Solved on Mar 07, 2024

Approximate $\sqrt{e}$ using a Maclaurin polynomial for $e^{x}$ with a maximum error of 0.01.

STEP 1

Assumptions
1. The Maclaurin polynomial for $e^x$ is given by the series $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ .
2. We want to approximate $\sqrt{e}$ , which means we are looking for the value of $e^{1/2}$ .
3. The maximum error allowed in the approximation is 0.01.
4. The Maclaurin series for $e^x$ converges for all $x$ , so we can use it to approximate $e^{1/2}$ .

STEP 2

The Maclaurin polynomial for $e^x$ up to the $n$ -th term is given by:
$P_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}$
We need to find $n$ such that the error in approximating $e^{1/2}$ using $P_n(1/2)$ is less than 0.01.

STEP 3

The error in using the Maclaurin polynomial is given by the remainder term of the Taylor series, which for the function $e^x$ at $x = 1/2$ is:
$R_n(1/2) = \frac{e^c}{(n+1)!} \left(\frac{1}{2}\right)^{n+1}$
where $c$ is some number between 0 and 1/2.

STEP 4

Since $e^c$ is always positive and increases with $c$ , the maximum error occurs when $c$ is largest, that is, when $c = 1/2$ . Therefore, we can estimate the error by setting $c = 1/2$ :
$R_n(1/2) \leq \frac{e^{1/2}}{(n+1)!} \left(\frac{1}{2}\right)^{n+1}$

STEP 5

We want the error to be less than 0.01:
$\frac{e^{1/2}}{(n+1)!} \left(\frac{1}{2}\right)^{n+1} < 0.01$

STEP 6

Now we need to find the smallest $n$ such that the inequality in STEP_5 holds. We can do this by trial and error, starting with $n=0$ and increasing $n$ until the inequality is satisfied.

STEP 7

Let's start with $n=0$ :
$\frac{e^{1/2}}{(0+1)!} \left(\frac{1}{2}\right)^{0+1} = \frac{\sqrt{e}}{2}$
Since $\sqrt{e} > 1$ , this is clearly greater than 0.01, so we need to try a higher value of $n$ .

STEP 8

Now let's try $n=1$ :
$\frac{e^{1/2}}{(1+1)!} \left(\frac{1}{2}\right)^{1+1} = \frac{\sqrt{e}}{8}$
Again, since $\sqrt{e} > 1$ , this is still greater than 0.01, so we need to try a higher value of $n$ .

STEP 9

Continue with $n=2$ :
$\frac{e^{1/2}}{(2+1)!} \left(\frac{1}{2}\right)^{2+1} = \frac{\sqrt{e}}{48}$
This might be less than 0.01, but we need to check. Since $e \approx 2.71828$ , we have $\sqrt{e} \approx 1.64872$ .

STEP 10

Calculate the approximate value of the error for $n=2$ :
$\frac{1.64872}{48} \approx 0.03435$
This is still greater than 0.01, so we need to try a higher value of $n$ .

STEP 11

Now let's try $n=3$ :
$\frac{e^{1/2}}{(3+1)!} \left(\frac{1}{2}\right)^{3+1} = \frac{\sqrt{e}}{384}$

STEP 12

Calculate the approximate value of the error for $n=3$ :
$\frac{1.64872}{384} \approx 0.00429$
This is less than 0.01, so $n=3$ is sufficient for the approximation.

STEP 13

Now we can write the Maclaurin polynomial for $e^x$ up to the third term to approximate $e^{1/2}$ :
$P_3(1/2) = 1 + \frac{1}{2} + \frac{(1/2)^2}{2!} + \frac{(1/2)^3}{3!}$

STEP 14

Calculate the approximation:
$P_3(1/2) = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{48}$

STEP 15

Add the terms to find the approximate value of $\sqrt{e}$ :
$P_3(1/2) = 1 + 0.5 + 0.125 + 0.0208333 \approx 1.6458333$
The Maclaurin polynomial approximation for $\sqrt{e}$ with a maximum error of 0.01 is approximately 1.6458333.

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Approximate $\sqrt{e}$ using a Maclaurin polynomial for $e^{x}$ with a maximum error of 0.01.

