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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,713

Calculus (General) Formulas

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

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**Ricky****Moderator**- Registered: 2005-12-04
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*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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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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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

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Ricky****Moderator**- Registered: 2005-12-04
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"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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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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**Fundamental limits**

a>0, n∈N

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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

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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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,379

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

**X'(y-Xβ)=0**

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**Ricky****Moderator**- Registered: 2005-12-04
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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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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,828

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

**Theorems on limits**

where p and q are integers.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 47,828

**Some Important Expansions:-**

for -1<x<1

for every positive value of x<1.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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

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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**Ricky****Moderator**- Registered: 2005-12-04
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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)*

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

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

**Maclaurin/Taylor Series**

**Exponentials and Logarithms**

**Trigonometric Functions**

**Hyperbolic Functions**

**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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**glenn101****Member**- Registered: 2008-04-02
- Posts: 108

**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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