Taylor/Maclaurin Series

## Answer 1: A

Using the Maclaurin series for e^x, substitute 5x for the x. Then solve the equation and simplify.

## Answer 2: C

Just like in question 1, use the Maclaurin series for e^x to solve this problem. Instead of substituting, multiply the whole equation by x and solve.

## Answer 3: C

Find the values of f(a), f'(a), f''(a), and so on, until there is only a value (no more variables to plug a into). Put these values into the Taylor series, but remember to add in the (x - 2) terms!

## Answer 4: C

Again, find the values of f(a), f'(a), and so on. Because cos(x) is an oscillating function, use the Maclaurin series for cos(x) to create the Taylor series expression.

## Answer 5: A

Knowing that 1/x is the derivative of ln(x), simply take the derivative of each term in Maclaurin series for ln(x), and then create an expressions for each general term.

## Answer 6: D

The third-degree Maclaurin series polynomial includes the terms n = 0, 1, and 2. Plug in the n values into the Maclaurin series for sin(x), and then simplify to find the first three terms. Remember to watch your positive and negative signs!

## Answer 7: B

Plug the n values into the Maclaurin series for cos(x), but remember to include the terms (x - 8) to make it a Taylor series.

## Answer 8: A

Try to change the Maclaurin series into a similar series, in this case cos(x). By knowing how to manipulate the problem, an x can be taken out to make the function xcos(x).

## Answer 9: D

Knowing the general expression for a Maclaurin series, all that is needed is the n and a value, then plug that into the expression.

## Answer 10: C

Use the fifth-degree Maclaurin series polynomial for e^x to calculate this value. Plug in the x value for each x in the series and calculate. Be sure to round to the appropriate number of decimal places.