Lagrange Form of the Remainder Term in a Taylor Series
Key Questions
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By taking the derivatives,
f(x)=e^{4x}f(x)=e4x
f'(x)=4e^{4x}
f''(x)=4^2e^{4x}
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f^{(n+1)}(x)=4^{n+1}e^{4x} So,
R_n(x;3)={f^{(n+1)}(z)}/{(n+1)!}(x-3)^{n+1}={4^{n+1}e^{4z}}/{(n+1)!}(x-3)^{n+1} ,
wherez is betweenx and3 . -
Remainder Term of Taylor Series
R_n(x;c)={f^{(n+1)}(z)}/{(n+1)!}(x-c)^{n+1} ,where
z is a number betweenx andc .
Let us find
R_3(x;1) forf(x)=sin2x .By taking derivatives,
f'(x)=2cos2x
f''(x)=-4sin2x
f'''(x)=-8cos2x
f^{(4)}(x)=16sin2x So, we have
R_3(x;1)={16sin2z}/{4!}(x-1)^4 ,where
z is a number betweenx and1 .
I hope that this was helpful.
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Taylor remainder term
R_n(x;c)={f^{(n+1)}(z)}/{(n+1)!}(x-c)^{n+1} ,where
z is betweenx andc .
Questions
Power Series
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Introduction to Power Series
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Differentiating and Integrating Power Series
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Constructing a Taylor Series
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Constructing a Maclaurin Series
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Lagrange Form of the Remainder Term in a Taylor Series
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Determining the Radius and Interval of Convergence for a Power Series
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Applications of Power Series
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Power Series Representations of Functions
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Power Series and Exact Values of Numerical Series
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Power Series and Estimation of Integrals
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Power Series and Limits
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Product of Power Series
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Binomial Series
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Power Series Solutions of Differential Equations