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Author: Subject: Calculus! For beginners, with a ‘no theorems’ approach!
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[*] posted on 27-4-2016 at 10:38


Doh ! After all that i pressed the wrong calculator buttons at the easy step !

How did you get a number out of the 'transcendental equation' ?

Is there a faster method than Meditiation or Reincarnation ?




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[*] posted on 27-4-2016 at 10:50


Quote: Originally posted by aga  
Doh ! After all that i pressed the wrong calculator buttons at the easy step !

How did you get a number out of the 'transcendental equation' ?

Is there a faster method than Meditiation or Reincarnation ?


Reincarnation is very slow and takes guts. And then you end up in Cleopatra's body and it all gets very messy!

For 1 variable equations that have no analytical solutions, there exists a myriad of numerical (aka iteration methods), the oldest of which is probably the Regula Falsi method.

[Edited on 27-4-2016 by blogfast25]




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[*] posted on 27-4-2016 at 11:50


Pretty much maths-saturated at the mo, so i'll file that for Future reading.

Thanks for the reference tho.




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[*] posted on 27-4-2016 at 12:37


I'll leave the partials till tomorrow then.



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[*] posted on 28-4-2016 at 06:06
Partial Derivatives:


Earlier we saw that the first derivative of a function f(x):
$$f’(x)=\frac{df(x)}{dx}$$
... is the gradient (slope) of that function. So how does that work for a function of two variables, f(x,y), or even three or four? Will there be 2, 3 or 4 gradients?

Let’s look at a simple function f(x,y):

$$z=ax+by+c$$
This is the description of a (flat) plane in three dimensions:

Partial derivatives.png - 7kB

z is the black plane in the pic.

Now we add another plane, this one red:

$$y=B$$

The intersection of the black and red planes is the green line with formula:

$$z=ax+bB+c$$
The first derivative to x of this line is:
$$\frac{dz}{dx}=a+0+0=a$$

This is the gradient of the black plane in the x-direction.

Analogously, if we add another plane, this one light blue:
$$x=A$$

The intersection of the black and light blue planes is the yellow line with formula:

$$z=aA+by+c$$
The first derivative to y of this line is:
$$\frac{dz}{dy}=0+b+0=b$$

This is the gradient of the black plane in the y-direction.

We call these gradients partial derivatives:
$$\frac{\partial z}{\partial x}\:\text{and}\:\frac{\partial z}{\partial y}$$
The partial derivative of z to x is computed by ‘pretending’ y is a constant:
$$\frac{\partial z}{\partial x}=a+0+0=a$$
The partial derivative of z to y is computed by ‘pretending’ x is a constant:
$$\frac{\partial z}{\partial y}=0+b+0=b$$
But this isn’t limited to simple functions like z. It works for ALL functions f(x,y), as a few examples will show:

1.
$$z=3x^2-5y^3$$
$$\frac{\partial z}{\partial x}=6x$$
$$\frac{\partial z}{\partial y}=-15y^2$$
2.
$$u=x^2y+3xy^3-xy$$
$$\frac{\partial u}{\partial x}=2xy+3y^3-y$$
$$\frac{\partial u}{\partial y}=x^2+9xy^2-x$$
3.
$$v=\frac{x}{y} -2\frac{y}{x}$$
$$\frac{\partial v}{\partial x}=\frac1y+2\frac{y}{x^2}$$
$$\frac{\partial v}{\partial y}=-\frac{x}{y^2}-\frac2x$$
4.
$$y=ze^{2x}$$
$$\frac{\partial y}{\partial x}=2ze^{2x}$$
$$\frac{\partial y}{\partial z}=e^{2x}$$
Rules of Partial Derivation:

The rules of partial derivation are exactly the same as for ‘normal’ derivation, including the chain rule. Two examples of the latter:

1.
$$z=\sqrt{x^2-3xy}$$
$$\frac{\partial z}{\partial x}=\frac12 (x^2-3xy)^{-\frac12}\frac{\partial}{\partial x}(x^2-3xy)$$
$$=\frac12 (x^2-3xy)^{-\frac12}(2x-3y)$$
and:
$$\frac{\partial z}{\partial y}=\frac12 (x^2-3xy)^{-\frac12}\frac{\partial}{\partial y}(x^2-3xy)$$
$$=\frac12 (x^2-3xy)^{-\frac12}(-3x)=-\frac32 x(x^2-3xy)^{-\frac12}$$
2.
$$z=e^{x^3-4xy^2}$$
$$\frac{\partial z}{\partial x}=e^{x^3-4xy^2}\frac{\partial}{\partial x}(x^3-4xy^2)$$
$$=e^{x^3-4xy^2}(3x^2-4y^2)$$
and:
$$\frac{\partial z}{\partial y}=e^{x^3-4xy^2}\frac{\partial}{\partial y}(x^3-4xy^2)$$
$$=e^{x^3-4xy^2}(-8xy)=-8xye^{x^3-4xy^2}$$

