User:Egm6341.s11.team4/HW2

Homework 2 =Problem 2.1: Taylor Series Expansion Proof=

As assigned in lecture 6

Given
Given the remainder term after the first integration by parts upon a function


 * $$ \int_{x_{0}}^{x} (x-t) f^{(2)}(t) dt   $$    (1.1)

Objectives
1.    Repeat the integration by parts method used for 3 more degrees of the Taylor Series Approximation


 * $$ \frac{(x-x_{0})^{2}}{2!}f^{(2)}(x_{0}) +   \frac{(x-x_{0})^{3}}{3!}   f^{(3)}(x_{0})   +     \frac{(x-x_{0})^{4}}{4!}   \!$$  (1.2)

2. Use the Intermediate Mean Value Theorem to express the remainder in terms of


 * $$ f^{(5)}(\xi) \!$$ (1.3)
 * $$ \xi \epsilon [x_{0},x] \!$$

3. Do integration by parts a fourth time on the equation to verify the Taylor Series remainder equations for (n+1) expansion with $$ R_{n+2}(x) $$

4. Use the Intermediate Mean Value Theorem on the Remainder term to show


 * $$ R_{n+1}(x) = \frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}(\xi) \!$$ (1.4)

Solution
1. First we will start with the "remainder" term from the Taylor series expansion as derived in class lecture [].


 * $$ R_{1}=\int_{x_{0}}^{x}(x-t)f^{(2)}(t)dt \!$$   (1.5)

The same integration by parts method used in class will be used on this remainder to further expand the Taylor series.


 * $$\int v(t) u'(t)\, dt = v(t) u(t) - \int v'(t) u(t)\, dt\!$$ (1.6)

Here we will now define the parts in our integration by parts scheme.


 * $$ u^{\prime}=(x-t) \!$$ (1.7)


 * $$ v=f^{(2)}(t) \!$$ (1.8)


 * $$  u=(xt-\frac{t^{2}}{2})       \!$$  (1.9)

Now, substituting into our integration by parts equation.


 * $$  R_{1}=\left [ (xt-\frac{t^{2}}{2}) f^{(2)}(t)\right ]_{t=x_{0}}^{t=x}-\int_{x_{0}}^{x}(xt-\frac{t^{2}}{2}) f^{(3)}(t)dt       \!$$  (1.10)

Expanding out the term in brackets we obtain


 * $$ R_{1}=(\frac{1}{2}x^{2})f^{(2)}(x)-(xx_{0}-\frac{1}{2}x_{0}^{2})f^{(2)}(x_{0}) -\int_{x_{0}}^{x}(xt-\frac{t^{2}}{2}) f^{(3)}(t)dt  \!$$


 * $$+ [ (\frac{1}{2}x^{2})f^{(2)}(x_{0})-(\frac{1}{2}x^{2})f^{(2)}(x_{0}) ]     \!$$  (1.11)

The second line in this equation sums to zero and is added so that we can rearrange the resulting equation into an integral term and a non-integral term.


 * $$     R_{1}=[ \frac{1}{2}x^{2}f^{(2)}(x)-\frac{1}{2}x^{2}f^{(2)}(x_{0}) ] +(\frac{1}{2}x^{2}-xx_{0}+\frac{1}{2}x_{0}^{2})f^{(2)}(x_{0})  -\int_{x_{0}}^{x}(xt-\frac{t^{2}}{2}) f^{(3)}(t)dt                     \!$$  (1.12)

The bracketed term can be re-expressed as an integral. Additionally, the term $$ (\frac{1}{2}x^{2}-xx_{0}+\frac{1}{2}x_{0}^{2}) \!$$ can be expressed as $$ \frac{1}{2}(x-x_{0})^{2} \!$$. This gives us the following.


 * $$   R_{1}= \int_{x_{0}}^{x}\frac{1}{2}x^{2}f^{(3)}(t)+\frac{1}{2}(x-x_{0})^{2}f^{(2)}(x_{0})-\int_{x_{0}}^{x}(xt-\frac{t^{2}}{2}) f^{(3)}(t)dt                                 \!$$  (1.13)

Now regrouping the integral terms under one integral we obtain the following.


 * $$     R_{1}=  \int_{x_{0}}^{x}\frac{1}{2}(x-t)^{2}f^{(3)}(t)+\frac{1}{2}(x-x_{0})^{2}f^{(2)}(x_{0})                       \!$$  (1.14)

This can be grouped back into our Taylor series expansion to give us the following.


 * {| style="width:70%" border="0" align="center"

$$\displaystyle f(x)=f(x_{0})+\frac{(x-x_{0})^{1}}{1!}f^{(1)}(x_{0})+\frac{(x-x_{0})^{2}}{2!}f^{(2)}(x_{0})+\int_{x_{0}}^{x}\frac{1}{2}(x-t)^{2}f^{(3)}(t) $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }. (1.15)
 * }.<p style="text-align:right"> (1.15)

Our new remainder term, the term in the integral, will now be further expanded.


 * $$      R_{2} =   \int_{x_{0}}^{x}\frac{1}{2}(x-t)^{2}f^{(3)}(t)                       \!$$<p style="text-align:right">  (1.16)

Our integration by parts terms are now the following.
 * $$          u'=  \frac{1}{2}(x-t)^{2}                  \!$$<p style="text-align:right">  (1.17)
 * $$             v=f^{(3)}(t)                  \!$$<p style="text-align:right">  (1.18)
 * $$             u=\frac{1}{6}t^{3}-.5t^{2}x+.5tx^{2}                 \!$$<p style="text-align:right">  (1.19)

Substituting into our integration by parts equation we obtain the following.


 * $$ R_{2} =   [(\frac{1}{6}t^{3}-.5t^{2}x+.5tx^{2})f^{(3)}(t)]_{t=x_{0}}^{t=x} - \int_{x_{0}}^{x}  (\frac{1}{6}t^{3}-.5t^{2}x+.5tx^{2} )  f^{(4)}(t) dt            \!$$<p style="text-align:right">  (1.20)

For clarity, the term in brackets will now be expressed as $$ \alpha \!$$. Expanding the terms in bracket we obtain the following.
 * $$  R_{2} =(\frac{1}{6}x^{3}) f^{(4)}(x)- (\frac{1}{6}x_{0}^{3}-\frac{1}{2}x_{0}^{2}x+\frac{1}{2}x_{0}x^{2}) f^{(4)}(x_{0}) - \alpha             \!$$


 * $$ +(\frac{1}{6}x^{3}) f^{(4)}(x_{0}) - (\frac{1}{6}x^{3}) f^{(4)}(x_{0})                    \!$$<p style="text-align:right">  (1.21)

The second line of the equation adds to zero as before and is added to help rearrange the expressions into an integral later on.

Now we can group together the terms appropriately so that we can simplify the equation.


 * $$ R_{2} =  \int_{x_{0}}^{x} \frac{1}{6}x^{3} f^{(4)} dt   + (\frac{1}{6}x^{3} - \frac{1}{6}x_{0}^{3}-\frac{1}{2}x_{0}^{2}x+\frac{1}{2}x_{0}x^{2})f^{(3)}(x_{0}) - \alpha                 \!$$<p style="text-align:right">  (1.22)

Pulling out the fractional coefficients of the terms outside of the integrals.


 * $$ R_{2} =  \int_{x_{0}}^{x} \frac{1}{6}x^{3} f^{(4)} dt +\frac{1}{3!}(x^{3}-3x_{0}^{2}x+3x_{0}x^{2}-x_{0}^{3})  f^{(3)}(x_{0}) -\alpha                           \!$$<p style="text-align:right">  (1.23)

We can further simplify the term outside of the integral to be


 * $$ R_{2} =  \int_{x_{0}}^{x} \frac{1}{6}x^{3} f^{(4)} dt +\frac{1}{3!} (x-x_{0})  f^{(3)}(x_{0}) -\alpha                           \!$$<p style="text-align:right">  (1.24)

Now combining the terms under the integrals to eliminate our $$ \alpha \!$$ expression.


