Series convergence doubt

[VGNDISCPLN]

Hello everyone i am currently studying transport phenomena by my own and i have a doubt about how the author makes a simplification of an equation by using the taylor series, i have like 4 days trying to understand how does he gets the result. Can someone please explain me what he does and why. i will really apreciate it a lot.
This is the simplification:

How does he determines that this reduces to this:
i apreciate your help a lot.

 

[Cubist]

[math] \left(-\frac{1}{a^2}\right)-\left(-\frac{1}{a^2}\right)=0[/math]
[math] \left(\frac{1}{2}\right)-\left(\frac{1}{2}\frac{x^2}{\delta^2}\right) = \frac{1}{2}\left[ 1-\left(\frac{x}{\delta}\right)^2\right][/math]...you missed the author's "half" outside the square bracket (the denominator becomes multiplied by two)

Then all the remaining addends are multiplied by \(a\) which tends towards 0 in the limit

 

[VGNDISCPLN]

i still don´t understand
thanks anyway

 

[Singleton]

You simply plug in a=0. The only place you can't plug in is when a (actually a²) is in the denominator. But you've got two of those and they happen to cancel.

 

[Dr.Peterson]

i still don´t understand
thanks anyway

Can you state what you don't understand, to help others help you?

The part you are asking about is

[MATH]\lim_{a\rightarrow 0}\left[\left(-\frac{1}{a^2} + \frac{1}{2} +\frac{1}{3}a + \dots\right) - \left(-\frac{1}{a^2} + \frac{1}{2}\frac{x^2}{\delta^2} +\frac{1}{3}\frac{x^3}{\delta^3}a + \dots\right)\right] =[/MATH][MATH]\lim_{a\rightarrow 0}\left[-\frac{1}{a^2} + \frac{1}{2} +\frac{1}{3}a + \dots + \frac{1}{a^2} - \frac{1}{2}\frac{x^2}{\delta^2} -\frac{1}{3}\frac{x^3}{\delta^3}a - \dots\right] =[/MATH][MATH]\lim_{a\rightarrow 0}\left[\frac{1}{2} +\frac{1}{3}a + \dots - \frac{1}{2}\frac{x^2}{\delta^2} -\frac{1}{3}\frac{x^3}{\delta^3}a - \dots\right] =[/MATH][MATH]\lim_{a\rightarrow 0}\left[\left(\frac{1}{2} - \frac{1}{2}\frac{x^2}{\delta^2}\right) + a\left(\frac{1}{3} -\frac{1}{3}\frac{x^3}{\delta^3} + \dots\right)\right] =[/MATH][MATH]\left[\frac{1}{2} - \frac{1}{2}\frac{x^2}{\delta^2} \right] =[/MATH][MATH]\frac{1}{2}\left[ 1-\left(\frac{x}{\delta}\right)^2\right][/MATH]
Which step do you not follow?

 

[JeffM]

Again the notation gets in the way. When you are given expressions with lots of symbols, it is hard for the human mind to keep track. So SIMPLIFY.

[MATH]A = \dfrac{\rho g \delta^2 cos ( \beta )}{\mu_0} \left \{ \left ( 1 + \alpha + \dfrac{\alpha ^2}{2} + \dfrac{\alpha^3}{3! }\ ... \right ) \\ * \left ( \dfrac{1}{a} - \dfrac{1}{a^2} \right ) \\ - \left ( 1 + \dfrac{\alpha x }{\delta} + \dfrac{\alpha^2 x^2}{\delta^2 * 2!} + \dfrac{\alpha^3 x^3}{\delta^3 * 3!}\ ... \ \right ) \\ * \left ( \dfrac{x}{\alpha \delta} - \dfrac{1}{\alpha^2} \right ) \right \}.[/MATH]What a mess! We can start by getting rid of that initial fraction and using summation notation.

[MATH]A = B \left \{ \left ( \sum_{j=0}^{\infty} \dfrac{\alpha ^j}{j!} \right) \left (\dfrac{1}{\alpha} - \dfrac{1}{\alpha ^2}\right ) - \left ( \sum_{j=0}^{\infty} \dfrac{\alpha^j x^j}{\delta^j * j!} \right ) \left ( \dfrac{x}{\alpha \delta} - \dfrac{1}{\alpha^2} \right ) \right \}.[/MATH]
Now that still looks quite ugly, but notice that there may be opportunities for simplifying by telescoping.

