Can someone explain to me how I go about solving this?

- Thread starter Preston019
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Can someone explain to me how I go about solving this?

A **terms** of the Taylor Series \(\displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x-c)^n\) are those \(\displaystyle \frac{f^{(n)}(c)}{n!}(x-c)^n\) that are being added.

The "general term" is \(\displaystyle \frac{f^{(n)}(c)}{n!}(x-c)^n\), so say exactly what that is for this f and c, in terms of n.

The "first three terms" are:

\(\displaystyle \frac{ f^{(0)} (c) }{ 0! } (x-c)^0 = f(c)\) (but numerically computed from f and c),

\(\displaystyle \frac{ f^{(1)} (c) } { 1! } (x-c)^1 = f'(c)(x-c)\) (but numerically computed from f and c),

\(\displaystyle \frac{ f^{(2)} (c) } {2!} (x-c)^2 = \frac{f''(c)}{2}(x-c)^2\) (but numerically computed from f and c).

However, those might not be the first three**non-zero** terms.

The answer wants those f and c references replaced with values as determined by what's given for f and c.

For instance, the first term is: \(\displaystyle f(c) = \sin(2c) = \sin\left(2\left(\frac{\pi}{4}\right)\right) = \sin\left(\frac{\pi}{2}\right) = 1.\)

The second term is: \(\displaystyle f'(c)(x-c) = [2\cos(2c)](x-c) = \left[ 2\cos \left( 2 \left( \frac{\pi}{4} \right) \right) \right] \left( x - \frac{\pi}{4}\right)\)

\(\displaystyle = \left[2\cos\left(\frac{\pi}{2}\right)\right] \left(x - \frac{\pi}{4}\right) = 0.\) (Notice that this is NOT a "non-zero term").

The 3rd term is: \(\displaystyle \frac{f''(c)}{2}(x-c)^2 = \left[ \frac{-4\sin(2c)}{2} \right] (x-c)^2 = -2 \sin \left( 2 \left( \frac{\pi}{4} \right) \right) \left( x - \frac{\pi}{4}\right)^2\)

\(\displaystyle = -2 \sin \left(\frac{\pi}{2}\right) \left(x - \frac{\pi}{4}\right)^2 = -2(1)\left(x - \frac{\pi}{4}\right)^2 = -2\left(x - \frac{\pi}{4}\right)^2.\)

The "general term" is \(\displaystyle \frac{f^{(n)}(c)}{n!}(x-c)^n\), so say exactly what that is for this f and c, in terms of n.

The "first three terms" are:

\(\displaystyle \frac{ f^{(0)} (c) }{ 0! } (x-c)^0 = f(c)\) (but numerically computed from f and c),

\(\displaystyle \frac{ f^{(1)} (c) } { 1! } (x-c)^1 = f'(c)(x-c)\) (but numerically computed from f and c),

\(\displaystyle \frac{ f^{(2)} (c) } {2!} (x-c)^2 = \frac{f''(c)}{2}(x-c)^2\) (but numerically computed from f and c).

However, those might not be the first three

The answer wants those f and c references replaced with values as determined by what's given for f and c.

For instance, the first term is: \(\displaystyle f(c) = \sin(2c) = \sin\left(2\left(\frac{\pi}{4}\right)\right) = \sin\left(\frac{\pi}{2}\right) = 1.\)

The second term is: \(\displaystyle f'(c)(x-c) = [2\cos(2c)](x-c) = \left[ 2\cos \left( 2 \left( \frac{\pi}{4} \right) \right) \right] \left( x - \frac{\pi}{4}\right)\)

\(\displaystyle = \left[2\cos\left(\frac{\pi}{2}\right)\right] \left(x - \frac{\pi}{4}\right) = 0.\) (Notice that this is NOT a "non-zero term").

The 3rd term is: \(\displaystyle \frac{f''(c)}{2}(x-c)^2 = \left[ \frac{-4\sin(2c)}{2} \right] (x-c)^2 = -2 \sin \left( 2 \left( \frac{\pi}{4} \right) \right) \left( x - \frac{\pi}{4}\right)^2\)

\(\displaystyle = -2 \sin \left(\frac{\pi}{2}\right) \left(x - \frac{\pi}{4}\right)^2 = -2(1)\left(x - \frac{\pi}{4}\right)^2 = -2\left(x - \frac{\pi}{4}\right)^2.\)

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When it says "non-zero terms" is it wanting (f^n(c)/n!)(x-c)^n when it doesn't equal 0 when you plug in f and c?A "term" of the Taylor Series \(\displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x-c)^n\) are those \(\displaystyle \frac{f^{(n)}(c)}{n!}(x-c)^n\) that are being added.

The "general term" is \(\displaystyle \frac{f^{(n)}(c)}{n!}(x-c)^n\), so say exactly what that is for this f and c, in terms of n.

The "first three terms" are:

\(\displaystyle \frac{ f^{(0)} (c) }{ 0! } (x-c)^0 = f(c)\) (but numerically computed from f and c),

\(\displaystyle \frac{ f^{(1)} (c) } { 1! } (x-c)^1 = f'(c)(x-c)\) (but numerically computed from f and c),

\(\displaystyle \frac{ f^{(2)} (c) } {2!} (x-c)^2 = \frac{f''(c)}{2}(x-c)^2\) (but numerically computed from f and c).

However, those might not be the first threenon-zeroterms.

The answer wants those f and c references replaced with values as determined by what's given for f and c.

For instance, the first term is: \(\displaystyle f(c) = \sin(2c) = \sin\left(2\left(\frac{\pi}{4}\right)\right) = \sin\left(\frac{\pi}{2}\right) = 1.\)

The second term is: \(\displaystyle f'(c)(x-c) = [2\cos(2c)](x-c) = \left[ 2\cos \left( 2 \left( \frac{\pi}{4} \right) \right) \right] \left( x - \frac{\pi}{4}\right)\)

\(\displaystyle = \left[2\cos\left(\frac{\pi}{2}\right)\right] \left(x - \frac{\pi}{4}\right) = 0.\)

The 3rd term is: \(\displaystyle \frac{f''(c)}{2}(x-c)^2 = \left[ \frac{-4\sin(2c)}{2} \right] (x-c)^2 = -2 \sin \left( 2 \left( \frac{\pi}{4} \right) \right) \left( x - \frac{\pi}{4}\right)^2\)

\(\displaystyle = -2 \sin \left(\frac{\pi}{2}\right) \left(x - \frac{\pi}{4}\right)^2 = -2(1)\left(x - \frac{\pi}{4}\right)^2 = = -2\left(x - \frac{\pi}{4}\right)^2.\)

The first two non-zero terms are: \(\displaystyle 1 \text{ and } -2\left(x - \frac{\pi}{4} \right)^2.\)

You now need only find the 3rd non-zero term.

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