STEP 1

STEP 2

The Maclaurin polynomial for $e^x$ up to the $n$ -th term is given by:
$P_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}$
We need to find $n$ such that the error in approximating $e^{1/2}$ using $P_n(1/2)$ is less than 0.01.

STEP 3

The error in using the Maclaurin polynomial is given by the remainder term of the Taylor series, which for the function $e^x$ at $x = 1/2$ is:
$R_n(1/2) = \frac{e^c}{(n+1)!} \left(\frac{1}{2}\right)^{n+1}$
where $c$ is some number between 0 and 1/2.

STEP 4

Since $e^c$ is always positive and increases with $c$ , the maximum error occurs when $c$ is largest, that is, when $c = 1/2$ . Therefore, we can estimate the error by setting $c = 1/2$ :
$R_n(1/2) \leq \frac{e^{1/2}}{(n+1)!} \left(\frac{1}{2}\right)^{n+1}$

STEP 5

We want the error to be less than 0.01:
$\frac{e^{1/2}}{(n+1)!} \left(\frac{1}{2}\right)^{n+1} < 0.01$

STEP 6

Now we need to find the smallest $n$ such that the inequality in STEP_5 holds. We can do this by trial and error, starting with $n=0$ and increasing $n$ until the inequality is satisfied.

STEP 7

Let's start with $n=0$ :
$\frac{e^{1/2}}{(0+1)!} \left(\frac{1}{2}\right)^{0+1} = \frac{\sqrt{e}}{2}$
Since $\sqrt{e} > 1$ , this is clearly greater than 0.01, so we need to try a higher value of $n$ .

STEP 8

Now let's try $n=1$ :
$\frac{e^{1/2}}{(1+1)!} \left(\frac{1}{2}\right)^{1+1} = \frac{\sqrt{e}}{8}$
Again, since $\sqrt{e} > 1$ , this is still greater than 0.01, so we need to try a higher value of $n$ .

STEP 9

Continue with $n=2$ :
$\frac{e^{1/2}}{(2+1)!} \left(\frac{1}{2}\right)^{2+1} = \frac{\sqrt{e}}{48}$
This might be less than 0.01, but we need to check. Since $e \approx 2.71828$ , we have $\sqrt{e} \approx 1.64872$ .

STEP 10

Calculate the approximate value of the error for $n=2$ :
$\frac{1.64872}{48} \approx 0.03435$
This is still greater than 0.01, so we need to try a higher value of $n$ .

STEP 11

Now let's try $n=3$ :
$\frac{e^{1/2}}{(3+1)!} \left(\frac{1}{2}\right)^{3+1} = \frac{\sqrt{e}}{384}$

STEP 12

Calculate the approximate value of the error for $n=3$ :
$\frac{1.64872}{384} \approx 0.00429$
This is less than 0.01, so $n=3$ is sufficient for the approximation.

STEP 13

Now we can write the Maclaurin polynomial for $e^x$ up to the third term to approximate $e^{1/2}$ :
$P_3(1/2) = 1 + \frac{1}{2} + \frac{(1/2)^2}{2!} + \frac{(1/2)^3}{3!}$

STEP 14

Calculate the approximation:
$P_3(1/2) = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{48}$

STEP 15

Add the terms to find the approximate value of $\sqrt{e}$ :
$P_3(1/2) = 1 + 0.5 + 0.125 + 0.0208333 \approx 1.6458333$
The Maclaurin polynomial approximation for $\sqrt{e}$ with a maximum error of 0.01 is approximately 1.6458333.

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Approximate e\sqrt{e}e​ using a Maclaurin polynomial for exe^{x}ex with a maximum error of 0.01.

STEP 1

STEP 2

STEP 3

STEP 4

STEP 5

We want the error to be less than 0.01:e1/2(n+1)!(12)n+1<0.01\frac{e^{1/2}}{(n+1)!} \left(\frac{1}{2}\right)^{n+1} < 0.01(n+1)!e1/2​(21​)n+1<0.01

STEP 6

Now we need to find the smallest nnn such that the inequality in STEP_5 holds. We can do this by trial and error, starting with n=0n=0n=0 and increasing nnn until the inequality is satisfied.

