What should i do after using Taylor series?

[Outrageous]

Homework Statement

The first equation on the uploaded paper converts to the last equation.

Homework Equations

when i substitute ln (1-u)=-u-(1/2)(u^2) into the first equation, i can get the first term in (3rd equation).
but the second term of the 3rd equation ?

The Attempt at a Solution

I tried ln(1-u)= -u-(1/2)(u^2)-(u^3)/3, then by completing the square. yet can't get. i wonder is that any way that i not yet learn.
please guide what to do. Thank you.

 

[Mandelbroth]

Homework Statement

The first equation on the uploaded paper converts to the last equation.

Homework Equations

when i substitute ln (1-u)=-u-(1/2)(u^2) into the first equation, i can get the first term in (3rd equation).
but the second term of the 3rd equation ?

The Attempt at a Solution

I tried ln(1-u)= -u-(1/2)(u^2)-(u^3)/3, then by completing the square. yet can't get. i wonder is that any way that i not yet learn.
please guide what to do. Thank you.

I don't understand your question. Could you rephrase what you are asking?

Though, I'll point out (as a fun fact) that, analytically, $x_{\text{max}}=\dot{x}_0\frac{g \operatorname{W}\left(\frac{-e^{\frac{-\gamma \dot{z}_0}{g}-1} (g+\gamma \dot{z}_0)}{g}\right)+g+\gamma \dot{z}_0}{\gamma (g+\gamma \dot{z}_0)}$, where $\text{W}$ is the Lambert W function. It's pretty.

 

[Seydlitz]

I don't understand your question. Could you rephrase what you are asking?

Though, I'll point out (as a fun fact) that, analytically, $x_{\text{max}}=\dot{x}_0\frac{g \operatorname{W}\left(\frac{-e^{\frac{-\gamma \dot{z}_0}{g}-1} (g+\gamma \dot{z}_0)}{g}\right)+g+\gamma \dot{z}_0}{\gamma (g+\gamma \dot{z}_0)}$, where $\text{W}$ is the Lambert W function. It's pretty.

He's asking why he didn't get the second term in $x_{max}$ equation after he had substituted the natural logarithm with its Taylor expansion. He managed only to recreate the first term. Is there anything that he missed?

 

[Mandelbroth]

Let's see what I can do:

$$\left(\frac{\dot{z}_0}{\gamma}+\frac{g}{\gamma^2}\right)\frac{\gamma x_{\text{max}}}{\dot{x}_0}+\frac{g}{\gamma^2}\ln\left(1-\frac{\gamma x_{\text{max}}}{\dot{x}_0}\right)=0 \\ \left(\frac{\dot{z}_0}{\gamma}+\frac{g}{\gamma^2}\right)\frac{\gamma x_{\text{max}}}{\dot{x}_0}=\frac{g}{\gamma^2}\left(\frac{\gamma x_{\text{max}}}{\dot{x}_0}+\frac{1}{2}\left(\frac{\gamma x_{\text{max}}}{\dot{x}_0}\right)^2+\cdots\right)\\ \text{For clarity and ease of notation, I now substitute the variable } u \text{ for}\frac{\gamma x_{\text{max}}}{\dot{x}_0}. \\ \left(\frac{\dot{z}_0}{\gamma}+\frac{g}{\gamma ^2}\right)u=\frac{g}{\gamma^2}\left(u+\frac{u^2}{2}+\frac{u^3}{3}\cdots\right)\\ 0=\frac{\dot{z}_0 u}{\gamma}+\frac{g}{\gamma^2}\left(\frac{u^2}{2}+\frac{u^3}{3}+\cdots \right)$$

Can you take it from there, Outrageous?

 

[Outrageous]

Let's see what I can do:

$$\left(\frac{\dot{z}_0}{\gamma}+\frac{g}{\gamma^2}\right)\frac{\gamma x_{\text{max}}}{\dot{x}_0}+\frac{g}{\gamma^2}\ln\left(1-\frac{\gamma x_{\text{max}}}{\dot{x}_0}\right)=0 \\ \left(\frac{\dot{z}_0}{\gamma}+\frac{g}{\gamma^2}\right)\frac{\gamma x_{\text{max}}}{\dot{x}_0}=\frac{g}{\gamma^2}\left(\frac{\gamma x_{\text{max}}}{\dot{x}_0}+\frac{1}{2}\left(\frac{\gamma x_{\text{max}}}{\dot{x}_0}\right)^2+\cdots\right)\\ \text{For clarity and ease of notation, I now substitute the variable } u \text{ for}\frac{\gamma x_{\text{max}}}{\dot{x}_0}. \\ \left(\frac{\dot{z}_0}{\gamma}+\frac{g}{\gamma ^2}\right)u=\frac{g}{\gamma^2}\left(u+\frac{u^2}{2}+\frac{u^3}{3}\cdots\right)\\ 0=\frac{\dot{z}_0 u}{\gamma}+\frac{g}{\gamma^2}\left(\frac{u^2}{2}+\frac{u^3}{3}+\cdots \right)$$

Can you take it from there, Outrageous?

Thanks for replies.
My question is exactly like what Seydlitz said.
Then after that, what should I do next?

 

[Mandelbroth]

Thanks for replies.
My question is exactly like what Seydlitz said.
Then after that, what should I do next?

Isolate $u$, substitute back $u=\frac{\gamma x_{\text{max}}}{\dot{x}_0}$, isolate $x_{\text{max}}$.

 

[Outrageous]

Isolate $u$, substitute back $u=\frac{\gamma x_{\text{max}}}{\dot{x}_0}$, isolate $x_{\text{max}}$.

Sorry. Can I ask how to isolate u? By using?

 

[dirk_mec1]

There are three solutions (if you negect terms higher than O(u^4) ). One of them is u=0.

 

[Outrageous]

There are three solutions (if you negect terms higher than O(u^4) ). One of them is u=0.

One will be u= 0 , then what should I going to do next in order to get ...last equation?

 

[CompuChip]

If u is not 0 you can divide it out and you are left with a quadratic equation.
Do you know how to solve something of the form ax² + bx + c = 0?

 

[Outrageous]

If u is not 0 you can divide it out and you are left with a quadratic equation.
Do you know how to solve something of the form ax² + bx + c = 0?

Use http://www.sosmath.com/algebra/quadraticeq/quadraformula/quadraformula.html formula or by completing the square...
This are two i know but I can't get the last equation I uploaded . Please guide. Thanks TT