 * $$  R_{2} =  \int_{x_{0}}^{x} (\frac{1}{3!}x^{3}-\frac{1}{6}t^{3}-\frac{1}{2}t^{2}x+\frac{1}{2}tx^{2})f^{(4)} + \frac{1}{3!} (x-x_{0})  f^{(3)}(x_{0})                           \!$$<p style="text-align:right">  (1.25)

Further simplifying, this becomes


 * $$ R_{2} =  \frac{1}{3!} (x-x_{0})  f^{(3)}(x_{0})   +\int_{x_{0}}^{x} (\frac{1}{3!}(x-t)^{3})f^{(4)}(t)dt                         \!$$<p style="text-align:right">  (1.26)

Substituting this back into our original equation we obtain the following.


 * {| style="width:70%" border="0" align="center"

$$\displaystyle f(x)=f(x_{0})+\frac{(x-x_{0})^{1}}{1!}f^{(1)}(x_{0})+\frac{(x-x_{0})^{2}}{2!}f^{(2)}(x_{0})+  \frac{(x-x_{0})^{3}}{3!}   f^{(3)}(x_{0})   +\int_{x_{0}}^{x} (\frac{1}{3!}(x-t)^{3})f^{(4)}(t)dt                                    $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle
 * style= |
 * }.<p style="text-align:right"> (1.27)
 * }.<p style="text-align:right"> (1.27)

Now we can use the same method for the next term in the Taylor series.


 * $$       \int_{x_{0}}^{x} \frac{1}{3!}(x-t)^{3}f^{(4)}(t)dt                             \!$$<p style="text-align:right">  (1.28)
 * $$      u'=     \frac{1}{3!}(x-t)^{3}                   \!$$<p style="text-align:right">  (1.29)
 * $$       v= f^{(4)}(t)                       \!$$<p style="text-align:right">  (1.30)
 * $$       u=(-\frac{t^{4}}{24}+\frac{1}{6}t^{3}x-\frac{1}{4}t^{2}x^{2}+\frac{1}{6}tx^{3} )                      \!$$<p style="text-align:right">  (1.31)

Now substituting into our integration by parts equation
 * $$   R_{3}= [(-\frac{t^{4}}{24}+\frac{1}{6}t^{3}x-\frac{1}{4}t^{2}x^{2}+\frac{1}{6}tx^{3} ) f^{(4)}(t)  ]_{x_{0}}^{x} -\int_{x_{0}}^{x}  (-\frac{t^{4}}{24}+\frac{1}{6}t^{3}x-\frac{1}{4}t^{2}x^{2}+\frac{1}{6}tx^{3} )  f^{(5)}(t)  dt                           \!$$<p style="text-align:right">  (1.32)

Expanding the bracket and substituting $$ \alpha \!$$ for the integral term, we obtain the following equation.


 * $$ R_{3}= (-\frac{x^{4}}{24}+\frac{1}{6}x^{4}-\frac{1}{4}x^{4}+\frac{1}{6}x^{4} )f^{(4)}(x)  -(-\frac{x_{0}^{4}}{24}+\frac{1}{6}x_{0}^{3}x-\frac{1}{4}x_{0}^{2}x^{2}+\frac{1}{6}x_{0}x^{3} )f^{(4)}(x_{0}) -\alpha                            \!$$<p style="text-align:right">  (1.33)
 * $$ +\frac{1}{24}x^{4}  f^{(4)}(x_{0})-    \frac{1}{24}x^{4}   f^{(4)}(x_{0})                         \!$$<p style="text-align:right">  (1.34)

The second line sums to zero and is added in order to re-arrange the terms into an integral.

Now we can express the equation as the following.


 * $$ R_{3}=   \int_{x_{0}}^{x}   \frac{1}{24}x^{4}f^{(5)}(t)dt -\alpha +[\frac{1}{24}x^{4} +\frac{x_{0}^{4}}{24}-\frac{1}{6}x_{0}^{3}x+\frac{1}{4}x_{0}^{2}x^{2}-\frac{1}{6}x_{0}x^{3}  ]             \!$$<p style="text-align:right">  (1.35)

Now, combining the integral term with our $$ \alpha \!$$ term and then simplifying the terms in brackets we can express the equation as follows


 * $$R_{3}=  \frac{(x-x_{0})^{4}}{4!}   +  \int_{x_{0}}^{x}   \frac{1}{4!}(x-t)^{4}f^{(5)}(t)dt  \!$$<p style="text-align:right">  (1.36)

Substituting this into our Taylor series polynomial expression, we obtain


 * {| style="width:70%" border="0" align="center"

$$\displaystyle f(x)=f(x_{0})+\frac{(x-x_{0})^{1}}{1!}f^{(1)}(x_{0})+\frac{(x-x_{0})^{2}}{2!}f^{(2)}(x_{0})        \!$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ +  \frac{(x-x_{0})^{3}}{3!}   f^{(3)}(x_{0})   +     \frac{(x-x_{0})^{4}}{4!}   +  \int_{x_{0}}^{x}   \frac{1}{4!}(x-t)^{4}f^{(5)}(t)dt  $$

$$ 2.
 * <p style="text-align:right;">$$\displaystyle
 * style= |
 * }.<p style="text-align:right"> (1.37)
 * }.<p style="text-align:right"> (1.37)

First we state the Intermediate Mean Value Theorem (IMVT)
 * $$ \int_{a}^{b} w(t)f(t)dt = f(\xi) \int_{a}^{b} w(t)dt \!$$<p style="text-align:right">  (1.38)

This was previously proven in assignment 1 [] Now, we know the remainder for our equation is the following
 * $$ R_{4}=\int_{x_{0}}^{x}  \frac{1}{4!}(x-t)^{4}f^{(5)}(t)dt \!$$<p style="text-align:right">  (1.39)

Now, if we identify w(t) and f(t) we can use the IMVT for our problem
 * $$ W(t)= \frac{1}{4!}(x-t) \!$$<p style="text-align:right"> (1.40)
 * $$ f(t) = f^{(5)}(t) \!$$<p style="text-align:right"> (1.41)

Since we are integrating from our constants of integration x and $$ x_{0} $$ we can use the IMVT, substituting our equations, to obtain the following.


 * {| style="width:70%" border="0" align="center"

$$\displaystyle \int_{x_{0}}^{x}   \frac{1}{4!}(x-t)^{4}f^{(5)}(t)dt = f^{(5)}(\xi)\int_{x_{0}}^{x}  \frac{1}{4!}(x-t)^{4} dt    $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle
 * style= |
 * }.<p style="text-align:right"> (1.42)
 * }.<p style="text-align:right"> (1.42)

Where

$$ \xi \in [x_{0},x] \!$$<p style="text-align:right"> (1.43)

3.

Here, we must do integration by parts for a fourth time to verify equations 3 and 4 on lecture slide 3-3 for (n+1) expansion with $$ R_{n+2}(x) $$. In essence, this requires us to see if


 * $$ R_{n+2}(x) = \frac{1}{(n+1)!} \int_{x_{0}}^{x} (x-t)^{(n+1)} f^{n+2}(t)dt \!$$<p style="text-align:right"> (1.44)

Where, in our case, n=4.

We will start with the remainder we obtained in part 1.
 * $$ R_{5}(x)=\frac{1}{4!}\int_{x_{0}}^{x} (x-t)^{4} f^{(5)}(t)dt \!$$<p style="text-align:right">  (1.45)

Now identifying our integration by parts terms.


