Maclaurin Series Problem

[Tesla3]

Could anybody help me with this Maclaurin Series question? I would be really grateful for your help. I have attached the question as in image.

 

[PlanetMaster]

We know Maclaurin Expansion for $e^{t}$ is
$e^{t}=1+\frac{1}{1!}t+\frac{1}{2!}t^2+\frac{1}{3!}t^3+\frac{1}{4!}t^4+\ldots \:$
(should be in formula sheet)

For $e^{-kt}$, this becomes
$e^{-kt}=1+\frac{1}{1!}(-k)t+\frac{1}{2!}t^2(-k)^2+\frac{1}{3!}t^3(-k)^3+\frac{1}{4!}t^4(-k)^4-\ldots \:$
(If you are not sure how we got this, have a look at Taylor series since Maclaurin series is a special case of Taylor series)

Simplifying this, we get
$e^{-kt}=1-{kt}+\frac{t^2k^2}{2!}-\frac{t^3k^3}{3!}+\frac{t^4k^4}{4!}-\ldots \:$

Now for Model 2, we have
${x}=\frac{u}{k}(1-e^{-kt})$

Substituting our expansion for $e^{-kt}$ in this, we get
${x}=\frac{u}{k}(1-[1-{kt}+\frac{t^2k^2}{2!}-\frac{t^3k^3}{3!}+\frac{t^4k^4}{4!}-\ldots \: ])$

And simplifying this, we get
${x}=\frac{u}{k}(1-1+{kt}-\frac{t^2k^2}{2!}+\frac{t^3k^3}{3!}-\frac{t^4k^4}{4!}+\ldots \: )$
${x}=\frac{u}{k}({kt}-\frac{t^2k^2}{2!}+\frac{t^3k^3}{3!}-\frac{t^4k^4}{4!}+\ldots \: )$
${x}={ut}-\frac{ut^2k}{2!}+\frac{ut^3k^2}{3!}-\frac{ut^4k^3}{4!}+\ldots \:$

Substituting $k=0$, we get
${x}={ut}$
which is same as the result given by Model 1.


Now for ${\gamma}$ in Model 2, we have
${\gamma}=\frac{g+kv}{k^2}(1-e^{-kt})-\frac{gt}{k}$

Substituting our expansion for $e^{-kt}$ again, we get
${\gamma}=\frac{g+kv}{k^2}(1-[1-{kt}+\frac{t^2k^2}{2!}-\frac{t^3k^3}{3!}+\frac{t^4k^4}{4!}-\ldots \: ])-\frac{gt}{k}$

And simplifying this, we get
${\gamma}=\frac{g+kv}{k^2}(1-1+{kt}-\frac{t^2k^2}{2!}+\frac{t^3k^3}{3!}-\frac{t^4k^4}{4!}+\ldots \: )-\frac{gt}{k}$
${\gamma}=\frac{g+kv}{k^2}({kt}-\frac{t^2k^2}{2!}+\frac{t^3k^3}{3!}-\frac{t^4k^4}{4!}+\ldots \: )-\frac{gt}{k}$

Taking LCM and further simplifying, we get
${\gamma}=\frac{1}{k^2}[({g+kv})({kt}-\frac{t^2k^2}{2!}+\frac{t^3k^3}{3!}-\frac{t^4k^4}{4!}+\ldots \: )-{gtk}]$

Now we can use a little bit of logic to make things simpler (we could have done this above as well to save time).
We know that $k=0$, so any fraction with a $k$ in numerator is going to be equal to $0$.

So we only need terms where $k$ eliminates completely.
In our equation, these would be the terms with $k^{2}$ in numerator since we have a $k^{2}$ in denominator.

Simplifying further to only fractions that would get us a $k^{2}$ in numerator, we get
${\gamma}=\frac{1}{k^2}[{-\frac{gt^2k^2}{2!}+{k^2vt}}+\ldots \: ]$
${\gamma}={vt}-\frac{1}{2}{gt^2}+\ldots \:$

And substituting $k=0$ get us
${\gamma}={vt}-\frac{1}{2}{gt^2}$
which is the same as the result given by Model 1.

 

[Tesla3]

Thanks for help PlanetMaster There is only one thing that I do not understand in your explanation. Thing is, the part where you say that " We know that k=0, so any fraction with a k in either numerator or denominator is going to be equal to 0. ", is a bit confusing because If any fraction has k in denominator and we put k = 0 it is going to be undefined while if it is in numerator then the fraction will be equal to zero. I have attached my steps. Sorry for late reply, as at the time I posted the question no one replied.

${y}=\frac{g+kv}{k^2}(1-e^{-kt})-\frac{gt}{k}$

${y}=\frac{g+kv}{k^2}(1-(1-{kt}+\frac{k^{2}t^{2}}{2!}-\frac{k^{3}t^{3}}{3!}+\ldots \: ))-\frac{gt}{k}$

${y}=\frac{g+kv}{k^2}(0+{kt}-\frac{k^{2}t^{2}}{2!}+\frac{k^{3}t^{3}}{3!}+\ldots \: )-\frac{gt}{k}$

${y}=(g+kv)(0+\frac{t}{k}-\frac{t^2}{2!}+\frac{kt^3}{3!}+\ldots \: )-\frac{gt}{k}$

${y}=(\frac{gt}{k}+vt-\frac{gt^2}{2!}-\frac{kvt^2}{2!}+\frac{kgt^3}{3!}+\frac{k^{2}vt^{3}}{3!}+\ldots \: )-\frac{gt}{k}$

${y}=(vt-\frac{gt^2}{2!}-\frac{kvt^2}{2!}+\frac{kgt^3}{3!}+\frac{k^{2}vt^{3}}{3!}+\ldots \: )$

Finally we put k = 0 and get the desired result

 

[PlanetMaster]

Ah yeah my bad. I meant to say that any fraction with a ${k}$ in numerator is going to be equal to ${0}$.
Dividing by 0 would take us to infinity (and beyond!).

P.S. Fixed this in the answer as well.

 

[Tesla3]

Could you help me with that other question as well? PlanetMaster

I have attached the link to the other question:

Integration involving Inverse Hyperbolic Functions

Hi, could anybody help me with this integration question? Thing is, I understand the integration process through substitution x = 2sin(u) but there is a notice box in my book suggesting that this could be done without using the substitution x = 2sin(u). I have attached the question from my book...

xtremepape.rs

 

[PlanetMaster]

Sorry for the delay but here you go:

Integration involving Inverse Hyperbolic Functions

Hi, could anybody help me with this integration question? Thing is, I understand the integration process through substitution x = 2sin(u) but there is a notice box in my book suggesting that this could be done without using the substitution x = 2sin(u). I have attached the question from my book...

xtremepape.rs