Simplification of an Equation.

A

AdamD

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Hello, I've been attempting this question for a while now. I've read the forum terms, and realise that I'm supposed to show you how far I've got, and point out a specific issue with the question, however I've got no where with the problem, so there is no working to show you appart from 123 pages of frustrating scribbles, and the loss of hair on my carpet. I would appreciate any help on this problem if possible!!image

(Click image for larger version)

copyTpic.jpg


I hope this picture is adequate enough for which ever genius can actually solve this problem.

Thank you in advance!!!!!

Adam
 
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AdamD said:
Hello, I've been attempting this question for a while now. I've read the forum terms, and realise that I'm supposed to show you how far I've got, and point out a specific issue with the question, however I've got no where with the problem, so there is no working to show you appart from 123 pages of frustrating scribbles, and the loss of hair on my carpet. I would appreciate any help on this problem if possible!!

The equation is extremely long so I've taken a picture and uploaded it to the following link - <Link removed>

The question is stating the equation and asking how it can be simplified the the other equation stated.

I hope this picture is adequate enough for which ever genius can actually solve this problem.

Thank you in advance!!!!!

Adam
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Thank you for reading READ BEFORE POSTING: so few do.

I presume that U(x) is some function defined somewhere else. What is that definition?

\(\displaystyle g(t) = \dfrac{2Et}{T}\left\{U(t) - U\left(t - \dfrac{T}{4}\right)\right\} + \left(\dfrac{Et}{T} + \dfrac{E}{4}\right)\left\{U\left(t - \dfrac{T}{4}\right) - U\left(t - \dfrac{3T}{4}\right)\right\} + \left(4E - \dfrac{4Et}{T}\right)\left\{U\left(t - \dfrac{3T}{4}\right) - U(t - T)\right\}.\)

I hope I transcribed that correctly. To work with that mess, let

\(\displaystyle a = U(t),\ b = U\left(t - \dfrac{T}{4}\right),\ c = U\left(t - \dfrac{3T}{4}\right),\ and\ d = U(t - T) \implies\)

\(\displaystyle g(t) = \dfrac{2Et}{T} * (a - b) + \left(\dfrac{Et}{T} + \dfrac{E}{4}\right) * (b - c) + \left(4E - \dfrac{4Et}{T}\right) * (c - d) \implies\)

\(\displaystyle g(t) = \dfrac{8Et}{4T} * (a - b) + \dfrac{4Et + ET}{4T} * (b - c) + \dfrac{16ET - 16Et}{4T} * (c - d) \implies\)

\(\displaystyle g(t) = \dfrac{E}{4T} \{8t(a - b) + (4t + T)(b - c) + (16T - 16t)(c - d)\} \implies\)

\(\displaystyle g(t) = \dfrac{E}{4T} \{8at - 8bt + 4bt + bT - 4ct - cT - 16ct + 16cT - 16dT + 16dt\} \implies\)

\(\displaystyle g(t) = \dfrac{E}{4T} \{8at - 4bt + bT - 20ct + 15cT + 16d(t - T)\} \implies\)

\(\displaystyle g(t) = \dfrac{E}{T}\left\{t(2a - b - 5c + 4d) + T\left(\dfrac{b + 15c}{4} - 4d\right)\right\}.\)

Assuming I have not erred in MY algebra, we need to have some information about U(x) to proceed.
 
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JeffM said:
Thank you for reading READ BEFORE POSTING: so few do.

I presume that U(x) is some function defined somewhere else. What is that definition?

\(\displaystyle g(t) = \dfrac{2Et}{T}\left\{U(t) - U\left(t - \dfrac{T}{4}\right)\right\} + \left(\dfrac{Et}{T} + \dfrac{E}{4}\right)\left\{U\left(t - \dfrac{T}{4}\right) - U\left(t - \dfrac{3T}{4}\right)\right\} + \left(4E - \dfrac{4Et}{T}\right)\left\{U\left(t - \dfrac{3T}{4}\right) - U(t - T)\right\}.\)
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the way it's being used U(t) is probably the unit step function.
 
Romsek said:
the way it's being used U(t) is probably the unit step function.
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That's a neat hypothesis. But I think it may require additional specification because that "function" is really a family of functions defined by a parameter, isn't it? And it makes me wonder whether there is an unspecified relationship between T and t as well. That numerator of 2 in the answer seems to come from nowhere.
 
JeffM said:
That's a neat hypothesis. But I think it may require additional specification because that "function" is really a family of functions defined by a parameter, isn't it? And it makes me wonder whether there is an unspecified relationship between T and t as well. That numerator of 2 in the answer seems to come from nowhere.
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as it's commonly used in electrical engineering

\(\displaystyle u(t)=\begin{array}{r}&1:t \geq 0\\&0: t<0\end{array}\)

you can choose to delay it via \(\displaystyle u(t-T)\) but there are no other parameters.

But you are certainly correct in calling it a hypothesis. I have no reason other than the use of U, t, and T as variables to suspect it's the unit step.

It's looks like the typical sort of mess that comes out of using Laplace transforms to find transient responses of systems.
 