 * $$  u'=\frac{(x-t)^{4}}{4!}       \!$$<p style="text-align:right">  (1.46)
 * $$   v=f^{(5)}(t)      \!$$<p style="text-align:right">  (1.47)
 * $$    u=(\frac{t^{5}}{5!}-\frac{t^{4}x}{4!}+\frac{t^{3}x^{2}}{12}-\frac{t^{2}x^{3}}{12}+\frac{tx^{4}}{4!})     \!$$<p style="text-align:right">  (1.48)

Substituting into our integration by parts equation.


 * $$ R_{5}(x)= [(\frac{t^{5}}{5!}-\frac{t^{4}x}{4!}+\frac{t^{3}x^{2}}{12}-\frac{t^{2}x^{3}}{12}+\frac{tx^{4}}{4!}) (f^{(5)}(t) )]_{x_{0}}^{x} - \int_{x_{0}}^{x}  (\frac{t^{5}}{5!}-\frac{t^{4}x}{4!}+\frac{t^{3}x^{2}}{12}-\frac{t^{2}x^{3}}{12}+\frac{tx^{4}}{4!}) f^{(6)}(t) dt     \!$$<p style="text-align:right">  (1.49)

Now referring to the term to the integral term as $$ \alpha $$ as in the previous several iterations, expanding the bracketed term as before, and adding two terms that sum to zero in order to aid in re-arranging the equation


 * $$ R_{5}(x)=(\frac{x^{5}}{5!}-\frac{x^{5}}{4!}+\frac{x^{5}}{12}-\frac{x^{5}}{12}+\frac{x^{5}}{4!}) f^{(5)}(x)   - (\frac{x_{0}^{5}}{5!}-\frac{x_{0}^{4}x}{4!}+\frac{x_{0}^{3}x^{2}}{12}-\frac{x_{0}^{2}x^{3}}{12}+\frac{x_{0}x^{4}}{4!}) f^{(5)}(x_{0})   -\alpha \!$$
 * $$ +\frac{x^{5}}{5!}f^{(5)}(x_{0}) - \frac{x^{5}}{5!}f^{(5)}(x_{0})        \!$$<p style="text-align:right">  (1.50)

The second line of the above equation sums to zero and is added so that we can rearrange the equation into a simpler form.


 * $$  R_{5}(x)= [\frac{x^{5}}{5!}f^{(5)}(x) -\frac{x^{5}}{5!}f^{(5)}(x_{0}) ]_{x_{0}}^{x} -(\frac{x_{0}^{5}}{5!}-\frac{x_{0}^{4}x}{4!}+\frac{x_{0}^{3}x^{2}}{12}-\frac{x_{0}^{2}x^{3}}{12}+\frac{x_{0}^{4}x}{4!}-\frac{x^{5}}{5!}) f^{(5)}(x)  -\alpha    \!$$<p style="text-align:right">  (1.51)

Now we can express the bracketed term as an integral. Meanwhile, the complex polynomial simplifies to $$ \frac{(x-x_{0})^{5}}{5!} $$.
 * $$  R_{5}(x)= \int_{x_{0}}^{x}\frac{x^{5}}{5!}f^{(6)}(t)dt+\frac{(x-x_{0})^{5}}{5!}f^{(5)}(x_{0})-\alpha      \!$$<p style="text-align:right">  (1.52)

Now, combining the integral term with the $$ \alpha $$ term we end up with the following equation


 * $$ R_{6}= \frac{(x-x_{0})^{5}}{5!}f^{(5)}(x_{0}) +  \int_{x_{0}}^{x}\frac{(x-t)^{5}}{5!}f^{(6)}(t)dt     \!$$<p style="text-align:right">  (1.53)

Since n=4 in our case, this verifies the expansion with $$ R_{n+2}(x) $$

4.

We can use the IMVT on the Remainder term, given as equation 4 lecture slide 3-3  in order to show equation 5 lecture slide 3-3. The IMVT can be stated as follows.


 * $$  \int_{a}^{b}w(t)g(t)dt=g(\xi)\int_{a}^{b}w(t)dt       \!$$<p style="text-align:right">  (1.54)

When applied to equation 4, the Taylor series remainder, we can evaluate the individual terms of the IMVT to be
 * $$  a=x_{0}          \!$$<p style="text-align:right">  (1.55)
 * $$  b=x        \!$$<p style="text-align:right">  (1.56)
 * $$  w(t)= \frac{(x-t)^{n}}{n!}       \!$$<p style="text-align:right">  (1.57)
 * $$     g(t)=f^{(n+1)}(t)       \!$$<p style="text-align:right">  (1.58)

Simple substitution of these equations into the IMVT gives us the following


 * $$  \int_{x_{0}}^{x}\frac{(x-t)^{n}}{n!}f^{(n+1)}(t)dt=f^{(n+1)}(\xi)  \int_{x_{0}}^{x}  \frac{(x-t)^{n}}{n!} dt        \!$$<p style="text-align:right">  (1.59)

Now, evaluating the integral present on the right hand side, we can further simplify the remainder to the following
 * $$   R_{n+1}(x) = \frac{(x-x_{0})^{(n+1)}}{(n+1)!}f^{(n+1)}(\xi)      \!$$<p style="text-align:right">  (1.60)

And this result is simply equation 5, so we have effectively shown that equation 5 arises naturally out of equation 4 in conjunction with the Intermediate Mean Value Theorem.

Problem solved by Brendan Mahon.

Given
The function f(x)
 * $$ f(x)=sin(x) \!$$

and the span of x
 * $$ x \epsilon [0, \pi] \!$$

Objective
Evaluate the Taylor Series about the point
 * $$ x_{0}=\frac{3\pi}{8} \!$$

For Taylor Series degrees of
 * $$ n=0,1,2,...,10 \!$$

We must then plot these series for each value of n in order to visually see how the Taylor Series improves with each degree of approximation.

Next we must estimate the error in our solution at the point $$ x= \frac{3\pi}{4} \!$$ for each degree of Taylor Series approximation.

Solution
Evaluating the Taylor Series of the sin(x) is a simple matter of following the general Taylor Series expansion formula for the case where f(x) is sin(x).

The general Taylor Series expansion as presented in the lecture slides is


 * $$ P_{n}(x) = f(x_{0}) + \frac{(x-x_{0})}{1!} f^{(1)}(x_{0}) +...+\frac{(x-x_{0})^{n}}{n!}f^{(n)}(x_{0}) \!$$<p style="text-align:right">  (2.1)

For the case of sin(x), the equation then becomes


 * $$ f(x) = sin(x_{0}) + \frac{(sin(x)-sin(x_{0}))^{1}}{1!}cos(x_{0}) - \frac{(sin(x)-sin(x_{0}))^{2}}{2!}sin(x_{0}) - ...\!$$<p style="text-align:right"> (2.2)

The periodicity of the sine function shows up in our evaluation of the Taylor Series and can be capitalized upon. Since the derivative of sine is the sine function itself every 4 derivatives evaluating the nth derivative is a simple matter of determining what the remainder is when n is divided by 4 and matching the derivative with the appropriate trigonometric function based on what the remainder is.

This method is implemented in code about the point $$ x_{0} = \frac{3\pi}{8}$$ up to degree n=10 and then plotted. The result is a plot that graphically shows how the Taylor Series comes closer to the true function it evaluates with each successive degree evaluated.

The maximum error of the Taylor Series can be expressed mathematically using equation 5 lecture slide 3-3
 * $$ \left | R_{n+1}(x) \right | \leq \frac{(x-x_{0})^{n+1}}{(n+1)!}(\alpha ) \!$$<p style="text-align:right"> (2.3)

Where $$ \alpha \!$$ represents the maximum of the maximum value of the n+1 derivative over the relevant range.


