Symmetric Difference Qutotient

  • Thread starter Juwad
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  • #1
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Show agrebraically that symmetric difference quotient produces the exact derivative f'(x) = 2ax+b for the quadractic function f(x) = ax^2+bx+c


i know that:

f(a + h) - f(a - h)
Symmetric Difference Quotient = -------------------
2h

I really have no clue on this problem..
 

Answers and Replies

  • #2
Hurkyl
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Have you done any work at all on this? Even a tiny amount?
 
  • #3
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i seem not to get the question can u tell me ?
 
  • #4
Hurkyl
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It wants you to prove

{symmetric quotient of f} = {derivative of f}


Oh, I apologize: you have at least written down the definition of the symmetric quotient of f, evaluated at a, so you have done some work. But my reaction was due to there being a very obvious next thing to try with that expression.


(P.S. the letter a is already defined to be one of the coefficients of your polynomial. So, you should be using a different variable to denote the point at which you're evaluating the symmetric difference)
 
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  • #5
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Hurkyl,

1st thanks for replying, could you help me a little bit here...What's the 1st step i should do? (Prolly not finding the derivative of {symmetric quotient})..umm some help

symmetric difference quotient:
f(n + h) - f(n - h)
-------------------
2h
 
Last edited:
  • #6
Hurkyl
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You've already invoked the definition of symmetric quotient, to get:

[tex]\{\, \text{symmetric quotient of f at n}\, \} = \frac{f(n + h) - f(n - h)}{2h}[/tex]

Are there any other definitions you can invoke to rewrite that expression? Or maybe an algebraic manipulation you can apply to it?
 
  • #7
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definition of derivative?
 
  • #8
Hurkyl
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The definition of derivative has already been applied: (you know f'(x) = 2ax + b). How about the definition of f?
 
  • #9
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definition of f? isn't f a function of x? :confused:
 
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  • #10
Office_Shredder
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What IS f(n+h)?
 
  • #11
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the distance in a point
 
  • #12
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it is said that this can make an invalid approximation..any ideas?
 
  • #13
Hurkyl
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isn't f a function of x?
No. f(x) is a function of x.

But you know a lot more about f than "f is a function"... you know exactly what function it is.
 
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  • #14
HallsofIvy
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Hurkyl said:
No. f(x) is a function of x.

Ouch! f is a function. f(x) is a number: the value of that function at x.
 
  • #15
HallsofIvy
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Hurkyl said:
No. f(x) is a function of x.

Ouch! f is a function. f(x) is a number: the value of that function at x.

Juwan, Hurkyl is trying to get you to DO what the problem itself tells you to do! f isn't just "a" function, it is a specific function and you know what it is. In your first post you said "f(x)= " and then gave the formula. If you've forgotten what it was go back and read your first post!

Now, do the calculation! Exactly what is f(x+h) for that f? Exactly what is f(x-h) for that f?
 
  • #16
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now i see..i got the first part done..

[tex]\L\\\lim_{h\to\0}\frac{[a(x+h)^{2}+b(x+h)+c]-[a(x-h)^{2}+b(x-h)+c]}{2h}[/tex]

[tex]\L\\\lim_{h\to\0}\frac{ax^{2}+2axh+ah^{2}+bx+c-ax^{2}+2axh-ah^{2}-bx+bh-c}{2h}[/tex]

[tex]\L\\\lim_{h\to\0}\frac{4axh+2bh}{2h}[/tex]

[tex]\L\\\lim_{h\to\0}\frac{2\sout{4}ax\sout{h}}{\sout{2h}}+\lim_{h\to\0}\frac{\sout{2}b\sout{h}}{\sout{2h}}=2ax+b[/tex]
explain why symmetric difference quotient usually gives a good approximation of the numerical derivative at a point on the graph of a function?
 
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  • #17
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One last question

For what types of problems is the symmetric difference quotient useful?
 
  • #18
Hurkyl
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There is no limit in the symmetric difference quotient.
 
  • #19
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oh cool!!, so this is right?
 
  • #20
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Just on question, is this for Benson's class? of is another teacher out there using the exact same wording and extra credit projects?
 

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