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State and prove Maclaurin's theorem​

Answer 1

Answer:

Maclaurin's theorem is: The Taylor's theorem provides a way of determining those values of x for which the Taylor series of a function f converges to f(x). ... Maclaurin series are a type of series expansion in which all terms are nonnegative integer powers of the variable.

Step-by-step explanation:

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Answer 2

Answer:

The Maclaurin series is a special case of the Taylor series for a continuous function at x = 0 . It is a summation of all the derivatives of a function at x = 0, and gives an approximation of the function for points close to the origin. It is generally a very close representation to the original function. It might possibly be used in engineering, statistics, physics, or actuarial work, though the Taylor series is more likely to be used.

Step-by-step explanation:

A function f(x) can be expanded in ascending power of x like

f(x) = f(0) + x/1! f'(0) + x^2/2! f''(0) + … + x^n /n! f^n (0)

Where,

f(0) = value of function at x=0

f'(0) = value of first derivative of f(x) at x= 0

f''(0) =value of second derivative of f(x) at x= 0

f^n (0) = value of nth derivative of f(x) at x= 0

Proof→

Let f(x) = A0 + A1x + A2X^2 + … An X^n

Where, A0, A1, A2, …An = constant

Consider the function of the form

$\[f\left( x \right) = {e^x}\]$

Using $x = 0$, the given equation function becomes

$\[f\left( 0 \right) = {e^0} = 1\]$

Now taking the derivatives of the given function and using $$x = 0$$, we have

$\[\begin{gathered} f’\left( x \right) = {e^x},\,\,\,\,\,\,\,\,\,\,f’\left( 0 \right) = {e^0} = 1 \\ f”\left( x \right) = {e^x},\,\,\,\,\,\,\,\,\,\,f”\left( 0 \right) = {e^0} = 1 \\ f”’\left( x \right) = {e^x},\,\,\,\,\,\,\,\,\,\,f”’\left( 0 \right) = {e^0} = 1 \\ {f^{\left( {{\text{iv}}} \right)}}\left( x \right) = {e^x},\,\,\,\,\,\,\,\,\,\,{f^{\left( {{\text{iv}}} \right)}}\left( 0 \right) = {e^0} = 1 \\ \cdots \cdots \cdots \cdots \cdots \cdots \ \end{gathered} \]$

Now using Maclaurin’s series expansion function, we have

$\[f\left( x \right) = f\left( 0 \right) + xf’\left( 0 \right) + \frac{{{x^2}}}{{2!}}f”\left( 0 \right) + \frac{{{x^3}}}{{3!}}f”’\left( 0 \right) + \cdots \]$

Putting the values in the above series, we have

$\[\begin{gathered} {e^x} = 1 + x\left( 1 \right) + \frac{{{x^2}}}{{2!}}\left( 1 \right) + \frac{{{x^3}}}{{3!}}\left( 1 \right) + \cdots \\ {e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \cdots \\ \end{gathered} \]$

Reference Link

• https://brainly.in/question/5718330