1/(x+1) + 1/(x-1), there was mistake in #2..

But: Integration and following derivation got it! I know some techniques how to do it by hand, but it is too slow and erroneous, especially in school exams.

So, thank You very much

But: Integration and following derivation got it! I know some techniques how to do it by hand, but it is too slow and erroneous, especially in school exams.

So, thank You very much

**BiTwhise**!### #8

Posted 27 November 2002 - 07:33 PM

btw..

I don't see why there isn't a seperate function for this on the calc... I mean.. it has to do this transformation anyway to perform integration

I don't see why there isn't a seperate function for this on the calc... I mean.. it has to do this transformation anyway to perform integration

### #9

Posted 28 November 2002 - 08:52 AM

I had a look as well and I didn't find out something like that (g100+ or fx2+)..

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