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Using Taylor's theorem express 40 53(x - 2) 19(x - 2)2 2(x - 2) power of x.​?
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Using Taylor's theorem express 40 53(x - 2) 19(x - 2)2 2(x - 2) ...
Taylor's Theorem

Taylor's theorem is a mathematical concept used to represent a given function as an infinite sum of terms. It involves using the derivatives of the function at a specific point to approximate the value of the function at nearby points.

Expressing the given function using Taylor's Theorem

The given function is 40 + 53(x - 2) + 19(x - 2)^2 + 2(x - 2)^x. We can use Taylor's theorem to express this function as an infinite sum of terms.

Step 1: Find the derivatives of the function

The first step is to find the derivatives of the function at the point x = 2. We can find the derivatives by using the power rule and the chain rule.

f(x) = 40 + 53(x - 2) + 19(x - 2)^2 + 2(x - 2)^x

f'(x) = 53 + 38(x - 2) + 2(x - 2)^(x-1) + 2(x - 2)^x * ln(x - 2)

f''(x) = 38 + 4(x - 2)^(x-1) + 2(x - 2)^(x-1) * ln(x - 2) + 2(x - 2)^x * ln(x - 2)^2 + 2(x - 2)^x / (x - 2)

f'''(x) = 4(x - 2)^(x-1) * ln(x - 2) + 6(x - 2)^(x-1) * ln(x - 2)^2 + 2(x - 2)^(x-1) * ln(x - 2)^3 + 2(x - 2)^x * (ln(x - 2)^3 + 3ln(x - 2) / (x - 2)^2 + 1 / (x - 2)^2)

Step 2: Write the Taylor series

Using the derivatives we found in step 1, we can write the Taylor series for the given function as:

f(x) = f(2) + f'(2)(x - 2) + f''(2)(x - 2)^2 / 2! + f'''(2)(x - 2)^3 / 3! + ...

Simplifying this expression gives us:

f(x) = 40 + 53(x - 2) + 19(x - 2)^2 + 2(x - 2)^x + (53/1! + 38/2!)(x - 2)^2 + (2(x - 2)^x * ln(x - 2) + 4/3!)(x - 2)^3 + ...

Step 3: Evaluate the Taylor series

We can use the Taylor series to approximate the value of the function at any point x near x = 2. The accuracy of the approximation depends on how many terms we include in the series.

For example, if we want to approximate the value of the function at x = 2.1, we can plug in the value of x and evaluate the series up to the fourth term:

f
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Using Taylor's theorem express 40 53(x - 2) 19(x - 2)2 2(x - 2) ...
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