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#1 2006-03-29 10:07:17

MathsIsFun
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Registered: 2005-01-21
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Calculus (General) Formulas

Calculus (General) Formulas


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#2 2006-03-29 11:24:17

Ricky
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Registered: 2005-12-04
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Re: Calculus (General) Formulas

Last edited by Ricky (2006-04-04 04:25:21)


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2006-04-01 19:13:48

ganesh
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Registered: 2005-06-28
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Re: Calculus (General) Formulas

If x=f(t), then dy/dx = (dy/dt)/(dx/dt)

dy/dx*dx/dy=1 or dx/dy = 1/(dy/dx)

d(logx)/dx = 1/x

d(Sinx)/dx = Cosx

d(Cosx)/dx = -Sinx

d(tanx)/dx = Sec²x

d(cotx)/dx = -Cosec²x

d(secx)/dx = secxtanx

d(cosecx)/dx = -CosecxCotx


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#4 2006-04-04 04:23:46

Ricky
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Re: Calculus (General) Formulas





"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#5 2006-04-06 02:19:24

krassi_holmz
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Registered: 2005-12-02
Posts: 1,908

Re: Calculus (General) Formulas

In the last Ricky post there is a little mistake:
1. lim a f(x)=a m, and
4. lim f(x)/g(x)= m/n , n != 0.
There won't be any eroors if, instead of:
"Let lim f(x)=m and lim g(x)=n",
we write:
"Let lim f(x)=n and lim g(x) = m".

//After the correction remove this


IPBLE:  Increasing Performance By Lowering Expectations.

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#6 2006-04-08 18:25:42

ganesh
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Registered: 2005-06-28
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Re: Calculus (General) Formulas

Fundamental limits

, a>0

,a>0

, a>0

,
a>0, n∈N


Character is who you are when no one is looking.

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#7 2006-04-09 04:52:13

mikau
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Registered: 2005-08-22
Posts: 1,504

Re: Calculus (General) Formulas

Hold it, the limit of tan 1/x as x approaches zero is pi/2? That doesn't seem right.

Shouldn't it be, limit of arctan 1/x as x approaches zero equals pi/2?


A logarithm is just a misspelled algorithm.

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#8 2006-04-10 00:42:20

George,Y
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Registered: 2006-03-12
Posts: 1,306

Re: Calculus (General) Formulas

and another fomula say it's -x/2 ???


X'(y-Xβ)=0

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#9 2006-04-10 02:07:58

Ricky
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Re: Calculus (General) Formulas

limit as x approaches 0 of tan(1/x) is undefined.

Same goes with arctan.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#10 2006-04-10 02:41:49

ganesh
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Registered: 2005-06-28
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Re: Calculus (General) Formulas

[Thank you, Mikau, George and Ricky. I have incorporated the correction.]

Theorems on limits


where p and q are integers.


Character is who you are when no one is looking.

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#11 2006-04-10 03:07:55

ganesh
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Registered: 2005-06-28
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Re: Calculus (General) Formulas

Some Important Expansions:-


for -1<x<1


for every positive value of x<1.


Character is who you are when no one is looking.

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#12 2006-04-10 05:54:26

mikau
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Registered: 2005-08-22
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Re: Calculus (General) Formulas

Ricky wrote:

limit as x approaches 0 of tan(1/x) is undefined.

Same goes with arctan.

I don't think thats correct. Yes the limit of tan (1/x) as x approaches zero is undefined, but the limit of arctan (1/x) as x approaches zero IS pi/2. Think about it, if  θ is an angle in a right triangle, and the side opposite  θ is 1 and the side adjacent to θ is x, as x approaches zero,  θ aproaches 90 from the left.


A logarithm is just a misspelled algorithm.

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#13 2006-04-10 07:22:58

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Calculus (General) Formulas

This isn't rigourous, but more of a way to think about it.  limit 1/x as x goes to 0 from the positive side is positive infinity.  So what we really want to find is the limit of arctan(x) as x goes to infinity, which is pi/2.

But what about when we approach 1/x from the negative side?  Then it's negative infinity, and thus, we want to find the limit of arctan(x) as x approaches negative infinity.  That's -pi / 2. 

Since the limit from the left is not the limit from the right, the limit does not exist.

For a clear way to see it, just use a graphing calculator.  Or just do arctan(1/.000000001) and arctan(1/-000000001).

Last edited by Ricky (2006-04-10 07:25:59)


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#14 2006-04-10 10:35:56

mikau
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Registered: 2005-08-22
Posts: 1,504

Re: Calculus (General) Formulas

Oh, right. Sorry, I forgot no one specified which side it approaches zero from so the left hand limit has to equal the right hand limit, for a limit to exist.

My bad.


A logarithm is just a misspelled algorithm.

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#15 2006-08-06 19:20:29

Zhylliolom
Real Member
Registered: 2005-09-05
Posts: 412

Re: Calculus (General) Formulas

Maclaurin/Taylor Series

Exponentials and Logarithms

Trigonometric Functions

See bottom of post for definition of B[sub]n[/sub].

See bottom of post for definition of B[sub]n[/sub].

See bottom of post for definition of E[sub]n[/sub].

See bottom of post for definition of B[sub]n[/sub].

Hyperbolic Functions

See bottom of post for definition of B[sub]n[/sub].

See bottom of post for definition of B[sub]n[/sub].

See bottom of post for definition of E[sub]n[/sub].

See bottom of post for definition of B[sub]n[/sub].

B[sub]n[/sub]: The Bernoulli Numbers

The Bernoulli numbers B[sub]n[/sub] are a sequence of special rational numbers. Their derivation is outside the scope of this thread, but an explanation may be offered elsewhere upon sufficient demand. The first few Bernoulli numbers are(note that for odd n other than 1, B[sub]n[/sub] = 0):








E[sub]n[/sub]: The Euler Numbers

The Euler numbers E[sub]n[/sub] are a sequence of special numbers. Their derivation is outside the scope of this thread, but an explanation may be offered elsewhere upon sufficient demand. The first few Euler numbers are(note that for all odd n, E[sub]n[/sub] = 0):







Last edited by Zhylliolom (2006-08-06 22:51:43)

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#16 2008-12-22 13:14:20

glenn101
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Registered: 2008-04-02
Posts: 108

Re: Calculus (General) Formulas

Finding Derivatives By First Principles

f'(x)=lim    f(x+h)-f(x)
        h->0 -------------
                      h


"If your going through hell, keep going."

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