Solved on Mar 07, 2024

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

STEP 1

Assumptions
1. The Maclaurin polynomial for exe^x is given by the series ex=1+x+x22!+x33!+e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.
2. We want to approximate e\sqrt{e}, which means we are looking for the value of e1/2e^{1/2}.
3. The maximum error allowed in the approximation is 0.01.
4. The Maclaurin series for exe^x converges for all xx, so we can use it to approximate e1/2e^{1/2}.

STEP 2

The Maclaurin polynomial for exe^x up to the nn-th term is given by:
Pn(x)=1+x+x22!+x33!++xnn!P_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}
We need to find nn such that the error in approximating e1/2e^{1/2} using Pn(1/2)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 exe^x at x=1/2x = 1/2 is:
Rn(1/2)=ec(n+1)!(12)n+1R_n(1/2) = \frac{e^c}{(n+1)!} \left(\frac{1}{2}\right)^{n+1}
where cc is some number between 0 and 1/2.

STEP 4

Since ece^c is always positive and increases with cc, the maximum error occurs when cc is largest, that is, when c=1/2c = 1/2. Therefore, we can estimate the error by setting c=1/2c = 1/2:
Rn(1/2)e1/2(n+1)!(12)n+1R_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:
e1/2(n+1)!(12)n+1<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 nn such that the inequality in STEP_5 holds. We can do this by trial and error, starting with n=0n=0 and increasing nn until the inequality is satisfied.

STEP 7

Let's start with n=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}
Since e>1\sqrt{e} > 1, this is clearly greater than 0.01, so we need to try a higher value of nn.

STEP 8

Now let's try n=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}
Again, since e>1\sqrt{e} > 1, this is still greater than 0.01, so we need to try a higher value of nn.

STEP 9

Continue with n=2n=2:
e1/2(2+1)!(12)2+1=e48\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 e2.71828e \approx 2.71828, we have e1.64872\sqrt{e} \approx 1.64872.

STEP 10

Calculate the approximate value of the error for n=2n=2:
1.64872480.03435\frac{1.64872}{48} \approx 0.03435
This is still greater than 0.01, so we need to try a higher value of nn.

STEP 11

Now let's try n=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}

STEP 12

Calculate the approximate value of the error for n=3n=3:
1.648723840.00429\frac{1.64872}{384} \approx 0.00429
This is less than 0.01, so n=3n=3 is sufficient for the approximation.

STEP 13

Now we can write the Maclaurin polynomial for exe^x up to the third term to approximate e1/2e^{1/2}:
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!}

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}

STEP 15

Add the terms to find the approximate value of e\sqrt{e}:
P3(1/2)=1+0.5+0.125+0.02083331.6458333P_3(1/2) = 1 + 0.5 + 0.125 + 0.0208333 \approx 1.6458333
The Maclaurin polynomial approximation for e\sqrt{e} with a maximum error of 0.01 is approximately 1.6458333.

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