- #1

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i know that:

f(a + h) - f(a - h)

Symmetric Difference Quotient = -------------------

2h

I really have no clue on this problem..

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- Thread starter Juwad
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- #1

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i know that:

f(a + h) - f(a - h)

Symmetric Difference Quotient = -------------------

2h

I really have no clue on this problem..

- #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)

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

Oh, I apologize: you have at least written down the definition of the symmetric quotient of

(P.S. the letter

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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

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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[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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- #9

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definition of f? isn't f a function of x?

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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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No.isn't f a function of x?

But you know a lot more about

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- #14

HallsofIvy

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Hurkyl said:No.f(x)is a function ofx.

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 ofx.

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?

[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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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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