 * $$ \alpha = max \left | f^{(n+1)}(t) \right | \!$$
 * $$ t \epsilon [x_{0},x] \!$$<p style="text-align:right"> (2.4)

Now we will apply these equations to our specific function; the sine function and an evaluation point of $$ x_{0} = \frac{3\pi}{8} $$ and measurement point of $$ x = \frac{3\pi}{4} $$. First, we can say that


 * $$ \alpha = max \left | f^{(n+1)}(t) \right | = 1 \!$$
 * $$ t \epsilon [x_{0},x] \!$$<p style="text-align:right"> (2.5)

This arises because of the nature of the sine function. Every nth order derivative of sine is going to be positive or negative sine or cosine. Sine and cosine functions are bounded above and below by 1 and -1. This means that the maximum absolute value of any derivative of sine is 1.

Now we can use this result to simplify our remainder equation.


 * $$ \left | R_{n+1}(x) \right | \leq \frac{(x-x_{0})^{n+1}}{(n+1)!} \!$$<p style="text-align:right"> (2.6)

Since, for the problem we are evaluating $$ x_{0} = \frac{3\pi}{8} $$ and $$ x = \frac{3\pi}{4} $$ we can say that $$ (x-x_{0})=\frac{3\pi}{8} $$


 * $$ \left | R_{n+1}(\frac{3\pi}{4}) \right | \leq \frac{(\frac{3\pi}{8})^{n+1}}{(n+1)!} \!$$<p style="text-align:right"> (2.7)

Now we have an upper bound on the error for any degree Taylor Series approximation of our function. This equation is included in the code used to plot the sine function and evaluated for all degree Taylor Series approximations from 0 to 10.

Matlab Code:

The code as printed above gives us several outputs, the plot of the Taylor Series and the error for each degree of the Taylor series about the point $$ \frac{3\pi}{4} $$.

Plots:



As can be seen, the higher order Taylor Series functions do an excellent job of approximating the sine function. In fact, every approximation above the 5th degree approximation cannot even be seen in the plot since they coincide with and are obscured by the plotting of the true value of sin(x). Thus, we would expect to see a very small error associated with these higher order approximations. The code used above gives us the following error estimations.


 * $$ \left | R_{1}(\frac{3\pi}{4}) \right | \leq 1.1781 \!$$<p style="text-align:right">  (2.8)
 * $$ \left | R_{2}(\frac{3\pi}{4}) \right | \leq 0.6940 \!$$<p style="text-align:right">  (2.9)
 * $$ \left | R_{3}(\frac{3\pi}{4}) \right | \leq 0.2725 \!$$<p style="text-align:right">  (2.10)
 * $$ \left | R_{4}(\frac{3\pi}{4}) \right | \leq 0.0803 \!$$<p style="text-align:right">  (2.11)
 * $$ \left | R_{5}(\frac{3\pi}{4}) \right | \leq 0.0189 \!$$<p style="text-align:right">  (2.12)
 * $$ \left | R_{6}(\frac{3\pi}{4}) \right | \leq 0.0037 \!$$<p style="text-align:right">  (2.13)
 * $$ \left | R_{7}(\frac{3\pi}{4}) \right | \leq 6.2494*10^{-4} \!$$<p style="text-align:right">  (2.14)
 * $$ \left | R_{8}(\frac{3\pi}{4}) \right | \leq 9.2030*10^{-5} \!$$<p style="text-align:right">  (2.15)
 * $$ \left | R_{9}(\frac{3\pi}{4}) \right | \leq 1.2047*10^{-5} \!$$<p style="text-align:right"> (2.16)
 * $$ \left | R_{10}(\frac{3\pi}{4}) \right | \leq 1.4192*10^{-6} \!$$<p style="text-align:right">  (2.17)
 * $$ \left | R_{11}(\frac{3\pi}{4}) \right | \leq 1.5200*10^{-7} \!$$<p style="text-align:right">  (2.18)

As expected, the higher order approximations from about degree six onward produce an almost negligible error.

Problem solved by Brendan Mahon.

Given

 * $$f(x)=\frac{e^{x}-1}{x}$$

Objectives
To examine the Taylor series expansions of both $$\displaystyle e^x$$ and $$\displaystyle f(x)$$ and their remainders to verify that:


 * $$R_{n+1}[f(x);x] = \frac{(x-0)^{n+1}}{x(n+1)!}e^\xi = \frac{R_{n}[e^{x};x]}{n+1}$$

Solution
For this problem we will examine Taylor series expansions about the point $$ x_0 = 0$$.

In general, functions for which the (n+1)th derivative exists and is continuous, can be expressed in the form

where $$p_{n}(x)$$ is a polynomial of order n and $$R_{n+1}(x)$$ is the corresponding remainder.

During lecture 3, it was shown that these polynomials and remainders take a specific form for Taylor series expansions.

Equation 3 from lecture slide [[media:nm1.s11.mtg3.djvu|3-3]]:

where $$\displaystyle f^{(n)}(x_0)$$ denotes the n-th derivative of f evaluated at $$\displaystyle x_0$$.

Equation 5 from lecture slide [[media:nm1.s11.mtg3.djvu|3-3]]:

For the Taylor series expansion of $$\displaystyle e^x$$ about $$\displaystyle x_0=0$$, all derivatives of $$\displaystyle e^x$$ are also $$\displaystyle e^x$$ and are simplified to 1 when evaluated at $$\displaystyle x_0=0$$.

The result of (3.4) will be manipulated to solve for the expansion and remainder of $$\displaystyle f(x)$$:

(3.8) verifies equation 2 from lecture slide [[media:nm1.s11.mtg6.djvu|6-5]]. Comparing (3.5) and (3.8), it is obvious that the remainders of $$e^x$$ and $$f(x)$$ are similar. If the Taylor series expansion for $$e^x$$ were expanded to (n-1) terms rather than n, then the remainders of both $$e^x$$ and $$f(x)$$ would be of the same degree.

(3.10) confirms equation 1 from lecture slide [[media:nm1.s11.mtg7.djvu|7-1]]!

Problem solved by William Kurth.

=Problem 2.4: Comparing Methods of Numerical Integration= Refer to lecture slide [[media:nm1.s11.mtg7.djvu|7-3]] for problem statement.

Objectives
Plot f(x). Then, integrate f(x) using the following methods:
 * (1) Taylor series expansion
 * (2) Composite trapezoidal rule
 * (3) Composite Simpson rule
 * (4) Gauss-Legendre quadrature

for $$n=2,4,8...$$, until the error reaches the order $$10^{-6}$$.

Solution
The following MATLAB code was used to plot f(x) from -1 to 1.



(1) Integration by Taylor Series Expansion

To evaluate (4.1) by Taylor series expansion, the function f(x) will be approximated by an expansion about the point $$\displaystyle x_0 = 0$$. In homework problem 1.2, this expansion was shown to be

To solve for n, we must first examine the nature of the error.

On lecture slide [[media:nm1.s11.mtg7.djvu|7-1]], this error was shown to have the form that follows (adjusted for different interval):

When observing $$\displaystyle e^{\xi(\alpha)}$$ over the interval $$\displaystyle\alpha\;\in\;[-1,1]$$ it is seen that the function is bounded on both the top and bottom by

This can be extended to say that the error, $$\displaystyle E_n(f)$$, is also bounded as follows:

Now we will calculate the integration of f(x) and the upper bound of the error for increasing values of n until the error reaches the order of $$10^{-6}$$. The following table shows these values.

(2) Integration by Composite Trapezoidal Rule

Equation (1) from lecture slide [[media:nm1.s11.mtg7.djvu|7-4]]:

where $$\displaystyle h=\frac{b-a}{n}$$ and  $$\displaystyle f_n=f(x_n)$$.

The error associated with the composite trapezoidal rule is (Atkinson, p.253)

For this problem (4.10) becomes

Values of In and En are calculated from equations (4.9) and (4.13), respectively, for increasing values of n and shown in the table below.

(3) Integration by Composite Simpson Rule

Equation (4) from lecture slide [[media:nm1.s11.mtg7.djvu|7-4]]:

where $$\displaystyle h=\frac{b-a}{n}$$ and  $$\displaystyle f_n=f(x_n)$$ and n must be even.

