Geometric Derivation of Maclauren series?
I had a couple of questions regarding what might be a called a "geometric" derivation. What have a derived, and is a geometric approach possible?
I'm familiar with how to derive Maclauren's Series for a function.
$f(x)=sum_{i=0}^infty c_ix^i$ then find relationships between the coefficients and the value of f and its derivatives at 0.
1) My current approach.
Given the value of a function at 0, $f(0)$, and its derivative there $f'(0)$, the derivative functions as the tangent of a triangle. So multiply by a horizontal distance, $Delta x$, I get the increase in height, the other leg, $f'(x)Delta x$.
Well the derivative isn't the same over $[0,Delta x]$. It will be $f'(0)$ at 0 and approximately $f'(0)+Delta x f''(0)$ at $Delta x$. Average those two values together, we get $f'(0)+frac{Delta x}{2}f''(0)$
Putting it all together, we get $f(Delta x)=f(0)+Delta xf'(0)+frac{Delta x^2}{2}f''(0)$.
So far, we have agreenment with Maclauren's Series to second order. And here is where it breaks down.
Iterate the same principle for the second derivative. It has different values at (0) and $Delta x$ so take the average of $f''(0)$ and $f''(0)+Delta xf'''(0)$, getting $f''(0)+frac{Delta x}{2}f'''(0)$. Plug that in, and we get.
$f(Delta x)=f(0)+Delta xf'(0)+frac{Delta x^2}{2}f''(0)+frac{Delta x^3}{4}f'''(0)$
And so there's a deviation from Maclauren series from third order on. Instead of factorials in the denominator, I have powers of two.
This is clearly not the Maclauren Series. Is it anything? I'm not sure if the series produces any useful functions if carried out.
2) It looks like taking the average of the value of a function and its derivatives at two points, combined with geometric arguments suffices to get a polynomial series correct to second order. Beyond that it fails. One possible way of moving forward is Take the average of the second derivative in 3 places, the third derivative in 4, and so on. Not sure if this or something like it will get you there.
In short, is there away of "averaging tangents" taken at select intervals that will give yo u the Maclauren series?
Are there other such approaches?
taylor-expansion
asked Nov 21 at 19:06
TurlocTheRed
818311
add a comment |
I had a couple of questions regarding what might be a called a "geometric" derivation. What have a derived, and is a geometric approach possible?
I'm familiar with how to derive Maclauren's Series for a function.
$f(x)=sum_{i=0}^infty c_ix^i$ then find relationships between the coefficients and the value of f and its derivatives at 0.
1) My current approach.
Given the value of a function at 0, $f(0)$, and its derivative there $f'(0)$, the derivative functions as the tangent of a triangle. So multiply by a horizontal distance, $Delta x$, I get the increase in height, the other leg, $f'(x)Delta x$.
Well the derivative isn't the same over $[0,Delta x]$. It will be $f'(0)$ at 0 and approximately $f'(0)+Delta x f''(0)$ at $Delta x$. Average those two values together, we get $f'(0)+frac{Delta x}{2}f''(0)$
Putting it all together, we get $f(Delta x)=f(0)+Delta xf'(0)+frac{Delta x^2}{2}f''(0)$.
So far, we have agreenment with Maclauren's Series to second order. And here is where it breaks down.
Iterate the same principle for the second derivative. It has different values at (0) and $Delta x$ so take the average of $f''(0)$ and $f''(0)+Delta xf'''(0)$, getting $f''(0)+frac{Delta x}{2}f'''(0)$. Plug that in, and we get.
$f(Delta x)=f(0)+Delta xf'(0)+frac{Delta x^2}{2}f''(0)+frac{Delta x^3}{4}f'''(0)$
And so there's a deviation from Maclauren series from third order on. Instead of factorials in the denominator, I have powers of two.
This is clearly not the Maclauren Series. Is it anything? I'm not sure if the series produces any useful functions if carried out.
2) It looks like taking the average of the value of a function and its derivatives at two points, combined with geometric arguments suffices to get a polynomial series correct to second order. Beyond that it fails. One possible way of moving forward is Take the average of the second derivative in 3 places, the third derivative in 4, and so on. Not sure if this or something like it will get you there.
In short, is there away of "averaging tangents" taken at select intervals that will give yo u the Maclauren series?
Are there other such approaches?
taylor-expansion
asked Nov 21 at 19:06
TurlocTheRed
818311
add a comment |
I had a couple of questions regarding what might be a called a "geometric" derivation. What have a derived, and is a geometric approach possible?
