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**Maclaurin Series Calculator: **Do you feel solving Maclaurin Series of function is difficult? Then you are on the right place. You can get the step by step procedure to solve the maclaurin series function in a shot span of time. This Maclaurin Series Calculator gives the answer for your question immediately. Go through the following sections to get a clarity on the Maclaurin Series.

Below mentioned are the simple and easy steps that are helpful while solving the maclaurin series function. Follow these guidelines

- Take any function and its range to solve the Maclaurin Series.
- For any function f(x) the maclaurin series is given by f(x)=∑
_{k=0}^{∞}f^{(k)}(a)* x^{k}/ k! - Find f
^{(k)}(a) by calculating the function derivative and substituting the range values in the function. - Compute the k! for each step.
- Replace the values in the above formula.
- Apply sigma function and obtain the answer.

**Example**

**Question: Calculate Maclaurin Series of sin(x) up to n = 5?**

**Solution:**

Given function f(x)= Sin(x)

n = 0 to 5

Maclaurin series for the function is f(x)=∑_{k=0}^{∞} f^{(k)} (a)* x^{k}/ k!

f(x)≈ ∑_{k=0}^{5} f^{(k)} (a)* x^{k}/ k!

So, calculate the derivative and evaluate them at the given point to get the result into the given formula.

f^{0}(x) = f(x) = sin(x)

Evaluate function: f(0) = 0

Calculate first derivative f^{1}(x) = [f^{0}(x)]'

= [sin(x)]' = cos(x)

[f^{0}(x)]' = cos(x)

Evaluate first derivative (f(0))' = cos(0) = 1

Second Derivative: f^{2}(x) = [f^{1}(x)]' = [cos(x)]' = -sin(x)

(f(0))′′=0

Third derivative: f^{3}(x) = [f^{2}(x)]' = (-sin(x))' = -cos(x)

Evaluate third derivative (f(0))''' = -cos(0)

= -1

Fourth derivative: f^{4}(x) = [f^{3}(x)]' = [-cos(x)]' = sin(x)

Evaluate 4th derivative (f(0))'''' = sin(0) = 0

Fifth derivative: f^{5}(x) = [f^{4}(x)]' = [sin(x)]' = cos(x)

Evaluate 5th derivative (f(0)''''' = cos(0) = 1

Replace the derivative values in the above formula

f(x) ≈ 0/0! x^{0} + 1/1! x^{1} + 0/2! x^{2} + (-1)/3! x^{3} + 0/4! x^{4} + 1/5! x^{5}

f(x) ≈ 0 + x + 0 - 1/6 x^{3} + 0 + 1/120 x^{5}

sin(x) ≈ x-1/6 x^{3} + 1/120 x^{5}

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**1. How do you define the Maclaurin Series?**

Maclaurin Series is defined as the expanded series of the function. Here, approximate value of the function can be evaluated as the sum of derivatives of the function.

**2. What is the difference between Maclaurin and Taylor Series?**

Taylor Series is the representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. Maclaurin series is a special case of tayloe series which uses zero as a single point.

**3. How does Maclaurin series work?**

A maclaurin series is a power series that allows you to calculate an approximation of function f(x) for the input values close to zero, given that one knows the values of the successive derivatives of the function at zero.

**4. What is the formula of Maclaurin series?**

Maclaurin series for function is defined as f(x)=∑_{k=0}^{∞} f^{(k)} (a)* x^{k}/ k!