The error associated with the composite trapezoidal rule is (Atkinson, p.257)

For this problem (4.15) becomes

Values of In and En are calculated from equations (4.14) and (4.18), respectively, for increasing values of n and shown in the table below.

(4)Integration by Gauss-Legendre Quadrature

Equation (1) from lecture slide [[media:nm1.s11.mtg7.djvu|7-5]]:

where $$\displaystyle \{x_i, i=1,2,3,...,n\}$$ are roots of the Legendre polynomial of order n, $$\displaystyle \mathcal{P}_n(x)$$, and $$\displaystyle w_i$$ are their corresponding weights. Some of these values are shown in the table below supplied in lecture slide [[media:nm1.s11.mtg7.djvu|7-5]].

The error associated with using this method for approximating the value of an integral is given by equation (3) on lecture slide [[media:nm1.s11.mtg8.djvu|8-2]].

Equations (4.19) and (4.20) were solved for increasing values of n until the error reached the order of 10-6 and shown in the table below.

Solved by William Kurth.

=Problem 2.5: Verifying Legendre Polynomial Roots and Weights= Refer to lecture slide [[media:nm1.s11.mtg7.djvu|mtg-7]] for the problem statement.

Given
Gauss-legendre quadrature:

Objectives
Verify this table against NIST handbook (lecture plan).

Solution
According to the Legendre polynomial in lecture note [[media:nm1.s11.mtg8.djvu|mtg-8]], {  Xi, i=1….n }  is the root of Legendre Polynomial of order n, and the associated weight with Xi is :
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{W_i} = \frac

$$     (Eq 5.1 ) where the Legendre Polynomial is:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{P_n}(x) = \sum\limits_{i = 0}^{[n/2]} \frac

$$     (Eq 5.2 )
 * <p style="text-align:right">
 * }.
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{P_1}(x) = x

$$     (Eq 5.3 )
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{P_2}(x) = \frac{1}{2}(3{x^2} - 1)

$$     (Eq 5.4 )
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{P_3}(x) = \frac{1}{2}(5{x^3} - 3x)

$$     (Eq 5.5)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{P_4}(x) = \frac{8}{x^4} - \frac{4}{x^2} + \frac{3}{8}

$$     (Eq 5.6)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{P_5}(x) = \frac{8}{x^5} - \frac{8}{x^3} + \frac{8}x

$$     (Eq 5.7 )
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{P_6}(x) = \frac{x^6} - \frac{x^4} + \frac{x^2} - \frac{5}

$$     (Eq 5.8 )
 * <p style="text-align:right">
 * }

1). When n=1, the root of P1(x)=0 is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_{1}} = 0

$$     (Eq 5.9 )
 * <p style="text-align:right">
 * }

And the weight is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{1}}=\frac{-2}{(1+1)[P_{1}^{'}({{x}_{1}}){{P}_{2}}({{x}_{1}})]}=\frac{-2}{(1+1)[1\centerdot (3{{x}_{1}}^{2}-1)/2]}=\frac{-2}{(1+1)[1\centerdot (3\centerdot 0-1)/2]}=1

$$     (Eq 5.10 )
 * <p style="text-align:right">
 * }

2). When n=2, the root of P2(x)=0 is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_{1,2}} = \pm 1/\sqrt 3

$$     (Eq 5.11 )
 * <p style="text-align:right">
 * }

And the weight is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{1}}=\frac{-2}{(2+1)[P_{2}^{'}({{x}_{1}}){{P}_{3}}({{x}_{1}})]}=\frac{-2}{(2+1)[(3{{x}_{1}})\centerdot (5{{x}_{1}}^{3}-3{{x}_{1}})/2]}=1

$$     (Eq 5.12 )
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{2}}=\frac{-2}{(2+1)[P_{2}^{'}({{x}_{2}}){{P}_{3}}({{x}_{2}})]}=\frac{-2}{(2+1)[(3{{x}_{2}})\centerdot (5{{x}_{2}}^{3}-3{{x}_{2}})/2]}=1

$$     (Eq 5.13 )
 * <p style="text-align:right">
 * }

3). When n=3, the root of P3(x)=0 is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_1} = 0,\begin{array}{ccccccccccccccc} {}&{{x_{2,3}} = \pm \sqrt {15} /5} \end{array}

$$     (Eq 5.14 )
 * <p style="text-align:right">
 * }

And the weight is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{1}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{1}}){{P}_{4}}({{x}_{1}})]}=\frac{-2}{(3+1)[(15{{x}_{1}}^{2}-3)/2\centerdot (\frac{35}{8}{{x}_{1}}^{4}-\frac{15}{4}{{x}_{1}}^{2}+\frac{3}{8})]}=\frac{8}{9}

$$     (Eq 5.15 )
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{2}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{2}}){{P}_{4}}({{x}_{2}})]}=\frac{-2}{(3+1)[(15{{x}_{2}}^{2}-3)/2\centerdot (\frac{35}{8}{{x}_{2}}^{4}-\frac{15}{4}{{x}_{2}}^{2}+\frac{3}{8})]}=\frac{5}{9}

$$     (Eq 5.16)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{3}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{3}}){{P}_{4}}({{x}_{3}})]}=\frac{-2}{(3+1)[(15{{x}_{3}}^{2}-3)/2\centerdot (\frac{35}{8}{{x}_{3}}^{4}-\frac{15}{4}{{x}_{3}}^{2}+\frac{3}{8})]}=\frac{5}{9}

$$     (Eq 5.17)
 * <p style="text-align:right">
 * }

4). When n=4, the root of P4(x)=0 is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac{8}{x^4} - \frac{4}{x^2} + \frac{3}{8} = 0

$$     (Eq 5.18)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{x^2} = (3 \pm 2\sqrt {6/5} )/7

$$     (Eq 5.19)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_{1,2}} = \pm \sqrt {(3 + 2\sqrt {6/5} )/7}

$$     (Eq 5.20 )
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_{3,4}} = \pm \sqrt {(3 - 2\sqrt {6/5} )/7}

$$     (Eq 5.21 )
 * <p style="text-align:right">
 * }

And the weight is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{1}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{1}}){{P}_{5}}({{x}_{1}})]}=\frac{-2}{(4+1)[(\frac{35}{2}{{x}_{1}}^{3}-\frac{15}{2}{{x}_{1}})\centerdot (\frac{63}{8}{{x}_{1}}^{5}-\frac{70}{8}{{x}_{1}}^{3}+\frac{15}{8}{{x}_{1}})]}=(18-\sqrt{30})/36

$$     (Eq 5.22 )
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{2}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{2}}){{P}_{5}}({{x}_{2}})]}=\frac{-2}{(4+1)[(\frac{35}{2}{{x}_{2}}^{3}-\frac{15}{2}{{x}_{2}})\centerdot (\frac{63}{8}{{x}_{2}}^{5}-\frac{70}{8}{{x}_{2}}^{3}+\frac{15}{8}{{x}_{2}})]}=(18-\sqrt{30})/36

$$     (Eq 5.23 )
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{3}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{3}}){{P}_{5}}({{x}_{3}})]}=\frac{-2}{(4+1)[(\frac{35}{2}{{x}_{3}}^{3}-\frac{15}{2}{{x}_{3}})\centerdot (\frac{63}{8}{{x}_{3}}^{5}-\frac{70}{8}{{x}_{3}}^{3}+\frac{15}{8}{{x}_{3}})]}=(18+\sqrt{30})/36

$$     (Eq 5.24)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{4}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{4}}){{P}_{5}}({{x}_{4}})]}=\frac{-2}{(4+1)[(\frac{35}{2}{{x}_{4}}^{3}-\frac{15}{2}{{x}_{4}})\centerdot (\frac{63}{8}{{x}_{4}}^{5}-\frac{70}{8}{{x}_{4}}^{3}+\frac{15}{8}{{x}_{4}})]}=(18+\sqrt{30})/36