I'm familiar with how to derive Maclauren's Series for a function.
$f(x)=sum_{i=0}^infty c_ix^i$ then find relationships between the coefficients and the value of f and its derivatives at 0.
1) My current approach.
Given the value of a function at 0, $f(0)$, and its derivative there $f'(0)$, the derivative functions as the tangent of a triangle. So multiply by a horizontal distance, $Delta x$, I get the increase in height, the other leg, $f'(x)Delta x$.
Well the derivative isn't the same over $[0,Delta x]$. It will be $f'(0)$ at 0 and approximately $f'(0)+Delta x f''(0)$ at $Delta x$. Average those two values together, we get $f'(0)+frac{Delta x}{2}f''(0)$
Putting it all together, we get $f(Delta x)=f(0)+Delta xf'(0)+frac{Delta x^2}{2}f''(0)$.
So far, we have agreenment with Maclauren's Series to second order. And here is where it breaks down.
Iterate the same principle for the second derivative. It has different values at (0) and $Delta x$ so take the average of $f''(0)$ and $f''(0)+Delta xf'''(0)$, getting $f''(0)+frac{Delta x}{2}f'''(0)$. Plug that in, and we get.
$f(Delta x)=f(0)+Delta xf'(0)+frac{Delta x^2}{2}f''(0)+frac{Delta x^3}{4}f'''(0)$
And so there's a deviation from Maclauren series from third order on. Instead of factorials in the denominator, I have powers of two.
This is clearly not the Maclauren Series. Is it anything? I'm not sure if the series produces any useful functions if carried out.
2) It looks like taking the average of the value of a function and its derivatives at two points, combined with geometric arguments suffices to get a polynomial series correct to second order. Beyond that it fails. One possible way of moving forward is Take the average of the second derivative in 3 places, the third derivative in 4, and so on. Not sure if this or something like it will get you there.
In short, is there away of "averaging tangents" taken at select intervals that will give yo u the Maclauren series?
Are there other such approaches?
taylor-expansion
asked Nov 21 at 19:06
TurlocTheRed
818311
I had a couple of questions regarding what might be a called a "geometric" derivation. What have a derived, and is a geometric approach possible?
I'm familiar with how to derive Maclauren's Series for a function.
$f(x)=sum_{i=0}^infty c_ix^i$ then find relationships between the coefficients and the value of f and its derivatives at 0.
1) My current approach.
Given the value of a function at 0, $f(0)$, and its derivative there $f'(0)$, the derivative functions as the tangent of a triangle. So multiply by a horizontal distance, $Delta x$, I get the increase in height, the other leg, $f'(x)Delta x$.
Well the derivative isn't the same over $[0,Delta x]$. It will be $f'(0)$ at 0 and approximately $f'(0)+Delta x f''(0)$ at $Delta x$. Average those two values together, we get $f'(0)+frac{Delta x}{2}f''(0)$
Putting it all together, we get $f(Delta x)=f(0)+Delta xf'(0)+frac{Delta x^2}{2}f''(0)$.
So far, we have agreenment with Maclauren's Series to second order. And here is where it breaks down.
Iterate the same principle for the second derivative. It has different values at (0) and $Delta x$ so take the average of $f''(0)$ and $f''(0)+Delta xf'''(0)$, getting $f''(0)+frac{Delta x}{2}f'''(0)$. Plug that in, and we get.
$f(Delta x)=f(0)+Delta xf'(0)+frac{Delta x^2}{2}f''(0)+frac{Delta x^3}{4}f'''(0)$
And so there's a deviation from Maclauren series from third order on. Instead of factorials in the denominator, I have powers of two.
This is clearly not the Maclauren Series. Is it anything? I'm not sure if the series produces any useful functions if carried out.
2) It looks like taking the average of the value of a function and its derivatives at two points, combined with geometric arguments suffices to get a polynomial series correct to second order. Beyond that it fails. One possible way of moving forward is Take the average of the second derivative in 3 places, the third derivative in 4, and so on. Not sure if this or something like it will get you there.
In short, is there away of "averaging tangents" taken at select intervals that will give yo u the Maclauren series?
Are there other such approaches?
taylor-expansion
taylor-expansion
asked Nov 21 at 19:06
TurlocTheRed
818311
asked Nov 21 at 19:06
TurlocTheRed
818311
asked Nov 21 at 19:06
TurlocTheRed
818311
asked Nov 21 at 19:06
TurlocTheRed
818311
asked Nov 21 at 19:06
TurlocTheRed
818311
818311
add a comment |
add a comment |
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