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Maclaurin Series Problem

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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.

Q10.png
 
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PlanetMaster

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We know Maclaurin Expansion for ete^{t} is
et=1+11!t+12!t2+13!t3+14!t4+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 ekte^{-kt} , this becomes
ekt=1+11!(k)t+12!t2(k)2+13!t3(k)3+14!t4(k)4e^{-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
ekt=1kt+t2k22!t3k33!+t4k44!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=uk(1ekt){x}=\frac{u}{k}(1-e^{-kt})

Substituting our expansion for ekte^{-kt} in this, we get
x=uk(1[1kt+t2k22!t3k33!+t4k44!]){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=uk(11+ktt2k22!+t3k33!t4k44!+){x}=\frac{u}{k}(1-1+{kt}-\frac{t^2k^2}{2!}+\frac{t^3k^3}{3!}-\frac{t^4k^4}{4!}+\ldots \: )
x=uk(ktt2k22!+t3k33!t4k44!+){x}=\frac{u}{k}({kt}-\frac{t^2k^2}{2!}+\frac{t^3k^3}{3!}-\frac{t^4k^4}{4!}+\ldots \: )
x=utut2k2!+ut3k23!ut4k34!+{x}={ut}-\frac{ut^2k}{2!}+\frac{ut^3k^2}{3!}-\frac{ut^4k^3}{4!}+\ldots \:


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


Now for γ{\gamma} in Model 2, we have
γ=g+kvk2(1ekt)gtk{\gamma}=\frac{g+kv}{k^2}(1-e^{-kt})-\frac{gt}{k}

Substituting our expansion for ekte^{-kt} again, we get
γ=g+kvk2(1[1kt+t2k22!t3k33!+t4k44!])gtk{\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
γ=g+kvk2(11+ktt2k22!+t3k33!t4k44!+)gtk{\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}
γ=g+kvk2(ktt2k22!+t3k33!t4k44!+)gtk{\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
γ=1k2[(g+kv)(ktt2k22!+t3k33!t4k44!+)gtk]{\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=0k=0 , so any fraction with a kk in numerator is going to be equal to 00 .

So we only need terms where kk eliminates completely.
In our equation, these would be the terms with k2k^{2} in numerator since we have a k2k^{2} in denominator.

Simplifying further to only fractions that would get us a k2k^{2} in numerator, we get
γ=1k2[gt2k22!+k2vt+]{\gamma}=\frac{1}{k^2}[{-\frac{gt^2k^2}{2!}+{k^2vt}}+\ldots \: ]
γ=vt12gt2+{\gamma}={vt}-\frac{1}{2}{gt^2}+\ldots \:


And substituting k=0k=0 get us
γ=vt12gt2{\gamma}={vt}-\frac{1}{2}{gt^2}
which is the same as the result given by Model 1.
 
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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=g+kvk2(1ekt)gtk{y}=\frac{g+kv}{k^2}(1-e^{-kt})-\frac{gt}{k}

y=g+kvk2(1(1kt+k2t22!k3t33!+))gtk{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=g+kvk2(0+ktk2t22!+k3t33!+)gtk{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+tkt22!+kt33!+)gtk{y}=(g+kv)(0+\frac{t}{k}-\frac{t^2}{2!}+\frac{kt^3}{3!}+\ldots \: )-\frac{gt}{k}

y=(gtk+vtgt22!kvt22!+kgt33!+k2vt33!+)gtk{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=(vtgt22!kvt22!+kgt33!+k2vt33!+){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
 
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PlanetMaster

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

P.S. Fixed this in the answer as well.
 
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Could you help me with that other question as well? PlanetMaster

I have attached the link to the other question:
 

PlanetMaster

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Sorry for the delay but here you go:
 
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