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Maclaurin Series Problem
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PlanetMaster
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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.
\(e^{t}=1+\frac{1}{1!}t+\frac{1}{2!}t^2+\frac{1}{3!}t^3+\frac{1}{4!}t^4+\ldots \: \)
(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.
Last edited:
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
\({y}=\frac{g+kv}{k^2}(1-e^{-kt})-\frac{gt}{k} \)
\({y}=\frac{g+kv}{k^2}(1-(1-{kt}+\frac{k^{2}t^{2}}{2!}-\frac{k^{3}t^{3}}{3!}+\ldots \: ))-\frac{gt}{k} \)
\({y}=\frac{g+kv}{k^2}(0+{kt}-\frac{k^{2}t^{2}}{2!}+\frac{k^{3}t^{3}}{3!}+\ldots \: )-\frac{gt}{k} \)
\({y}=(g+kv)(0+\frac{t}{k}-\frac{t^2}{2!}+\frac{kt^3}{3!}+\ldots \: )-\frac{gt}{k} \)
\({y}=(\frac{gt}{k}+vt-\frac{gt^2}{2!}-\frac{kvt^2}{2!}+\frac{kgt^3}{3!}+\frac{k^{2}vt^{3}}{3!}+\ldots \: )-\frac{gt}{k} \)
\({y}=(vt-\frac{gt^2}{2!}-\frac{kvt^2}{2!}+\frac{kgt^3}{3!}+\frac{k^{2}vt^{3}}{3!}+\ldots \: ) \)
Finally we put k = 0 and get the desired result
Last edited by a moderator:
PlanetMaster
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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.
Dividing by 0 would take us to infinity (and beyond!).
P.S. Fixed this in the answer as well.
Could you help me with that other question as well? PlanetMaster
I have attached the link to the other question:
xtremepape.rs
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
XPRS Administrator
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Sorry for the delay but here you go:
xtremepape.rs
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