$$     (Eq 5.25)
 * <p style="text-align:right">
 * }

5). When n=5, the root of P5(x)=0 is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{P_5}(x) = \frac{8}{x^5} - \frac{8}{x^3} + \frac{8}x

$$     (Eq 5.26)
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{x_1} = 0,\begin{array}{ccccccccccccccc} {}&{{{({x_{2,3,4,5}})}^2} = (5 \pm 2\sqrt {10/7} )/9} \end{array}

$$     (Eq 5.27)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_{2,3}} = \pm \sqrt {(5 + 2\sqrt {10/7} )} /3

$$     (Eq 5.28)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_{4,5}} = \pm \sqrt {(5 - 2\sqrt {10/7} )} /3

$$     (Eq 5.29)
 * <p style="text-align:right">
 * }

And the weight is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{W}_{1}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{1}}){{P}_{6}}({{x}_{1}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{1}}^{4}-\frac{210}{8}{{x}_{1}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{1}}^{6}-\frac{315}{16}{{x}_{1}}^{4}+\frac{105}{16}{{x}_{1}}^{2}-\frac{5}{16})]} \\ & =128/225 \\ \end{align}

$$     (Eq 5.30)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{W}_{2}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{2}}){{P}_{6}}({{x}_{2}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{2}}^{4}-\frac{210}{8}{{x}_{2}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{2}}^{6}-\frac{315}{16}{{x}_{2}}^{4}+\frac{105}{16}{{x}_{2}}^{2}-\frac{5}{16})]} \\ & =(322-13\sqrt{70})/900 \\ \end{align}

$$     (Eq 5.31)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{W}_{3}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{3}}){{P}_{6}}({{x}_{3}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{3}}^{4}-\frac{210}{8}{{x}_{3}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{3}}^{6}-\frac{315}{16}{{x}_{3}}^{4}+\frac{105}{16}{{x}_{3}}^{2}-\frac{5}{16})]} \\ & =(322-13\sqrt{70})/900 \\ \end{align}

$$     (Eq 5.32)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{W}_{4}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{4}}){{P}_{6}}({{x}_{4}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{4}}^{4}-\frac{210}{8}{{x}_{4}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{4}}^{6}-\frac{315}{16}{{x}_{4}}^{4}+\frac{105}{16}{{x}_{4}}^{2}-\frac{5}{16})]} \\ & =(322+13\sqrt{70})/900 \\ \end{align}

$$     (Eq 5.33 )
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{W}_{5}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{5}}){{P}_{6}}({{x}_{5}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{5}}^{4}-\frac{210}{8}{{x}_{5}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{5}}^{6}-\frac{315}{16}{{x}_{5}}^{4}+\frac{105}{16}{{x}_{5}}^{2}-\frac{5}{16})]} \\ & =(322+13\sqrt{70})/900 \\ \end{align}

$$     (Eq 5.34)
 * <p style="text-align:right">
 * }

Problem solved by Shengfeng Yang.

=Problem 2.6: Verifying Legendre Polynomials=

Refer to lecture slide [[media:nm1.s11.mtg8.djvu|mtg-8]] for the problem statement.

Given
Legendre polynomial is given by:


 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{P}_{n}}(x)=\sum\limits_{i=0}^{[n/2]}\frac{(2n-2i)!{{x}^{n-2i}}}{{{2}^{n}}i!(n-2i)!(n-i)!}

$$     (Eq 6.1)
 * <p style="text-align:right">
 * }

Where [ n/2 ] denotes the integer part of n/2, e.g. n=5, [ n/2 ] =2.
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{P}_{0}}(x)=1

$$
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{P}_{1}}(x)=x

$$
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{P}_{2}}(x)=\frac{1}{2}(3{{x}^{2}}-1)

$$
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{P}_{3}}(x)=\frac{1}{2}(5{{x}^{3}}-3x)

$$
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{P_4}(x) = \frac{8}{x^4} - \frac{4}{x^2} + \frac{3}{8}

$$
 * <p style="text-align:right">
 * }

Objectives
Verify P0(x) – P4(x) using the Legendre polynomial.

Solutions
1). When n=0, [ n/2 ] =0, the Equation (6-1) becomes:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{P}_{0}}(x)={{(-1)}^{0}}\frac{0!{{x}^{0}}}{{{2}^{0}}\centerdot 0!\centerdot 0!}=1

$$     (Eq 6.2 ) 2). When n=1, [ n/2 ] =0, the Equation (6-1) becomes:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{P}_{1}}(x)={{(-1)}^{0}}\frac{2!{{x}^{1}}}{{{2}^{1}}\centerdot 0!\centerdot 0!}=x

$$     (Eq 6.3) 3). When n=2, [ n/2 ] =1, the Equation (6-1) becomes:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{P}_{2}}(x)={{(-1)}^{0}}\frac{4!{{x}^{2}}}{{{2}^{2}}\centerdot 2!\centerdot 2!}+{{(-1)}^{1}}\frac{2!{{x}^{0}}}{{{2}^{2}}\centerdot 1!\centerdot 0!}=\frac{1}{2}(3{{x}^{2}}-1)

$$     (Eq 6.4) 4). When n=3, [ n/2 ] =1, the Equation (6-1) becomes:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{P}_{3}}(x)={{(-1)}^{0}}\frac{6!{{x}^{3}}}{{{2}^{3}}\centerdot 3!\centerdot 3!}+{{(-1)}^{1}}\frac{4!{{x}^{0}}}{{{2}^{3}}\centerdot 1!\centerdot 1!\centerdot 2!}=\frac{1}{2}(5{{x}^{3}}-3x)

$$     (Eq 6.5) 5). When n=4, [ n/2 ] =2, the Equation (6-1) becomes:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{P}_{4}}(x)={{(-1)}^{0}}\frac{8!{{x}^{4}}}{{{2}^{4}}\centerdot 4!\centerdot 4!}+{{(-1)}^{1}}\frac{6!{{x}^{2}}}{{{2}^{4}}\centerdot 1!\centerdot 2!\centerdot 3!}+{{(-1)}^{2}}\frac{4!{{x}^{0}}}{{{2}^{4}}\centerdot 2!\centerdot 0!\centerdot 2!}=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8}

$$     (Eq 6.6)
 * <p style="text-align:right">
 * }

Problem solved by Shengfeng Yang.

=Problem 2.7: Constructing the Gram Matrix= Refer to lecture slide [[media:nm1.s11.mtg9.djvu|9-1]] for the problem statement.

Given
Legendre polynomials: Where $$[\frac{n}{2}]=$$ integer part of $$\frac{n}{2}$$.

Objectives
Construct (Gram matrix) $$\underline{\Gamma }$$; show $$\underline{\Gamma }$$ is diagonal or not; and compute $$ \det (\underline{\Gamma })$$.

Construct (Gram matrix) $$\underline{\Gamma }$$
By definition of Legendre polynomials: We can obtain: For $$n=0$$, For $$n=1$$, For $$n=2$$, For $$n=3$$, For $$n=4$$, For $$n=5$$, By definition of Gram matrix, we have

Hence, Check Result: Wolframe|Alpha. Check Result: Wolframe|Alpha. Check Result: Wolframe|Alpha. Check Result: Wolframe|Alpha. For $$[a,b]=[-1,1]$$, we have

compute $$ \det (\underline{\Gamma })$$
Problem solved by Hailong Chen.

Given
Newton-Cotes formula: WhereLagrange interpolating functions and $$f(x)$$ is the original function to be interpolated, with $$x\in [a,b]$$. Simple trapezoidal rule: Where $$f(x)$$ is the original function to be interpolated, with $$x\in [a,b]$$.

Objectives
Show when $$n=1$$, Newton-Cotes method implies Simple trapezoidal rule.

