Where Can I Find Help on Writing Series Expansion Formulas?

[kahlan]

how are u every one there
i am happy to be one of ur friends
just we have challenge because of this limit
so i hope that someone donate to solve it
thanx

 

[micromass]

I would suggest changing this in a 0/0 or an $$\infty / \infty$$ and applying L' Hopital.

 

[n.karthick]

But it looks obvious that the limit does not exist !

 

[micromass]

Ow yes, you're right. It obviously doesn't exist. I read one of the parantheses wrong.

 

[HallsofIvy]

Kahlan, the "wrong parentheses" micromass was seeing was
$$\lim_{x\to\infty}x[(1+ 1/x)^x- e]$$

It is well known that $(1+ 1/x)^x$ goes to e so that (1+ 1/x)^x- e would go to 0- the additional x outside the braces would give an indeterminant form of "infinity* 0".

However, you have the "x" inside the braces and -e outside. $\lim_{x\to\infty}x(1+ 1/x)^x$ is of the form "infinity*e" which does not converge.

 

[kahlan]

$$\lim_{x\to\infty}x[(1+ 1/x)^x- e]$$

 

[kahlan]

it sould be like $$\lim_{x\to\infty}x[(1+ 1/x)^x- e]$$
sorry guys
as well as answer will be 1

 

[kahlan]

how are u every one there
i am happy to be one of ur friends
just we have challenge because of this limit
so i hope that someone donate to solve it
thanx

sorry

 

[scichem]

The limit does exist.

You have:
x[(1 + 1/x)^x] - e = x[exp(x*ln(1+1/x))-e)
Then you use h=1/x, what gives:

1/h*[exp(ln(1+h)/h)-e)
=1/h*[exp(1/h*(h-h²/2+h^3/3+o(h))-e)]
=1/h*[exp(1-h/2+h²/3+o(h))-e)]
=1/h*[e(exp(-h/2+h²/3+o(h))-1)]
=1/h*[e(1-h/2+h²/3+o(h)-1)
=1/h*[e(-h/2+h²/3+o(h)]
=e/h*(-h/2+h²/3+o(h)]
=-e/2+h/3+o(h)

As x→∞, then h→0

Finally, you get:

lim x→∞ (x[(1 + 1/x)^x] - e) = lim h→0 (-e/2+h/3+o(h)) = -e/2

 

[D H]

scichem, you're being a bit sloppy with braces and parentheses, so it's hard to see where you went wrong. You should have obtained

$$x\bigl((1 + 1/x)^x - e\bigr) = -\frac e 2 + \frac{11 e}{24}\frac 1 x - O\bigl(1/x^2\bigr)$$
or
$$\bigl((1 + h)^{1/h} - e\bigr)/h = -\frac e 2 + \frac{11 e}{24}h - O\bigl(h^2\bigr)$$

In the limit x→∞ or h→0, these become -e/2.


Somehow you did obtain the right limit.

 

[scichem]

scichem, you're being a bit sloppy with braces and parentheses, so it's hard to see where you went wrong. You should have obtained

$$x\bigl((1 + 1/x)^x - e\bigr) = -\frac e 2 + \frac{11 e}{24}\frac 1 x - O\bigl(1/x^2\bigr)$$
or
$$\bigl((1 + h)^{1/h} - e\bigr)/h = -\frac e 2 + \frac{11 e}{24}h - O\bigl(h^2\bigr)$$

In the limit x→∞ or h→0, these become -e/2.Somehow you did obtain the right limit.

You're right, thanks for noticing, I've forgotten something in the series expansion of exp(u), actually the u²/2 part with u=-h/2+h²/3+o(h²), which actually gives the element h²/8:

1/h*[exp(ln(1+h)/h)-e)
=1/h*[exp(1/h*(h-h²/2+h^3/3+o(h²))-e)]
=1/h*[exp(1-h/2+h²/3+o(h²))-e)]
=1/h*[e(exp(-h/2+h²/3+o(h²))-1)]
=1/h*[e(1-h/2+h²/3+o(h²)-1)
=1/h*[e(-h/2+h²/3+o(h²)]
=e/h*(-h/2+h²/3+h²/8+o(h²)]
=e/h*(-h/2+11h²/24+o(h²)
=-e/2+11h/24+o(h)

And then you find the limit -e/2.

PS: Where can I find a tutorial/help section on this forum to know how to write series expansion formulas like what you've done?

 

[D H]

PS: Where can I find a tutorial/help section on this forum to know how to write series expansion formulas like what you've done?

This thread: [thread=546968]LaTeX Guide: Include mathematical symbols and equations in a post[/thread].Edit
One last comment about this thread:

Look at the dates when posting. This thread was last active over a year ago. In general it isn't a good idea to raise old threads from the dead. In this case it was OK because (a) there was no correct resolution, and (b) a year is old but not terribly old. On the other hand, if you come across a seven year old thread with no resolution, don't get the seven year itch to resurrect it. Just let it lie.