Math / Calculus

Question3. Show that $\mathcal{L}[t^n] = \frac{n!}{s^{n+1}}$ .
Note: A complete proof uses mathematical induction. Math majors may want to try this. Everyone else, just show that it holds for $n = 1$ using the definition of the Laplace transform.

Studdy Solution

STEP 1

What is this asking? Show that the Laplace Transform of $t^n$ is equal to $\frac{n!}{s^{n+1}}$ , specifically for the case where $n=1$ . Watch out! Don't forget to carefully apply integration by parts if needed!

STEP 2

1. Define the Laplace Transform
2. Compute for $n=1$

STEP 3

Alright, let's kick things off by reminding ourselves what the Laplace Transform actually is!
The Laplace Transform of a function $f(t)$ , written as $\mathcal{L}[f(t)]$ , is defined as: $\mathcal{L}[f(t)] = \int_0^\infty e^{-st} f(t) \, dt$ This formula takes our function $f(t)$ and transforms it into a new function of $s$ .
It's like giving our function a magical makeover!

STEP 4

Now, we want to find the Laplace Transform of $t^n$ when $n=1$ , which means we're looking at $f(t) = t$ .
Let's plug this into our Laplace Transform formula: $\mathcal{L}[t] = \int_0^\infty e^{-st} \cdot t \, dt$

STEP 5

Time for some integration by parts!
Remember, the formula for integration by parts is: $\int u \, dv = uv - \int v \, du$ Let's choose $u = t$ and $dv = e^{-st} \, dt$ .
Then, $du = dt$ and $v = \int e^{-st} \, dt = -\frac{1}{s}e^{-st}$ .

STEP 6

Substituting these values into the integration by parts formula, we get: $\mathcal{L}[t] = \left[ -\frac{1}{s} t e^{-st} \right]_0^\infty - \int_0^\infty -\frac{1}{s} e^{-st} \, dt$

STEP 7

Let's evaluate the first term: $\lim_{t \to \infty} \left( -\frac{t}{s} e^{-st} \right) - \left( -\frac{0}{s} e^{-s \cdot 0} \right) = 0 - 0 = 0$ The limit is zero because the exponential term $e^{-st}$ decays to zero much faster than $t$ grows as $t$ approaches infinity.

STEP 8

Now, let's tackle the second term: $\frac{1}{s} \int_0^\infty e^{-st} \, dt = \frac{1}{s} \left[ -\frac{1}{s} e^{-st} \right]_0^\infty = \frac{1}{s} \left( 0 - \left( -\frac{1}{s} \right) \right) = \frac{1}{s^2}$

STEP 9

So, putting it all together, we have: $\mathcal{L}[t] = 0 + \frac{1}{s^2} = \frac{1}{s^2}$ And since $1! = 1$ , we can write this as: $\mathcal{L}[t] = \frac{1!}{s^{1+1}} = \frac{1!}{s^2}$ This matches the formula $\frac{n!}{s^{n+1}}$ for $n=1$ !

STEP 10

We have shown that for $n=1$ , the Laplace Transform of $t^n$ is indeed $\frac{n!}{s^{n+1}}$ , which in this case is $\frac{1!}{s^2} = \frac{1}{s^2}$ .

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Question3. Show that $\mathcal{L}[t^n] = \frac{n!}{s^{n+1}}$ .
Note: A complete proof uses mathematical induction. Math majors may want to try this. Everyone else, just show that it holds for $n = 1$ using the definition of the Laplace transform.

Studdy Solution

STEP 1

What is this asking? Show that the Laplace Transform of $t^n$ is equal to $\frac{n!}{s^{n+1}}$ , specifically for the case where $n=1$ . Watch out! Don't forget to carefully apply integration by parts if needed!

STEP 2

1. Define the Laplace Transform
2. Compute for $n=1$

STEP 3

STEP 4

Now, we want to find the Laplace Transform of $t^n$ when $n=1$ , which means we're looking at $f(t) = t$ .
Let's plug this into our Laplace Transform formula: $\mathcal{L}[t] = \int_0^\infty e^{-st} \cdot t \, dt$

STEP 5

Time for some integration by parts!
Remember, the formula for integration by parts is: $\int u \, dv = uv - \int v \, du$ Let's choose $u = t$ and $dv = e^{-st} \, dt$ .
Then, $du = dt$ and $v = \int e^{-st} \, dt = -\frac{1}{s}e^{-st}$ .

