Dtjytyk

What is the term when two people sing in harmony, but they aren't singing the same notes?

Who must act to prevent Brexit on March 29th?

How will losing mobility of one hand affect my career as a programmer?

Superhero words!

How to check participants in at events?

Partial sums of primes

Invariance of results when scaling explanatory variables in logistic regression, is there a proof?

Why are all the doors on Ferenginar (the Ferengi home world) far shorter than the average Ferengi?

Would it be legal for a US State to ban exports of a natural resource?

A workplace installs custom certificates on personal devices, can this be used to decrypt HTTPS traffic?

Proof of Lemma: Every integer can be written as a product of primes

How do ultrasonic sensors differentiate between transmitted and received signals?

What should I use for Mishna study?

Bob has never been a M before

Calculating the number of days between 2 dates in Excel

Resetting two CD4017 counters simultaneously, only one resets

In Star Trek IV, why did the Bounty go back to a time when whales were already rare?

I2C signal and power over long range (10meter cable)

Latex for-and in equation

node command while defining a coordinate in TikZ

What does the "3am" section means in manpages?

Science Fiction story where a man invents a machine that can help him watch history unfold

What (else) happened July 1st 1858 in London?

Teaching indefinite integrals that require special-casing

Integrating lagrange polynomial with equispaced points

For n=3 Lagrange interpolation why is it equal to 1?Lagrange interpolating polynomials question?Lagrange Interpolation definition doubtLagrange InterpolationHow to obtain Lagrange interpolation formula from Vandermonde's determinantLagrange interpolation: Evaluation of error in interpolationVery confused with interpolating polynomialsFind the largest value for $x_1$ in (0,1) such that $f(0.5)-P_2(0.5) = -0.25$ (interpolation)Lagrange interpolation for ellipseFind the polynomial of at most 1 degree using Lagrange Polynomial

0

$begingroup$

Suppose we have some second order polynomial interpolant, $P_2$, defined on the equispaced points $x_0, x_1, x_2$, such that $x_{j+1}-x_j=h$. From $P_2$, we have Lagrange polynomials, $L_0, L_1, L_2$.

Suppose we want to integrate $L_0$ between $x_0$ and $x_2$. That would be,

$$int_{x_0}^{x_2}L_0(x)dx = frac{h}{3}$$

According to my professor, that is. However, I'm struggling to produce this result, so I'm hoping someone can help? Here is what I've done so far:

$$int_{x_0}^{x_2}L_0(x)dx=int_{x_0}^{x_2}frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}dx=int_{x_0}^{x_2}frac{(x-x_1)(x-x_2)}{(-h)(-2h)}dx$$

$$=frac{1}{2h^2}int_{x_0}^{x_2}(x-x_1)(x-x_2)dx=frac{1}{2h^2}bigg[int_{x_0}^{x_2}x^2dx-(x_2+x_1)int_{x_0}^{x_2}xdx+x_{1}x_{2}int_{x_0}^{x_2}dxbigg]$$

$$=frac{1}{2h^2}bigg[ frac{x_2^3-x_0^3}{3} - frac{(x_2+x_1)(x_2^2-x_0^2)}{2}+x_{1}x_{2}(x_2-x_0)bigg]=frac{1}{2h^2}bigg[ frac{x_2^3-x_0^3}{3} - frac{(x_2+x_1)(x_2^2-x_0^2)}{2}+2hx_{1}x_{2})bigg]$$

But this is a lot of messy algebra and I can't seem to tidy it up to the simple result of $h/3$. My professor said this would be a simple exercise, so it makes me think I am missing something that would make the whole thing a lot easier?

---

Edit: I realised you can do u-substitution. See here for solution: http://www.cs.uleth.ca/~holzmann/notes/simpsons.pdf

integration interpolation lagrange-interpolation simpsons-rule

share|cite|improve this question

edited Mar 14 at 22:16

HumptyDumpty

asked Mar 14 at 20:36

HumptyDumptyHumptyDumpty

358110

$endgroup$

add a comment | 

0

$begingroup$

Suppose we have some second order polynomial interpolant, $P_2$, defined on the equispaced points $x_0, x_1, x_2$, such that $x_{j+1}-x_j=h$. From $P_2$, we have Lagrange polynomials, $L_0, L_1, L_2$.