[MATH]\dfrac{1}{a} * \dfrac{\alpha^{k}}{k!} - \dfrac{1}{a^2} * \dfrac{\alpha^{(k+1)}}{(k + 1)!} = \dfrac{\alpha^{(k-1)}}{k!} - \dfrac{\alpha^{(k-1)}}{(k + 1)!} =\\ \dfrac{(k + 1)\alpha^{(k-1)}}{(k + 1)!} - \dfrac{\alpha^{(k-1)}}{(k + 1)!} = \dfrac{k\alpha^{(k-1)}}{(k + 1)!}.[/MATH]Now this is not consistent with your text. Maybe I made an error.

[MATH] \dfrac{x}{\alpha \delta} * \dfrac{\alpha^k x^k}{\delta^k * k!} - \dfrac{1}{\alpha^2} * \dfrac{\alpha^{(k+1)} x^{(k+1)}}{\delta^{(k+1)} * (k + 1)!} = [/MATH]
[MATH] \dfrac{\alpha^{(k-1)} x^{(k+1)}}{\delta^{(k+1)} * k!} - \dfrac{\alpha^{(k-1)} x^{(k+1)}}{\delta^{(k+1)} * (k + 1)!} = [/MATH]
[MATH] \dfrac{\alpha^{(k-1)} x^{(k+1)}}{\delta^{(k+1)} * k!} - \dfrac{\alpha^{(k-1)} x^{(k+1)}}{\delta^{(k+1)} * (k + 1)!} = [/MATH]
[MATH] \dfrac{(k + 1)\alpha^{(k-1)} x^{(k+1)}}{\delta^{(k+1)} * (k + 1)!} - \dfrac{\alpha^{(k-1)} x^{(k+1)}}{\delta^{(k+1)} * (k + 1)!} = [/MATH]
[MATH] \dfrac{k\alpha^{(k-1)} x^{(k+1)}}{\delta^{(k+1)} * (k + 1)!}.[/MATH]
Again this is not consistent with your text. Again I may have made an error. But let's assume I did not make an error.

[MATH]A = B \left \{ - \dfrac{1}{a^2} + \sum_{j=0}^{\infty} \dfrac{j\alpha^{(j-1)}}{(j + 1)!} - \left ( - \dfrac{1}{a^2} + \sum_{j=0}^{\infty} \dfrac{j\alpha^{(j-1)} x^{(j+1)}}{\delta^{(k+1)} * (k + 1)!} \right) \right \} =[/MATH]
[MATH]B \left \{ \left ( \sum_{j=0}^{\infty} \dfrac{j\alpha^{(j-1)}}{(j + 1)!} \right ) - \left ( \sum_{j=0}^{\infty} \dfrac{j\alpha^{(j-1)} x^{(j+1)}}{\delta^{(j+1)} * (j + 1)!} \right) \right \} =[/MATH]
[MATH]B \left \{\left ( \dfrac{0 * \alpha^{-1}}{1!} + \dfrac{1 * \alpha^0}{2!} + \sum_{j=2}^{\infty} \dfrac{j\alpha^{(j-1)}}{(j + 1)!} \right )\\ - \left ( \dfrac{0 * \alpha^{-1}x^1}{\delta^1 * 1!} + \dfrac{1 * \alpha^0x^2}{\delta^2 * 2!} + \sum_{j=2}^{\infty} \dfrac{j\alpha^{(j-1)} x^{(j+1)}}{\delta^{(j+1)} * (j + 1)!} \right) \right \} =[/MATH]
[MATH]B \left \{\left (0 + \dfrac{1}{2} + a *\sum_{j=2}^{\infty} \dfrac{j\alpha^{(j-2)}}{(j + 1)!} \right )\\ - \left (0 + \dfrac{1x^2}{2\delta^2} + a * \sum_{j=2}^{\infty} \dfrac{j\alpha^{(j-2)} x^{(j+1)}}{\delta^{(j+1)} * (j + 1)!} \right) \right \} =[/MATH]
[MATH]\dfrac{B}{2} \left (1 + \dfrac{x^2}{\delta^2} \right) - \alpha B \left \{\left ( \sum_{j=2}^{\infty} \dfrac{j\alpha^{(j-2)}}{(j + 1)!}\right ) - \left ( \sum_{j=2}^{\infty} \dfrac{j\alpha^{(j-2)} x^{(j+1)}}{\delta^{j+1}(j + 1)!} \right) \right \}.[/MATH]
But multiplying by alpha as it approaches zero means that ugly product approaches zero.

[MATH]\lim_{\alpha \rightarrow 0} A = \dfrac{B}{2} * \left ( 1 - \dfrac{x^2}{\delta^2} \right ).[/MATH]
So I get the same answer as the text.