STEP 7

Let's start with n=0n=0n=0:e1/2(0+1)!(12)0+1=e2\frac{e^{1/2}}{(0+1)!} \left(\frac{1}{2}\right)^{0+1} = \frac{\sqrt{e}}{2}(0+1)!e1/2​(21​)0+1=2e​​Since e>1\sqrt{e} > 1e​>1, this is clearly greater than 0.01, so we need to try a higher value of nnn.

STEP 8

Now let's try n=1n=1n=1:e1/2(1+1)!(12)1+1=e8\frac{e^{1/2}}{(1+1)!} \left(\frac{1}{2}\right)^{1+1} = \frac{\sqrt{e}}{8}(1+1)!e1/2​(21​)1+1=8e​​Again, since e>1\sqrt{e} > 1e​>1, this is still greater than 0.01, so we need to try a higher value of nnn.

STEP 9

STEP 10

Calculate the approximate value of the error for n=2n=2n=2:1.6487248≈0.03435\frac{1.64872}{48} \approx 0.03435481.64872​≈0.03435This is still greater than 0.01, so we need to try a higher value of nnn.

STEP 11

Now let's try n=3n=3n=3:e1/2(3+1)!(12)3+1=e384\frac{e^{1/2}}{(3+1)!} \left(\frac{1}{2}\right)^{3+1} = \frac{\sqrt{e}}{384}(3+1)!e1/2​(21​)3+1=384e​​

STEP 12

Calculate the approximate value of the error for n=3n=3n=3:1.64872384≈0.00429\frac{1.64872}{384} \approx 0.004293841.64872​≈0.00429This is less than 0.01, so n=3n=3n=3 is sufficient for the approximation.

STEP 13

Now we can write the Maclaurin polynomial for exe^xex up to the third term to approximate e1/2e^{1/2}e1/2:P3(1/2)=1+12+(1/2)22!+(1/2)33!P_3(1/2) = 1 + \frac{1}{2} + \frac{(1/2)^2}{2!} + \frac{(1/2)^3}{3!}P3​(1/2)=1+21​+2!(1/2)2​+3!(1/2)3​

STEP 14

Calculate the approximation:P3(1/2)=1+12+18+148P_3(1/2) = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{48}P3​(1/2)=1+21​+81​+481​

STEP 15

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Approximate $\sqrt{e}$ using a Maclaurin polynomial for $e^{x}$ with a maximum error of 0.01.

We want the error to be less than 0.01:
$\frac{e^{1/2}}{(n+1)!} \left(\frac{1}{2}\right)^{n+1} < 0.01$

Now we need to find the smallest $n$ such that the inequality in STEP_5 holds. We can do this by trial and error, starting with $n=0$ and increasing $n$ until the inequality is satisfied.

Let's start with $n=0$ :
$\frac{e^{1/2}}{(0+1)!} \left(\frac{1}{2}\right)^{0+1} = \frac{\sqrt{e}}{2}$
Since $\sqrt{e} > 1$ , this is clearly greater than 0.01, so we need to try a higher value of $n$ .

Now let's try $n=1$ :
$\frac{e^{1/2}}{(1+1)!} \left(\frac{1}{2}\right)^{1+1} = \frac{\sqrt{e}}{8}$
Again, since $\sqrt{e} > 1$ , this is still greater than 0.01, so we need to try a higher value of $n$ .

Calculate the approximate value of the error for $n=2$ :
$\frac{1.64872}{48} \approx 0.03435$
This is still greater than 0.01, so we need to try a higher value of $n$ .

Now let's try $n=3$ :
$\frac{e^{1/2}}{(3+1)!} \left(\frac{1}{2}\right)^{3+1} = \frac{\sqrt{e}}{384}$

Calculate the approximate value of the error for $n=3$ :
$\frac{1.64872}{384} \approx 0.00429$
This is less than 0.01, so $n=3$ is sufficient for the approximation.

Now we can write the Maclaurin polynomial for $e^x$ up to the third term to approximate $e^{1/2}$ :
$P_3(1/2) = 1 + \frac{1}{2} + \frac{(1/2)^2}{2!} + \frac{(1/2)^3}{3!}$

Calculate the approximation:
$P_3(1/2) = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{48}$