STEP 6

Substituting these values into the integration by parts formula, we get: $\mathcal{L}[t] = \left[ -\frac{1}{s} t e^{-st} \right]_0^\infty - \int_0^\infty -\frac{1}{s} e^{-st} \, dt$

STEP 7

Let's evaluate the first term: $\lim_{t \to \infty} \left( -\frac{t}{s} e^{-st} \right) - \left( -\frac{0}{s} e^{-s \cdot 0} \right) = 0 - 0 = 0$ The limit is zero because the exponential term $e^{-st}$ decays to zero much faster than $t$ grows as $t$ approaches infinity.

STEP 8

Now, let's tackle the second term: $\frac{1}{s} \int_0^\infty e^{-st} \, dt = \frac{1}{s} \left[ -\frac{1}{s} e^{-st} \right]_0^\infty = \frac{1}{s} \left( 0 - \left( -\frac{1}{s} \right) \right) = \frac{1}{s^2}$

STEP 9

So, putting it all together, we have: $\mathcal{L}[t] = 0 + \frac{1}{s^2} = \frac{1}{s^2}$ And since $1! = 1$ , we can write this as: $\mathcal{L}[t] = \frac{1!}{s^{1+1}} = \frac{1!}{s^2}$ This matches the formula $\frac{n!}{s^{n+1}}$ for $n=1$ !

STEP 10

We have shown that for $n=1$ , the Laplace Transform of $t^n$ is indeed $\frac{n!}{s^{n+1}}$ , which in this case is $\frac{1!}{s^2} = \frac{1}{s^2}$ .

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Question3. Show that L[tn]=n!sn+1\mathcal{L}[t^n] = \frac{n!}{s^{n+1}}L[tn]=sn+1n!​.Note: A complete proof uses mathematical induction. Math majors may want to try this. Everyone else, just show that it holds for n=1n = 1n=1 using the definition of the Laplace transform.

Studdy Solution

STEP 1

What is this asking? Show that the Laplace Transform of tnt^ntn is equal to n!sn+1\frac{n!}{s^{n+1}}sn+1n!​, specifically for the case where n=1n=1n=1. Watch out! Don't forget to carefully apply integration by parts if needed!

STEP 2

1. Define the Laplace Transform2. Compute for n=1n=1n=1

STEP 3

STEP 4

Now, we want to find the Laplace Transform of tnt^ntn when n=1n=1n=1, which means we're looking at f(t)=tf(t) = tf(t)=t.Let's **plug this into** our Laplace Transform formula: L[t]=∫0∞e−st⋅t dt\mathcal{L}[t] = \int_0^\infty e^{-st} \cdot t \, dtL[t]=∫0∞​e−st⋅tdt

STEP 5

STEP 6

**Substituting** these values into the integration by parts formula, we get: L[t]=[−1ste−st]0∞−∫0∞−1se−st dt\mathcal{L}[t] = \left[ -\frac{1}{s} t e^{-st} \right]_0^\infty - \int_0^\infty -\frac{1}{s} e^{-st} \, dtL[t]=[−s1​te−st]0∞​−∫0∞​−s1​e−stdt

STEP 7

STEP 8

STEP 9

STEP 10

We have shown that for n=1n=1n=1, the Laplace Transform of tnt^ntn is indeed n!sn+1\frac{n!}{s^{n+1}}sn+1n!​, which in this case is 1!s2=1s2\frac{1!}{s^2} = \frac{1}{s^2}s21!​=s21​.

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Question3. Show that $\mathcal{L}[t^n] = \frac{n!}{s^{n+1}}$ .
Note: A complete proof uses mathematical induction. Math majors may want to try this. Everyone else, just show that it holds for $n = 1$ using the definition of the Laplace transform.

What is this asking? Show that the Laplace Transform of $t^n$ is equal to $\frac{n!}{s^{n+1}}$ , specifically for the case where $n=1$ . Watch out! Don't forget to carefully apply integration by parts if needed!

1. Define the Laplace Transform
2. Compute for $n=1$

Now, we want to find the Laplace Transform of $t^n$ when $n=1$ , which means we're looking at $f(t) = t$ .
Let's plug this into our Laplace Transform formula: $\mathcal{L}[t] = \int_0^\infty e^{-st} \cdot t \, dt$

Substituting these values into the integration by parts formula, we get: $\mathcal{L}[t] = \left[ -\frac{1}{s} t e^{-st} \right]_0^\infty - \int_0^\infty -\frac{1}{s} e^{-st} \, dt$

We have shown that for $n=1$ , the Laplace Transform of $t^n$ is indeed $\frac{n!}{s^{n+1}}$ , which in this case is $\frac{1!}{s^2} = \frac{1}{s^2}$ .