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miskamm [114]
2 years ago
13

Your friend heard that functions are studied in intermediate and college algebra courses. He asks you what a function is. Provid

e him with a clear, relatively concise response.
Mathematics
1 answer:
hodyreva [135]2 years ago
3 0

The function is actually a relation between variables in which the one variable is dependent on the other variable.

<h3>What is a function?</h3>

A function is defined as the relationship between variables in which the alteration of one variable affects the other variable.

Suppose we are having a function:-

F(x)=2x^2-1

In the above function f(x) is an independent variable and the 2x^2-1 is a dependent variable so for the different values of x the value of the dependent function varies.

Hence the function is actually a relation between variables in which the one variable is dependent on the other variable.

To know more about Function follow

brainly.com/question/25638609

#SPJ1

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Answer:

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Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

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We first compute the n-th derivative of f(x)=\ln(1+2x), note that

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Now, if we compute the n-th derivative at 0 we get

f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!

and so the Maclaurin series for f(x)=ln(1+2x) is given by

f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2  \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}

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