Difference between revisions of "User:Saul/calculus"
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The initial '''x''' is just a guess.<br> | The initial '''x''' is just a guess.<br> | ||
Note: if this results in divergence ('''x''' does not converge at '''0''' or '''x → ∞''') then '''x<sub>1</sub>''' was a bad guess. | Note: if this results in divergence ('''x''' does not converge at '''0''' or '''x → ∞''') then '''x<sub>1</sub>''' was a bad guess. | ||
+ | |||
+ | ==== Approximating Roots ==== | ||
+ | Differentiation can be used to approximate roots.<br> | ||
+ | <br> | ||
+ | For example approximating the answer to:<br> | ||
+ | '''<sup>3</sup>√(120)'''<br> | ||
+ | <br> | ||
+ | First round to the nearest known value:<br> | ||
+ | '''<sup>3</sup>√(125) = 5'''<br> | ||
+ | <br> | ||
+ | Differentiate the original equation:<br> | ||
+ | '''ƒ(x) = <sup>3</sup>√x = x<sup>1/3</sup>'''<br> | ||
+ | '''ƒ<sup>′</sup>(x) = x<sup>-2/3</sup> / 3'''<br> | ||
+ | <br> | ||
+ | Substitute the approximation:<br> | ||
+ | '''ƒ<sup>′</sup>(120) = 120<sup>-2/3</sup> / 3'''<br> | ||
+ | '''120<sup>-2/3</sup> / 3 = 0.014'''<br> | ||
+ | <br> | ||
+ | Place this into the equation of gradient:<br> | ||
+ | '''(y - 5) / (x - 125) = 0.014'''<br> | ||
+ | <br> | ||
+ | Prearrange to the formulae of a line:<br> | ||
+ | '''y = 0.014x + 3.29'''<br> | ||
+ | <br> | ||
+ | Substitute for the number to approximate with '''x''':<br> | ||
+ | '''y = 0.014(120) + 3.29'''<br> | ||
+ | '''y = 4.934'''<br> | ||
+ | <br> | ||
+ | Therefore:<br> | ||
+ | '''<sup>3</sup>√(120) ≈ 4.934'''<br> | ||
+ | '''<sup>3</sup>√(120) = 4.932''' (calculated)<br> | ||
== Integration == | == Integration == |
Revision as of 00:46, 6 October 2019
Contents
Differentiation
We will use the function notation ƒ(x) which just applies some action to x like this function will double x: ƒ(x) = 2x
Differentiation of a function is taking a function (usually a curve) and finding the gradient at the single instant of x.
For example the function ƒ(x) = x2 will represent a 'U' shaped curve, the gradient at point x will be: ƒ′(x) = 2x
A derivative of a function will be notated with ƒ′, second derivatives are marked ƒ′′ and so on.
Various Notations
The following are various notations for derivatives.
Note that: y = ƒ(x) = ƒ
dy / dx
dƒ / dx
dƒ(x) / dx
y′
[y]′
ƒ′
[ƒ]′
Notation for second derivative:
ƒ′′
d2y / dx2
Various Rules
Note: n represents a real number, and a represents a constant.
ƒ(x) = xn
ƒ′(x) = nxn-1
ƒ(x) = axn
ƒ′(x) = anxn-1
ƒ(x) = a
ƒ′(x) = 0
ƒ(x) = x
ƒ′(x) = 1
ƒ(x) = fg
ƒ′(x) = f′g + g′f
ƒ(x) = f/g
ƒ′(x) = (f′g - g′f) / g2
ƒ(x) = sin(x)
ƒ′(x) = cos(x)
ƒ(x) = cos(x)
ƒ′(x) = -sin(x)
ƒ(x) = tan(x)
ƒ′(x) = 1 / cos2(x) = sec2(x)
Uses
Finding x-Intercept
Differentiation can be used to find the x-intercepts of a line.
The following formulae is used to find this:
xn+1 = xn - ƒ(xn) / ƒ′(xn)
Applying this function until ƒ(x) ≈ 0
The initial x is just a guess.
Note: if this results in divergence (x does not converge at 0 or x → ∞) then x1 was a bad guess.
Approximating Roots
Differentiation can be used to approximate roots.
For example approximating the answer to:
3√(120)
First round to the nearest known value:
3√(125) = 5
Differentiate the original equation:
ƒ(x) = 3√x = x1/3
ƒ′(x) = x-2/3 / 3
Substitute the approximation:
ƒ′(120) = 120-2/3 / 3
120-2/3 / 3 = 0.014
Place this into the equation of gradient:
(y - 5) / (x - 125) = 0.014
Prearrange to the formulae of a line:
y = 0.014x + 3.29
Substitute for the number to approximate with x:
y = 0.014(120) + 3.29
y = 4.934
Therefore:
3√(120) ≈ 4.934
3√(120) = 4.932 (calculated)
Integration
Integration is the reverse of differentiation, sometimes known as anti-derivative.
Integration is usually noted like so:
∫ƒ′(x)dx = ƒ(x) + c
Where dx shows that it will be the integral relative to x (there may be other variables in the equation).
c represents a unknown constant - this must exist because when a function is derived (differentiation) some information is lost so c is to make up for that loss.
Various Rules
k represents a constant.
∫k dx = k + c
∫kx dx = k∫x dx
∫xn dx = xn + 1 / (n + 1)
∫(xn + xm) dx = ∫xn dx + ∫xm dx
∫( ƒ(x) ± g(x) )dx = ∫ƒ(x)dx ± ∫g(x)dx
∫(1 / x) dx = Ln(x) + c
Note: Ln(x) = Loge(x)
Uses
Area
A use of integration is to find the area between a line and the x axis between two points on a graph.
The area between the two points a and b (where x = 0) can be found like so:
First find the derivative:
a∫bƒ′(x) dx = ƒ(x)
Say that the derivative is F, then find the difference of the values for F for each limit like so:
F(b) - F(a) = Area
Note that the area will be negative if it is under the x axis and positive if above.
Also if the line crosses the x axis more than twice then there will be more than two values where x = 0 and at least one positive area and a negative area, these will at least partially cancel each other out if only the left most and right most x values are taken, so to effectively calculate the area the absolute areas between every set of points (where x = 0) muse be calculated.
Solid of Revolution
A cool use for integration is to calculate a solid of revolution - the volume of a shape that is made from the area of two points spun around a axis, for example a triangle could create a cone.
The formulae for this is as follows:
V = Π a∫b ƒ2(x) dx
Note: ƒ2(x) = ƒ(x) * ƒ(x)