Solution
By definition of Lagrange interpolating functions, we can obtain: For $$n=1 $$,$$ {{x}_{0}}=a$$,$$ {{x}_{1}}=b$$ Hence, by definition of Newton-Cotes formula, we consequently have Thus, We get Simple trapezoidal rule from Newton-Cotes method when $$n=1$$. Problem solved by Hailong Chen.

Given

 * {| style="width:70%" border="0" align="center"



(9.1)
 * $$\displaystyle P_n (x) = \sum_{i=0}^{[n/2]} (-1)^i \frac{(2n-2i)! x^{n-2i}}{2^n i! (n-i)! (n-2i)!}$$
 * <p style="text-align:right">
 * }
 * }

Objectives
a.) Use 9.1 to generate P5(x) and compute the roots of P5(x) to check values in the table on lecture slide [[media:nm1.s11.mtg8.djvu|7-5]]. Plot the roots on [-1,1].

b.) Repear steps in a. for P10(x). Observe the location of the roots near endpoints -1 and +1.

Solution
a.)Using Wolfram Alpha to compute P5(x) we get:


 * {| style="width:70%" border="0" align="center"



(9.2)
 * $$\displaystyle P_5 (x) = \frac{63 x^5}{8} - \frac{35 x^3}{4} + \frac{15 x}{8}$$
 * <p style="text-align:right">
 * }
 * }

Matlab code to produce the roots of P5(x) and the plot of the roots:

>> C = [63/8 0 -35/4 0 15/8 0]; >> x = roots(C)

x =

0 -0.906179845938664  -0.538469310105683   0.906179845938664   0.538469310105683 >> y = [0;0;0;0;0]; >> plot(x,y,'.','markersize',15)

These roots compare well with the values in the table in lecture [[media:nm1.s11.mtg8.djvu|7-5]].


 * {| style="width:70%" border="0" align="center"



(9.3)
 * $$\displaystyle 0 = 0$$
 * <p style="text-align:right">
 * }
 * }


 * {| style="width:70%" border="0" align="center"



(9.4)
 * $$\displaystyle \pm0.538469310105683 \approx \pm \frac{1}{3} \sqrt{5 - 2\sqrt{10/7}}$$
 * <p style="text-align:right">
 * }
 * }


 * {| style="width:70%" border="0" align="center"



(9.5)
 * $$\displaystyle \pm0.906179845938664 \approx \pm \frac{1}{3} \sqrt{5 - 2\sqrt{10/7}}$$
 * <p style="text-align:right">
 * }
 * }

Below is a plot of the roots of P5(x):



b.)Using Wolfram Alpha to compute P10(x) we get:


 * {| style="width:70%" border="0" align="center"



(9.6)
 * $$\displaystyle P_{10} (x) = \frac{46189 x^{10}}{256} - \frac{109395 x^8}{256} + \frac{45045 x^6}{128} - \frac{15015 x^4}{128} + \frac{3465 x^2}{256} - \frac{63}{256}$$
 * <p style="text-align:right">
 * }
 * }

Matlab code to produce the roots of P10(x) and the plot of the roots:

>> C = [46189/256 0 -109395/256 0 45045/128 0 -15015/128 0 3465/256 0 -63/256]; >> x = roots(C)

x =

-0.973906528517167 -0.865063366688989  -0.679409568299026   0.973906528517167   0.865063366688989   0.679409568299024  -0.433395394129247   0.433395394129247  -0.148874338981631   0.148874338981631

>> y = [0;0;0;0;0;0;0;0;0;0]; >> plot(x,y,'.','markersize',15)

Below is a plot of the roots of P10(x):



We note that the roots are located disproportionately near the edges of our integration interval. Runge's Phenomenon is defined as a problem of oscillation near the end points of an interval. The more roots we have near the edges of our interval, the more severe this problem will be.

Problem solved by Erle Fields.

=Problem 2.10: Expression for {Ci}=

Refer to lecture slides [[media:nm1.s11.mtg10.djvu|10-2 and 10-4]] for the problem statement.

Given
Equations (2) and (3) on lecture slide [[media:nm1.s11.mtg10.djvu|10-2]]:


 * {| style="width:70%" border="0" align="center"



(10.1)
 * $$\displaystyle P_2 (x) = c_2x^2 + c_1x + c_0 $$
 * <p style="text-align:right">
 * }
 * }


 * {| style="width:70%" border="0" align="center"



(10.2)
 * $$\displaystyle P_2 (x_i) = f(x_i) \, for \, i = 0, 1, 2 $$
 * <p style="text-align:right">
 * }
 * }

Objectives
Use equations (10.1) and (10.2) to find an expression for {Ci} in terms of (xi,f(xi)) for i = 0, 1, 2.

Solution
From equation (4) on lecture slide [[media:nm1.s11.mtg10.djvu|10-2]] we know:


 * {| style="width:70%" border="0" align="center"



(10.3)
 * $$\displaystyle P_2 (x) = \sum_{i=0}^{n=2}l_{i,2}(x)f(x_i) $$
 * <p style="text-align:right">
 * }
 * }

Where, from equation (1) on lecture slide [[media:nm1.s11.mtg9.djvu|9-2]]


 * {| style="width:70%" border="0" align="center"



(10.4)
 * $$\displaystyle l_{i,2}(x) = \prod_{j=0,j \ne i}^{n=2}\frac{x-x_j}{x_i-x_j} $$
 * <p style="text-align:right">
 * }
 * }

Now we solve for $$ l_{0,2}(x),$$  $$l_{1,2}(x)$$  and  $$l_{2,2}(x) $$


 * {| style="width:70%" border="0" align="center"



(10.5)
 * $$\displaystyle l_{0,2}(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} $$
 * <p style="text-align:right">
 * }
 * }


 * {| style="width:70%" border="0" align="center"



(10.6)
 * $$\displaystyle l_{1,2}(x) = \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} $$
 * <p style="text-align:right">
 * }
 * }


 * {| style="width:70%" border="0" align="center"



(10.7)
 * $$\displaystyle l_{2,2}(x) = \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} $$
 * <p style="text-align:right">
 * }
 * }

Using equation (10.3) we get:


 * {| style="width:70%" border="0" align="center"



(10.8)
 * $$\displaystyle P_2 (x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}f(x_0) + \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}f(x_1) + \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}f(x_2) $$
 * <p style="text-align:right">
 * }
 * }

Combining terms of $$ x^2 $$  and  $$ x $$ we get:


 * {| style="width:70%" border="0" align="center"



(10.9)
 * $$\displaystyle P_2 (x) = \Big(\frac{f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{f(x_2)}{(x_2-x_0)(x_2-x_1)}\Big)x^2 $$
 * <p style="text-align:right">
 * }
 * }


 * {| style="width:70%" border="0" align="center"




 * $$\displaystyle - \Big(\frac{(x_1+x_2)f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{(x_0+x_2)f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{(x_0+x_1)f(x_2)}{(x_2-x_0)(x_2-x_1)}\Big)x $$


 * }
 * }


 * {| style="width:70%" border="0" align="center"




 * $$\displaystyle + \frac{(x_1x_2)f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{(x_0x_2)f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{(x_0x_1)f(x_2)}{(x_2-x_0)(x_2-x_1)} $$


 * }
 * }

After setting equation (10.1) equal to (10.9) we get:


 * {| style="width:70%" border="0" align="center"



(10.10)
 * $$\displaystyle C_2 = \frac{f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{f(x_2)}{(x_2-x_0)(x_2-x_1)} $$
 * <p style="text-align:right">
 * }
 * }


 * {| style="width:70%" border="0" align="center"



(10.11)
 * $$\displaystyle C_1 = - \Big(\frac{(x_1+x_2)f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{(x_0+x_2)f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{(x_0+x_1)f(x_2)}{(x_2-x_0)(x_2-x_1)}\Big) $$
 * <p style="text-align:right">
 * }
 * }


 * {| style="width:70%" border="0" align="center"



(10.12)
 * $$\displaystyle C_0 = \frac{(x_1x_2)f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{(x_0x_2)f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{(x_0x_1)f(x_2)}{(x_2-x_0)(x_2-x_1)} $$
 * <p style="text-align:right">
 * }
 * }

Problem solved by Erle Fields.