Partial derivatives of functions dependent on more than two independent variables:

Such a function can generically by represented by:
$$f(x_1,x_2,x_3,...,x_i,...,x_n)$$
For each variable x<sub>i</sub> a partial derivative:
$$\frac{\partial f}{\partial x_i}$$
... can be computed. It's done by considering all other variables (other than x<sub>i</sub>;) as constants.

Consider the following example:

$$u(x,y,z)=x^3y+x^2z^2-3zy^2+5xyz$$
$$\frac{\partial u}{\partial x}=3x^2y+2xz^2+5yz$$
$$\frac{\partial u}{\partial y}=x^3-6zy+5xz$$
$$\frac{\partial u}{\partial z}=2x^2z-3y^2+5xy$$
<hr>
I'll let that 'sink in' for a bit and take questions if there are any. Then some simple exercises will be posted.



[Edited on 28-4-2016 by blogfast25]




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[*] posted on 28-4-2016 at 09:33


Cool ! Let's Go !

Haven't gotten anything wrong for Hours now !





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[*] posted on 28-4-2016 at 11:27


Firstly, a word about notation. The still frequently used:
$$\frac{\partial f}{\partial x}$$
... is fairly clunky (even more so in LaTex), so I propose a more modern and shorter notation:
$$\frac{\partial f}{\partial x}=f_x$$
Example:
$$u=2x-6y^3$$
$$\implies \frac{\partial f}{\partial y}=f_y=-18y^2$$
Exercises: compute f<sub>x</sub> and f<sub>y</sub> for the following (x,y) functions. (I'll put in a trigger warning, if there's 'danger' ahead).

1.
$$x^2y^3-2xy^4$$
2.
$$\cos x+\sin y$$
3.
$$2xy+\frac{5y}{x}$$
4. (product rule)
$$(x^2-y)(1+y^2)$$
5.
$$x^3y^2-6xy+3x$$
6. (chain rule)
$$\cos(2x-y)$$
7.(product rule + chain rule)
$$x^2\sin (y-x)$$
8. (chain rule)
$$e^{3xy}$$

Enjoy!

[Edited on 28-4-2016 by blogfast25]




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[*] posted on 28-4-2016 at 12:21


Hmm. Best go cautiously then ...

1. $$x^2y^3-2xy^4$$

$$\frac{\partial f}{\partial x}= f_x = 2xy^3-2y^4$$
$$f_y = 3x^2y^2-8xy^3$$

Owzat ?




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[*] posted on 28-4-2016 at 13:12


Correct. One done, seven more to go.



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[*] posted on 28-4-2016 at 13:18


Hmm. Nothing Bad happened, but it could be a trap.

Let's try the next one
2.
$$\cos x + \sin y$$
$$f_x = sin y - sin x$$
$$f_y = \cos x + cos y$$




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[*] posted on 28-4-2016 at 13:30


3.
$$2xy+\frac{5y}{x}$$
$$f_x = 2y - \frac {5y}{x^2}$$
$$f_y = 2x +\frac 5x$$

it feels like i just stepped on a landmine that leapt 10m into the air after i took my foot off, yet failed to explode, just yet.




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[*] posted on 28-4-2016 at 13:36


4. (geting scared now)
$$(x^2-y)(1+y^2)$$
$$ = (x^2-y)' + (1+y^2)'$$
$$f_x = 2x + y^2$$
$$f_y = x^2 -1 + 2y$$

Gut instinct says it's time to start running.




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[*] posted on 28-4-2016 at 13:48


5.
$$x^3y^2-6xy+3x$$
$$f_x = 3x^2y^2 - 6y + 3$$
$$f_y = 2x^3y - 6x + 3x$$

Gravy Pressure is pretty much in the Danger Zone now.




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[*] posted on 28-4-2016 at 13:55


6.
$$\cos(2x-y)$$
$$f_{bleh} = \cos(2x-y)'(2x-y)'$$
$$f_x = -\sin(2x-y)(2-y)$$
$$f_y = -\sin(2x-y)(2x-1)$$

that does not look right, and i think the Gravy Seals are giving way.


