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Express the 4th Taylor polynomial in the a=1 neighborhood of this function.

f(x)=e^5x

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The nth term of the Taylor series for f(x) centered at a=1 is given by:

f^(n)(1)/n! * (x-1)^n

where f^(n)(1) is the nth derivative of f evaluated at x=1.

Since f(x) = e^(5x), we have:

f'(x) = 5e^(5x)
f''(x) = 25e^(5x)
f'''(x) = 125e^(5x)
f''''(x) = 625e^(5x)

Evaluating these at x=1, we get:

f'(1) = 5e^5
f''(1) = 25e^5
f'''(1) = 125e^5
f''''(1) = 625e^5

Substituting these into the formula for the 4th Taylor polynomial, we get:

P4(x) = f(1) + f'(1)(x-1) + f''(1)(x-1)^2/2 + f'''(1)(x-1)^3/6 + f''''(1)(x-1)^4/24

= e^5 + 5e^5(x-1) + 25e^5(x-1)^2/2 + 125e^5(x-1)^3/6 + 625e^5(x-1)^4/24

Simplifying, we get:

P4(x) = e^5 * (1 + 5(x-1) + 25(x-1)^2/2 + 125(x-1)^3/6 + 625(x-1)^4/24)
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