Given
Equation (4) on lecture slide [[media:nm1.s11.mtg10.djvu|10-2]]:


 * {| style="width:70%" border="0" align="center"



(11.1)
 * $$\displaystyle P_2 (x) = \sum_{i=0}^{2} l_{i,2}(x) f(x_i)$$
 * <p style="text-align:right">
 * }
 * }

Simple Simpson's Rule given in equations (2) and (3) on lecture slide [[media:nm1.s11.mtg7.djvu|7-4]]:


 * {| style="width:70%" border="0" align="center"



(11.2)
 * $$\displaystyle \int_a^bf(x)dx \approx I_2 = \int_a^bP_2(x)dx = \frac{h}{3}[f(x_0) + 4f(x_1) +f(x_2)]$$
 * <p style="text-align:right">
 * }
 * }

Where:
 * {| style="width:70%" border="0" align="center"



(11.3)
 * $$\displaystyle h = \frac{b-a}{2}$$
 * <p style="text-align:right">
 * }
 * }

And:
 * {| style="width:70%" border="0" align="center"




 * $$\displaystyle x_0 = a$$


 * }
 * }


 * {| style="width:70%" border="0" align="center"




 * $$\displaystyle x_1 = \frac{a+b}{2}$$


 * }
 * }


 * {| style="width:70%" border="0" align="center"




 * $$\displaystyle x_2 = b$$


 * }
 * }

Objectives
Use equation (11.1) to derive the Simple Simpson's Rule (11.2).

Solution
From equation (2) on lecture slide [[media:nm1.s11.mtg9.djvu|9-2]]:


 * {| style="width:70%" border="0" align="center"



(11.4)
 * $$\displaystyle \int_a^bP_2(x)dx = \sum_{i=0}^{2} \int_a^bl_{i,2}(x)dx f(x_i)$$
 * <p style="text-align:right">
 * }
 * }

From equation (1) on lecture slide [[media:nm1.s11.mtg9.djvu|9-2]]:


 * {| style="width:70%" border="0" align="center"



(11.5)
 * $$\displaystyle l_{i,2}(x) = \prod_{j=0,j \ne i}^{n=2}\frac{x-x_j}{x_i-x_j} $$
 * <p style="text-align:right">
 * }
 * }

Now we solve for $$ l_{0,2}(x),$$  $$l_{1,2}(x)$$  and  $$l_{2,2}(x) $$


 * {| style="width:70%" border="0" align="center"



(11.6)
 * $$\displaystyle l_{0,2}(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} $$
 * <p style="text-align:right">
 * }
 * }


 * {| style="width:70%" border="0" align="center"



(11.7)
 * $$\displaystyle l_{1,2}(x) = \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} $$
 * <p style="text-align:right">
 * }
 * }


 * {| style="width:70%" border="0" align="center"



(11.8)
 * $$\displaystyle l_{2,2}(x) = \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} $$
 * <p style="text-align:right">
 * }
 * }

Using Wolfram Alpha to solve the following integrals we get:

WA


 * {| style="width:70%" border="0" align="center"



(11.9)
 * $$\displaystyle \int_a^b l_{0,2}(x)dx = \int_{a}^{b} \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}dx = \frac{b-a}{6}$$
 * <p style="text-align:right">
 * }
 * }

WA


 * {| style="width:70%" border="0" align="center"



(11.10)
 * $$\displaystyle \int_a^b l_{1,2}(x)dx = \int_{a}^{b} \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}dx = -\frac{2}{3}(a-b)$$
 * <p style="text-align:right">
 * }
 * }

WA


 * {| style="width:70%" border="0" align="center"



(11.11)
 * $$\displaystyle \int_a^b l_{2,2}(x)dx = \int_{a}^{b} \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}dx = \frac{b-a}{6}$$
 * <p style="text-align:right">
 * }
 * }

Using Equation (11.4) and the results of equations (11.9), (11.10) and (11.11) we get:


 * {| style="width:70%" border="0" align="center"



(11.12)
 * $$\displaystyle \int_a^bP_2(x)dx = \frac{b-a}{6}f(x_0) + \frac{2}{3}(b-a)f(x_1) + \frac{b-a}{6}f(x_2)$$
 * <p style="text-align:right">
 * }
 * }

Simplifying equation (11.12) leaves us with the Simple Simpson's Rule:


 * {| style="width:70%" border="0" align="center"



(11.13)
 * $$\displaystyle I_2 = \int_a^bP_2(x)dx = \frac{b-a}{6}[f(x_0) + 4f(x_1) +f(x_2)] = \frac{h}{3}[f(x_0) + 4f(x_1) +f(x_2)]$$
 * <p style="text-align:right">
 * }
 * }

Where:
 * {| style="width:70%" border="0" align="center"




 * $$\displaystyle h = \frac{b-a}{2}$$


 * }
 * }

Problem solved by Erle Fields.

=Problem 2.12: Estimation by Lagrange Interpolation Functions=

Given
$$ f(x) = \frac{e^x-1}{x}, [-1,1], x_0 = -1, x_n = 1, n= 1,2,4,8,16 $$

Objectives
2-12-1) Construct

2-12-2) Plot

Solution
2-12-1) Plot

Matlab Code:



2-12-2) Plot Matlab Code:



Given
Trapezoidal rule: Simple rule: Composite rule: Where $$f(x)$$ is the original function to be interpolated, with $$x\in [a,b]$$. $$a={{x}_{0}}<{{x}_{1}}<\cdots <{{x}_{n-1}}<{{x}_{n}}=b$$. The distance among consecutive nodes is $$h=\frac{b-a}{n}$$. Simpson’s rule: Simple rule: Composite rule: Where $$f(x)$$ is the original function to be interpolated, with $$x\in [a,b]$$. >. $$a={{x}_{0}}<{{x}_{1}}<\cdots <{{x}_{n-1}}<{{x}_{n}}=b$$. The distance among consecutive nodes is $$h=\frac{b-a}{n}$$.

Objectives
Derive each composite rule from the simple rule.

Trapezoidal rule
Divide the interval into $$n$$ subintervals with equal length, hence $$h=\frac{b-a}{n}$$. Break the integral into $$n$$ subintegrals, Approximate each subintegrals using simple rule, then Hence, we get the composite trapezoidal rule.

Simpson’s rule
Using Newton-Cotes formula, we can get simple Simpson’s rule as follow: Where $$ c=\frac{b+a}{2}$$. In order to evaluate above integral directly, we use $$h=\frac{b-a}{2}$$ to make it  easy to solve. Then The complete evaluation of equation $$4$$ yields Like what we do in deriving composite trapezoidal rule, dividing interval $$[a,b]$$ into $$ n$$ subintervals, each containing three nodes. Thus Approximating each subintegrals by equation $$8$$ yields Hence, we get the composite Simpson’s rule. Problem solved by Hailong Chen.

Objective
Give details of how to obtain (14.1)

Solution
Let $$\displaystyle x=x(y)$$ be a function of y that satisfies (14.1). To satisfy the boundary conditions x(y) must satisfy the following conditions:

The solution to these conditions lies upon a line passing through the two ordered pairs.

Taking the partial derivative of each side yields an expression for dx:

Plugging these expression back into the middle term of (14.1) yields

Now, let

Combining equations (14.6) and (14.7) proves (14.1):


 * {| style="width:70%" border="0" align="center"

$$\displaystyle \therefore I(f) = \int_{a}^{b}f(x)dx = \int_{-1}^{1}\bar{f}(y)dy $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle
 * style= |
 * }
 * }

Solved by William Kurth and Brendan Mahon.