[Edited on 28-4-2016 by aga]

[Edited on 28-4-2016 by aga]




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[*] posted on 28-4-2016 at 13:57


It's no good Heuston. We have Leakage.

Best leave 7 and 8 till the 'morn.




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[*] posted on 28-4-2016 at 14:49


1. Correct

2. Incorrect.
$$\cos x+\sin y$$

If y is a constant, then so is siny, so:
$$f_x=-\sin x$$
If x is a constant, then so is cosx, so:
$$f_y=\cos y$$
3. Correct

4. (product rule)

Incorrect, you're applying the sum rule, not the product rule. Try:
$$(x^2-y)(1+y^2)$$
$$f_x=f_x(x^2-y) \times (1+y^2)+(x^2-y) \times f_x(1+y^2)$$
Similar for f<sub>y</sub>.

5. fx is correct but fy should be:
$$x^3y^2-6xy+3x$$
$$f_y=2x^3y-6x$$
6. (chain rule)
Almost but not quite:
$$\cos(2x-y)$$
$$f_x=-\sin(2x-y)f_x(2x-y)=2\sin(2x-y)$$
$$f_y=-\sin(2x-y)f_y(2x-y)=-\sin(2x-y)$$

Ok, so we've got some gremlins to deal with. :)

[Edited on 28-4-2016 by blogfast25]




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[*] posted on 29-4-2016 at 00:35


Doh. Chain rule is tricky.
7.
$$x^2\sin (y-x)$$
i think it goes like this for the chainy bit:
$$sin(y-x)'(y-x)'$$
$$f_x = (x^2)'sin(y-x)'(y-x)'$$
$$= 2x(-cos(y-x))(-1)$$
$$= 2xcos(y-x)$$
$$f_y = (x^2)(sin(y-x)')(y-x)'$$
$$= x^2(-cos(y-x))(1)$$
$$=-x^2cos(y-x)$$




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[*] posted on 29-4-2016 at 00:42


8.
Now these 'e' things really confuse the hell out of me.
$$e^{3xy}$$
the rule says e^n just ends up as e^n, but surely the 3xy needs messing with, so ignoring all the alarm bells, here goes:
$$f_x = e^{3(yx)'(x)'}$$
$$= e^{3y}$$
$$f_y = e^{3x}$$




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[*] posted on 29-4-2016 at 06:04


7. (chain rule + product rule)
$$x^2\sin(y-x)$$
You got the chain rule right and thus f<sub>y</sub>. But for f<sub>x</sub> you needed to apply also the product rule because x<sup>2</sup> is a function, so:
$$f_x=f_x(x^2) \times \sin(y-x)+x^2 \times f_x(\sin(y-x))$$
$$=2x\sin(y-x)+x^2[-\cos(y-x)\times f_x(y-x)]$$
$$=2x\sin(y-x)+\cos(y-x)$$

8. (chain rule)

$$e^{3xy}$$

Here you were a bit all over the place. Remember how to derive exponentials in one variable only, with the chain rule:

$$g=e^{f(x)}$$
$$g'=e^{f(x)}\times f'(x)$$
For an (x,y) exponential:
$$g=e^{f(x,y)}$$
$$g_x=e^{f(x,y)}\times f_x(x,y)$$
$$g_y=e^{f(x,y)}\times f_y(x,y)$$
Applied here:
$$f_x=e^{3xy} \times f_x(3xy)=e^{3xy}(3y)=3ye^{3xy}$$
$$f_y=e^{3xy} \times f_y(3xy)=e^{3xy}(3x)=3xe^{3xy}$$

So definitely some gremlins, not to mention a few missiles going down rather than up!

[Edited on 29-4-2016 by blogfast25]




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[*] posted on 29-4-2016 at 11:41


Quote: Originally posted by blogfast25  
So definitely some gremlins, not to mention a few missiles going down rather than up!

Beginning to realise that i'm That guy who gets you Rocket Men all the food, lodgings, material, and the booze & hookers every weekend, whilst earning handsomely from that plus side-rackets.

Then organises the team members' coffins when the Rocket goes sideways and some heads must roll (250% markup on coffins).

Then sells the little i can remember to the Enemy for a Vast fortune (and immunity) once they win the war.
(maybe before, depends on immunity from Whom).