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Romsek said:
as it's commonly used in electrical engineering

\(\displaystyle u(t)=\begin{array}{r}&1:t \geq 0\\&0: t<0\end{array}\)

you can choose to delay it via \(\displaystyle u(t-T)\) but there are no other parameters.

But you are certainly correct in calling it a hypothesis. I have no reason other than the use of U, t, and T as variables to suspect it's the unit step.

It's looks like the typical sort of mess that comes out of using Laplace transforms to find transient responses of systems.
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Beginning to sound as though a history major better bow out
 
JeffM said:
Beginning to sound as though a history major better bow out
Click to expand...

Thanks for all the responses regarding this problem!!

Sorry for my late reply, and for not describing the equation properly.

You are absolutely correct, the U(t) is the unit step function. t is the variable and the rest constants. This is my issue, if it was just algerbra I'd have no issues, however this equation is the minimum of crazy, no clue at all.

Thanks again!!

Adam
 
AdamD said:
Thanks for all the responses regarding this problem!!

Sorry for my late reply, and for not describing the equation properly.

You are absolutely correct, the U(t) is the unit step function. t is the variable and the rest constants. This is my issue, if it was just algerbra I'd have no issues, however this equation is the minimum of crazy, no clue at all.

Thanks again!!

Adam
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Is there any restriction on t such as \(\displaystyle t \ge \dfrac{T}{2}\)?
 
AdamD said:
Yes that would be a correct assumption, as the step function can't be negative. (I think)

Thanks.
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what? The argument of the unit step will never turn it negative.

T can be any value. It simply moves where the step "turns on". The problem may specify that T>0 so that the value is always 0 for t<0 but that would be specific to this problem.
 
The problem does not specify that T can only be >0, so I assume it can be any value like you have stated. I am newly introduced to the 'step function', and its implementation within equations.

I've tried taking laplace transforms of the both equations and working them through in the laplace domain, which seems easier. However, the laplace transform has used 4 pages and I'm still no further. Any suggestions on how to tackle this problem?

Thank You!
 
AdamD said:
The problem does not specify that T can only be >0, so I assume it can be any value like you have stated. I am newly introduced to the 'step function', and its implementation within equations.

I've tried taking laplace transforms of the both equations and working them through in the laplace domain, which seems easier. However, the laplace transform has used 4 pages and I'm still no further. Any suggestions on how to tackle this problem?

Thank You!
Click to expand...
As I explained before, I was (during ancient times) a history major and so tend to approach things mathematical in a very simple and perhaps inelegant way. The unit step function can take any argument as romsek explained. So there is no implied restriction on t.

\(\displaystyle g(t) = \dfrac{2Et}{T}\left\{U(t) - U\left(t - \dfrac{T}{4}\right)\right\} + \left(\dfrac{Et}{T} + \dfrac{E}{4}\right)\left\{U\left(t - \dfrac{T}{4}\right) - U\left(t - \dfrac{3T}{4}\right)\right\} + \left(4E - \dfrac{4Et}{T}\right)\left\{U\left(t - \dfrac{3T}{4}\right) - U(t - T)\right\}.\)

Is that properly transcribed?

\(\displaystyle Let\ h(t) = \dfrac{E}{T}\left\{2tU(t) - \left(t - \dfrac{T}{4}\right)U\left(t - \dfrac{T}{2}\right) - 5 \left(t - \dfrac{3T}{4}\right)U\left(t - \dfrac{3T}{4}\right) + 4(t -T)U(t - T)\right\}\)

Is that properly transcribed?

\(\displaystyle 0 \le \dfrac{T}{4} \le t < \dfrac{T}{2} \implies g(t) = \left(\dfrac{Et}{T} + \dfrac{E}{4}\right)(1 - 0) = \dfrac{E}{T}\left(t - \dfrac{T}{4}\right)\ and\ h(t) = \dfrac{2tE}{T}.\)

So it is not invariably true that g(t) = h(t).

EDIT: Mark kindly pointed out that the statement above may well be misconstrued and may even be wrong. There are domains where g(t) = h(t). So it can be shown that g(t) = h(t), provided that an appropriate restriction is placed on the domain of t. I was interpreting the problem to require a demonstration for all t.
 
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Thanks for all the input on this question I really appreciate it!

I've managed to finally prove it!!

Thanks, Adam
 
Denis said:
Oh ya?
Who shot Liberty Valance?
What was the "theme song" for Gillette Blue Blades?
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Sorry, denis.

Looked at my copies of Mommsen's Romische Geschicte, Fustel de Coulange's La Cite Antique, and Maitland's Domesday Book and Beyond. Not one word about who shot Liberty Valence or about blue blades in any of them. I tried, mon ami.
 
Denis said:
You mean them 2 questions were not on your final exam?
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History exams were usually along the lines of write two essays from this list of five questions. I probably skipped those two questions and chose a topic like what was the effect of Magyar nationalism on the Hapsburg state. I had a bent for the practical.

I'd really like to know how Adam proved something that is not true. That would be a very practical thing to know. EDIT: Mark kindly pointed out that the preceding statement above may well be misconstrued and may even be wrong. There are domains where Adam's proposition is true. I was interpreting the problem to require a demonstration for all real numbers. The proposition IS true for a broad domain of the reals.
 
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