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May 19th, 2016, 07:16 AM   #1
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Laplace transform

What is laplace transform?
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May 19th, 2016, 08:54 AM   #2
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What is laplace transform?
Let me google that for you
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May 23rd, 2016, 03:38 AM   #3
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Seriously? You are asking about "Laplace Transforms", a topic from post-Calculus "differential equations" and you just recently were asking about Roman Numerals, exponentials, and logarithms! Where are you getting all these?
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May 23rd, 2016, 03:40 AM   #4
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He has a degree in electrical engineering so i'm told. Just refreshing his memory.
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May 23rd, 2016, 04:56 AM   #5
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Seriously? You are asking about "Laplace Transforms", a topic from post-Calculus "differential equations" and you just recently were asking about Roman Numerals, exponentials, and logarithms! Where are you getting all these?
And he asked the question about Laplace transforms in the Elementary Math forum, too
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May 23rd, 2016, 05:51 AM   #6
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He has a degree in electrical engineering so i'm told. Just refreshing his memory.
... from a past life?
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May 23rd, 2016, 06:00 AM   #7
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hello,
The huge syllabus has place less memory.
so, i am trying to implement practically again.
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May 29th, 2016, 04:23 AM   #8
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The Laplace transform is the operator that "transforms" the function f(x) to the function $\displaystyle L(f(x))= \int_0^\infty f(t)e^{-st}dt$, a function of "s".

It has the property that $\displaystyle L(f'(x))= \int_0^\infty e^{-st} f'(t)dt= \left[e^{-st}f(t)\right]_0^\infty+ \frac{1}{s}\int_0^\infty e^{-st}f(t)dt= f(0)+ \frac{1}{s}L(f(t))$ (assuming f goes to 0 as x goes to infinity). That is, the Laplace transform of the derivative of f can be expressed in terms of the Laplace transform of f. That allows us to change differential equation in f (equations involving the derivatives of f) to an algebraic equation in the Laplace transform of f.
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