Chapter 8: Q . 8 (page 700)

If f is a function such that $f (0) = 1 a n d f^{'} (x) = f (x)$ for every value of x, find the Maclaurin series for f.

Short Answer

Expert verified

So, The Maclaurin series for the function is:

$f (x) = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots$

Or, it can be written as

$e^{x} = \sum_{i = 0}^{\infty} \frac{1}{k!} x^{k}$

Step by step solution

given information

Let us consider the function such that that $f (0) = I a n d f^{'} (x) = f (x)$

So, the function must be $f (x) = e^{x}$

calculation

Since, the general formula to calculate the Maclaurin series for the function is:

f (x) = f (0) + f^{'} (0) x + \frac{f^{''} (0)}{2!} x^{2} + \frac{f^{''} (0)}{3!} x^{3} + \dots

As $f (0) = 1,$ therefore $f^{'} (0), f^{''} (0) a n d f^{''} (0)$ are 1 .

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If f is a function such that $f (0) = 1 a n d f^{'} (x) = f (x)$ for every value of x, find the Maclaurin series for f.

Short Answer

Step by step solution

given information

calculation

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If f is a function such thatf(0)=1andf'(x)=f(x)for every value of x, find the Maclaurin series for f.

Short Answer

Step by step solution

given information

calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

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If f is a function such that $f (0) = 1 a n d f^{'} (x) = f (x)$ for every value of x, find the Maclaurin series for f.