Symmetric Difference Qutotient

[Johnwill]

how agrebraically that symmetric difference quotient produces the exact derivative f'(x) = 2ax+b for the quadractic function f(x) = ax^2+bx+c


explain why symmetric difference quotient usually gives a good approximation of the numerical derivative at a point on the graph of a function?


regards

John

 

[galactus]

The 'symmetric difference quotient'. Maybe I am out of touch, but that's some new terminology for me. Learn somethin' new all the time.

Anyway, this is the basis behind differentiation. How and why it works is what calc is about. I will help with the first part. You can answer part b. Okey-doke?.

\(\displaystyle \L\\\lim_{h\to\0}\frac{a(x+h)^{2}+b(x+h)+c-(ax^{2}+bx+c)}{h}\)

Simplify this limit and do away with what h's you can. The h's left tend to 0 and your limit remains. You know what it is, 2ax+b. Give 'er a shot.

 

[Johnwill]

yo galactus, can you explain me a bit...i know you replaced the (x=x+h)..did you use the definition of symmetric quotient


\(\displaystyle \{\, \text{symmetric quotient of f at n}\, \} = \frac{f(n + h) - f(n - h)}{2h}\)

 

[galactus]

I used:

\(\displaystyle \L\\\lim_{h\to\0}\frac{f(x+h)-f(x)}{h}\)

 

[galactus]

I used the forward difference quotient. You need the symmetric:

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

\(\displaystyle \L\\\lim_{h\to\0}\frac{ax^{2}+2axh+ah^{2}+bx+c-ax^{2}+2axh-ah^{2}-bx+bh-c}{2h}\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{4axh+2bh}{2h}\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{2\sout{4}ax\sout{h}}{\sout{2h}}+\lim_{h\to\0}\frac{\sout{2}b\sout{h}}{\sout{2h}}=2ax+b\)

 

[Johnwill]

I used the forward difference quotient. You need the symmetric:

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

\(\displaystyle \L\\\lim_{h\to\0}\frac{ax^{2}+2axh+ah^{2}+bx+c-ax^{2}+2axh-ah^{2}-bx+bh-c}{2h}\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{4axh+2bh}{2h}\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{2\sout{4}ax\sout{h}}{\sout{2h}}+\lim_{h\to\0}\frac{\sout{2}b\sout{h}}{\sout{2h}}=2ax+b\)

yea, your pretty good..but it's 4axh instead of 24

would this give an invalid approximation? any idea? :roll: thanks though

 

[galactus]

That's not 24, it's 4. Look closer. The strikethrough makes it look like a 24. It is 4 struck out replaced with the 2 beside it.

It's \(\displaystyle \L\\\frac{4axh+2bh}{2h}=2ax+b\)

 

[Johnwill]

That's not 24, it's 4. Look closer. The strikethrough makes it look like a 24. It is 4 struck out replaced with the 2 beside it.

It's \(\displaystyle \L\\\frac{4axh+2bh}{2h}=2ax+b\)

yup, thanx, would this make error or invalid approximation in some calculations?

 

[galactus]

The symmetric difference quotient is an average of the two other difference quotients.

That's the same answer you'd get if you just differentiated.

 

[Johnwill]

so symmetric difference is just the average of the two points and it gives an accurate measure...There was a question that asked where would it make an invalid approximation?

 

[Johnwill]

explain why symmetric difference quotient usually gives a good approximation of the numerical derivative at a point on the graph of a function?

 

[pka]

explain why symmetric difference quotient usually gives a good approximation of the numerical derivative at a point on the graph of a function?

I have stayed out of this tread because I was not sure that you actually understood this topic. Are you aware that if the derivative exists, that what you call the “symmetric difference quotient” actually equals the derivative? So in what sense do you mean “approximation of the numerical derivative”? Is this a question about numerical methods for finding derivatives? Clearly this would be easier to program. So it the function is differentiable, because the two are equivalent, it would be reasonable to use this definition.

 

[Johnwill]

explain why symmetric difference quotient usually gives a good approximation of the numerical derivative at a point on the graph of a function?

I have stayed out of this tread because I was not sure that you actually understood this topic. Are you aware that if the derivative exists, that what you call the “symmetric difference quotient” actually equals the derivative? So in what sense do you mean “approximation of the numerical derivative”? Is this a question about numerical methods for finding derivatives? Clearly this would be easier to program. So it the function is differentiable, because the two are equivalent, it would be reasonable to use this definition.

pka thanks for replying yes i understand that “symmetric difference quotient” is same as the derivative, what i've come to know is that the “symmetric difference quotient” is used by the calculator.(ti-83..)

do you know for what types of problems is the symmetric difference quotient useful?

I think for a curve function, maybe a cubic, quadratice or exponential function.

Do you agree?

 

[pka]

that the “symmetric difference quotient” is used by the calculator.(ti-83..)

Well I did say the it is easy to program that way.

I think for a curve function, maybe a cubic, quadratic or exponential function.

All of those are ‘well behaved’ functions. If fact, most function a user of ti-83’s would be analytic, that is have derivatives. Therefore, that method works quite well on that class of functions.

But look at the graph in the graphic:

Clearly, that function has no derivative at x=1.
However, symmetric derivative exists at x=1 and is itself 0.

 

[Johnwill]

thank you