Suppose we want to integrate $L_0$ between $x_0$ and $x_2$. That would be,

$$int_{x_0}^{x_2}L_0(x)dx = frac{h}{3}$$

According to my professor, that is. However, I'm struggling to produce this result, so I'm hoping someone can help? Here is what I've done so far:

$$int_{x_0}^{x_2}L_0(x)dx=int_{x_0}^{x_2}frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}dx=int_{x_0}^{x_2}frac{(x-x_1)(x-x_2)}{(-h)(-2h)}dx$$

$$=frac{1}{2h^2}int_{x_0}^{x_2}(x-x_1)(x-x_2)dx=frac{1}{2h^2}bigg[int_{x_0}^{x_2}x^2dx-(x_2+x_1)int_{x_0}^{x_2}xdx+x_{1}x_{2}int_{x_0}^{x_2}dxbigg]$$

$$=frac{1}{2h^2}bigg[ frac{x_2^3-x_0^3}{3} - frac{(x_2+x_1)(x_2^2-x_0^2)}{2}+x_{1}x_{2}(x_2-x_0)bigg]=frac{1}{2h^2}bigg[ frac{x_2^3-x_0^3}{3} - frac{(x_2+x_1)(x_2^2-x_0^2)}{2}+2hx_{1}x_{2})bigg]$$

But this is a lot of messy algebra and I can't seem to tidy it up to the simple result of $h/3$. My professor said this would be a simple exercise, so it makes me think I am missing something that would make the whole thing a lot easier?

---

Edit: I realised you can do u-substitution. See here for solution: http://www.cs.uleth.ca/~holzmann/notes/simpsons.pdf

integration interpolation lagrange-interpolation simpsons-rule

share|cite|improve this question

edited Mar 14 at 22:16

HumptyDumpty

asked Mar 14 at 20:36

HumptyDumptyHumptyDumpty

358110

$endgroup$

add a comment | 

$begingroup$

Suppose we have some second order polynomial interpolant, $P_2$, defined on the equispaced points $x_0, x_1, x_2$, such that $x_{j+1}-x_j=h$. From $P_2$, we have Lagrange polynomials, $L_0, L_1, L_2$.

Suppose we want to integrate $L_0$ between $x_0$ and $x_2$. That would be,

$$int_{x_0}^{x_2}L_0(x)dx = frac{h}{3}$$

According to my professor, that is. However, I'm struggling to produce this result, so I'm hoping someone can help? Here is what I've done so far:

$$int_{x_0}^{x_2}L_0(x)dx=int_{x_0}^{x_2}frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}dx=int_{x_0}^{x_2}frac{(x-x_1)(x-x_2)}{(-h)(-2h)}dx$$

$$=frac{1}{2h^2}int_{x_0}^{x_2}(x-x_1)(x-x_2)dx=frac{1}{2h^2}bigg[int_{x_0}^{x_2}x^2dx-(x_2+x_1)int_{x_0}^{x_2}xdx+x_{1}x_{2}int_{x_0}^{x_2}dxbigg]$$

$$=frac{1}{2h^2}bigg[ frac{x_2^3-x_0^3}{3} - frac{(x_2+x_1)(x_2^2-x_0^2)}{2}+x_{1}x_{2}(x_2-x_0)bigg]=frac{1}{2h^2}bigg[ frac{x_2^3-x_0^3}{3} - frac{(x_2+x_1)(x_2^2-x_0^2)}{2}+2hx_{1}x_{2})bigg]$$

But this is a lot of messy algebra and I can't seem to tidy it up to the simple result of $h/3$. My professor said this would be a simple exercise, so it makes me think I am missing something that would make the whole thing a lot easier?

---

Edit: I realised you can do u-substitution. See here for solution: http://www.cs.uleth.ca/~holzmann/notes/simpsons.pdf

integration interpolation lagrange-interpolation simpsons-rule

share|cite|improve this question

edited Mar 14 at 22:16

HumptyDumpty

asked Mar 14 at 20:36

HumptyDumptyHumptyDumpty

358110

$endgroup$

share|cite|improve this question

edited Mar 14 at 22:16

HumptyDumpty

asked Mar 14 at 20:36

HumptyDumptyHumptyDumpty

358110

share|cite|improve this question

edited Mar 14 at 22:16

HumptyDumpty

asked Mar 14 at 20:36

HumptyDumptyHumptyDumpty

358110

asked Mar 14 at 20:36

HumptyDumptyHumptyDumpty

358110

asked Mar 14 at 20:36

HumptyDumptyHumptyDumpty

358110