User:EGM6341.s10.Team1.toddmock/HW5

= Problem 3: Change dependent variable of a function =

Given
Given the equation


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle x(t)=t\dfrac{h}{2}+\dfrac{x_k+x_{k+1}}{2}
 * }.
 * }.

Find
Solve the above equation so that t is the dependent variable

Solution
It is simple enough to solve this equation for t giving


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle t(x)=\dfrac{2x}{h}-\dfrac{x_k+x_{k+1}}{h}
 * }.
 * }.

Thus because


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle h=x_{k+1}-x_k
 * }.
 * }.

When


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle x=x_k, t=-1
 * }.
 * }.

and when


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle x=x_{k+1}, t=1
 * }.
 * }.

= Problem 6: Taylor series expansion for Bernoulli numbers =

Given
Note that this function contradicts the one given in the notes, however it has been verified through added manipulation of equation 6 on wolfram


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ x\coth(x)= \sum_{r=0}^{\infty} \frac{2^{2r}B_{2r}}{(2r)!}x^{2r}
 * }
 * }

This equation can be used to obtain the Bernoulli numbers as well as the term


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ -d_{2r}=\frac{B_{2r}}{(2r)!}
 * }
 * }

Find
Using the above relations obtain the Richardson Extrapolation Coefficients
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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ d_{2},d_{4},d_{6},d_{8},d_{10}
 * }
 * }

Solution
The hyperbolic cotangent function can be defined as


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ \coth(x)=\frac{\cosh(x)}{\sinh(x)}=\frac{e^x+e^{-x}}{e^x-e^{-x}}
 * }
 * }

Furthermore we can write these terms as a series of summations


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ e^x=\sum_{j=0}^{\infty}\frac{x^j}{j!}
 * }
 * }

and


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ e^{-x}=\sum_{j=0}^{\infty}(-1)^j\frac{x^j}{j!}
 * }
 * }

Thus giving


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ e^x+e^{-x}=\sum_{j=0}^{\infty}\frac{x^j}{j!}(1+(-1)^j)
 * }
 * }

and


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ e^x-e^{-x}=\sum_{k=0}^{\infty}\frac{x^k}{k!}(1-(-1)^k)
 * }
 * }

or


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ x\coth(x)=\dfrac{\sum_{j=0}^{\infty}\dfrac{x^{j+1}}{j!}(1+(-1)^j)}{\sum_{k=j+1}^{\infty}\dfrac{x^k}{k!}(1-(-1)^k)}
 * }
 * }

With these formulations we can write the series out so that we can identify the Bernoulli numbers and the extrapolation terms


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$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ x\coth(x)=1+\dfrac{x^2}{3}-\dfrac{x^4}{45}+\dfrac{2x^6}{945}-\dfrac{x^8}{4725}+\dfrac{2x^{10}}{93555}
 * }
 * }

Thus pulling out the Bernoulli numbers

we can calculate the extrapolation coefficients to be