It's pretty much apparent that i was not designed to be a gifted theoretical mathematician.




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[*] posted on 29-4-2016 at 12:30


Quote: Originally posted by aga  

It's pretty much apparent that i was not designed to be a gifted theoretical mathematician.


For most (like me), math is hard work (but rewarding, IMO). So there's few with an excuse ("whoohoo, look at me, I'm sooooo gifted!"), so what? :D

In characteristic human style we'll now just stick our heads in the sand and plow on!

Tomorrow: simple multi-variate optima.




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[*] posted on 29-4-2016 at 13:17


Don't get me wrong :

i actually Like this algebra stuff, just that it's pretty clear that i'm not careful or thoughtful enough for such a Precise discipline.

Chemistry should be fine as i have Buckets to mix stuff up in ;)




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[*] posted on 30-4-2016 at 04:57
Multi-variate optima:


Higher up we saw that for a simple function f(x), an optimum exists, if:
$$\frac{df(x)}{dx}=0$$

optima.png - 5kB

Similarly, a multivariate function:
$$f(x_1,x_2,...,x_i,...,x_n)$$
... has an optimum if:
$$f_{x_1}=0,f_{x_n}=0,...,f_{x_n}=0,$$
The latter condition forms a system of simultaneous equations which when solved yields the numerical values x<sub>i</sub> of the optimum.

We won't concern ourselves here with the formal theory of multi-variate optima (which includes detection of absolute and relative extremes and saddle points), instead we'll use common sense to determine whether a found optimum is a maximum or minimum.

First example:

$$f(x,y)=x^2+y^2$$
$$f_x=2x$$
$$f_y=2y$$
An optimum exists for:
$$2x=0 \implies x=0$$
$$2y=0 \implies y=0$$
The optimum value is:
$$f(0,0)=0$$
For slightly larger values of x and y, e.g. (1,1), we get:
$$f(1,1)=2$$
... which indicates (0,0) is a minimum.

Second example:

$$x^2+2y^2-xy+14y$$
$$f_x=2x-y=0 \implies y=2x$$
$$f_y=4y-x+14=0 \implies 8x-x=-14$$
$$x=-2, y=-4$$

3D plot:

optima ex 2.png - 92kB

Third example:
$$f(x,y)=xy+\frac8x+\frac8y$$
$$f_x=y-\frac{8}{x^2}=0$$
$$f_y=x-\frac{8}{y^2}=0$$
Some reworking and substituting:
$$y=\frac{8}{x^2}$$
$$x-\frac{x^4}{8}=0$$
$$x(1-\frac{x^3}{8})=0$$
$$1-\frac{x^3}{8}=0$$
$$x=2$$
$$y=2$$
$$f(2,2)=4+4+4=12$$
$$f(4,4)=16+2+2=18$$
(2,2) is a minimum.

Exercise:
$$f(x,y)=(x-1)^2+y^3-3y^2-9y+5$$


[Edited on 30-4-2016 by blogfast25]




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[*] posted on 30-4-2016 at 09:19


Um. Er. Ok. Find the f(x,y) = 0 ?
$$f(x,y)=(x-1)^2+y^3-3y^2-9y+5$$
$$=x^2-2x+1+y^3-3y^2-9y+5$$
$$f_x = 2x-2+0-0-0+0$$
$$ = 2x - 2$$
$$2x-2 = 0 $$
$$\therefore x=1$$
$$f_y = 0-0+0+3y^2-6y-9+0$$
$$=y^2-2y-3$$
$$=y(y-2)-3$$
$$y(y-2)-3 = 0$$
$$y(y-2)=3$$
$$\therefore y = 3$$
$$\implies f(x,y) = 0 = f(1,3)$$
that very last bit was more common sense and trying some numbers than actual maths.

Did i just become Cleopatra ? If so, bathe your ass in milk immediately.




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[*] posted on 30-4-2016 at 11:30


aga:

Both partial derivatives were correct.

So is x = 1.

But your attempt at solving:

$$y^2-2y-3=0$$

... was very '#creative#' but not very correct! :D:D

No, it does not imply that if:
$$y(y-2)=3$$
... then:
$$y=3$$

For a general quadratic equation:

$$ax^2+bx+c=0$$
$$\implies x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
So we have two roots:
$$y=3,y=-1$$
So we have two optima, (1,-1), a maximum and (1,3), a minimum, as the 3D plot shows:

optima ex 1.png - 134kB


[Edited on 30-4-2016 by blogfast25]




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