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#1 2009-10-08 12:43:25

MathsIsFun
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Operations with Functions

Please have a look at Operations with Functions and tell me anything wrong or lacking, thanks.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman
 

#2 2009-10-08 22:48:40

mathsyperson
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Re: Operations with Functions

Looks mostly good, but in your 'But Not Always' section you've got an (x-1)² where there should be an (x-3)², and there's an 'ex' that looks like a typo.

I'd also say that if f(x) = (x-3)²/(x-3) then f(3) is undefined, and we can't just simplify the function to make it x-3. There would be a limit at that point, but not a value.


Why did the vector cross the road?
It wanted to be normal.
 

#3 2009-10-09 08:49:26

MathsIsFun
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Re: Operations with Functions

Thank you.

You are right about the "But Not Always" example ... if I can't think of a correct example I will abandon that section.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman
 

#4 2009-10-09 09:23:45

MathsIsFun
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Re: Operations with Functions

How about f(x) = x + 1/x and g(x) = x

(fg)(x) = x+1

???


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman
 

#5 2009-10-09 10:16:10

bobbym
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Re: Operations with Functions

Hi MathsisFun;

Some functions have a removable singularity.

Def: When the function is bounded in a neighbourhood around a singularity, the function can be redefined at the point to remove it; hence it is known as a removable singularity. In contrast, whe a function tends to infinity as z approaches 0; thus, it is not bounded and the singularity is not removable,

According to this and I found other sites that agree. Just google for removable singularity.

f(x) = sin(x) / x  , f(0) is defined and is equal to 1 This is the example they all give for a removable singularity.

Last edited by bobbym (2009-10-09 10:17:40)


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

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