Period 1

Question 1

How to calculate f(x) when close to 0?

Answer 1

Usually this can be solved by using the Taylor polynomial, or by L’Hopital.

Question 2

How to explain miscalculations for x is close to 0?

Answer 2

First show what the actual value is
Then show that the value is below 2^-53 or 2^-52 (if 1 + x)

Question 3

How to explain outliers when x close to 0?

Answer 3

Usually you need to check for 1 - 2^-53

Question 4

How to show an error in a magnitude?

Answer 4

O(x^n), this is important

Question 5

How to proof a unique root?

Answer 5

First show that a and b have a different sign.
Then show that f(x) is continuous.
Then show that f(x) is monotone (i.e. increasing or decreasing) by its derivative.

Question 6

What is Newtons formula for root finding?

Answer 6

x^(n+1) = x^n - f(x^n)/f’(x^n)

Question 7

When Newtons algorithm converge monotone?

Answer 7

If the function that is approximated is monotone convex, or monotone concave.

Question 8

What is the error formula for the bisection method?

Answer 8

n = floor(log_2((b-a)/epsilon))

Question 9

What is the formula for the False Position method?

Answer 9

c = (af(b) - bf(a))/(f(b)-f(a))

Question 10

What are advantages and disadvantages of the Bisection method?

Answer 10

+ Works always
+ Bounding interval
+ No derivatives
+ Error estimation
- Slow convergence
- No generalization in higher dimensions

Question 11

How to find the interpolating polynomial using a system of linear equation?

Answer 11

Create for every entry
p_0 + p_1 x + p_2 x^2 + p_3 x^3 + …. = y

Solve for p

Question 12

What is are the advantages and disadvantages of polynomials and splines?

Answer 12

Polynomials are just 1 function
The values are sensitive for small changes (ill-conditioned)
Big oscillations at the end of the interval (Runge-effect).

Splines usually give a better fit of the data.

Question 13

How to proof that you get a unique polynomial?

Answer 13

Proof by contradiction:
Let two polynomials of degree n p(x), q(x), with p(x) != q(x)
Create r(x) = p(x) - q(x), a polynomial of degree n
Because p(x) and q(x) go through the points, r(x) = 0 n + 1 times
However a polynomial of degree n at most n roots. So contradiction.

Question 14

How to get interpolating polynomial using Lagrange?

Answer 14

p(x) = y_0 * (x - x_1)(x-x_2)…/(x_0 - x_1)(x_0 - x_2)…
+ y_1 * (x-x_0)(x-x_2)…/(x_1-x_0)(x_1/x_2)…
+ …

Question 15

What is the formula of Newtons Divided Differences?

Answer 15

Table: (f(bottom) - f(top))/(x_bottom - x_top)
Final formula: y_0 + f(., .)(x - x_0) + f(., ., .)(x-x_0)(x-x_1) + ….

Question 16

What are the advantages and disadvantages of the Vandermonde matrix?

Answer 16

• needs solving a system of equations
• is illconditioned
  + gives explicitly the coefficients
  + very efficient evaluation

Question 17

What are the advantages and disadvantages of the Lagrange form?

Answer 17

+ no preprocessing
+ gives immediately a formula of the polynomial
- needs more computation for evaluation, which makes it less useful for many points

Question 18

What are advantages of Newtons Divided Differences?

Answer 18

• need to preprocess
  + Gives one line formula, which is useful for many x
  + Adding new datapoints leads to computing just a single new row

Question 19

How to interpolate f(x) using a polynomial?

Answer 19

Just create n points on the line within the interval, then let p(x) go trough those points.

Question 20

How to interpolate f(x) using a spline?

Answer 20

Just create n point on the function within the interval, then create a spline

Question 21

How to create a spline?

Answer 21

Create the following system of equations:
S_i(x) = a_i + b_i x + c_i x^2 + d_i x^3 (to degree n)
S’_i(x) = b_i + 2c_i x = 3d_i x^2
… to the n-1 derivative.

Question 22

What is the definition of a spline?

Answer 22

Spline S: R -> R is a cubic spline iff.

1. S is n-1 times differentiable
2. S, S’, S’’, .., S^(n-1) are continuous
3. S interpolates the data
4. S is at most of degree n at every interval
5. S’‘(x_0)=S’‘(X_n) = 0

Question 23

How to find the number of points needed for interpolating with a spline?

Answer 23

1. Use the formula on the test, however use f^(n+1)(x) with n is the degree of the spline.
2. Use for the product the other formula, with n the degree of the spline
3. Write this formula smaller then the error
4. Solve for n

Question 24

How to find the number of points needed for interpolating with a polynomial?

Answer 24

1. Use the formula on the test, rewrite f^(n+1)(x) to something that you know is bigger.
2. Use for the product the other formula on the test
3. Write this formula smaller then the error
4. Solve for n

Question 25

How to get the formula of a Gaussian integral?

Answer 25

Create the following system of equations:
A_0 + A_1 + A_2 + … + A_n = (b-a)/1
A_0 x_0 + A_1 x_1 + A_2 x_ 2+ … + A_n x_n = (b-a)^2/2
A_0 x_0^2+ A_1 x_1^2 + A_2 x_2^2 + … + A_n x_n^2 = (b-a)^3/3
…
Every A represents a size before f(x)

Question 26

What is the formula of an integral created using the Midpoint rule?

Answer 26

h * Sum from 0 to n -1 of f((x_i + x_(i+1))/2).
I.e. h*(f(x_0.5) + f(x_1.5) + f(x_2.5) + …)

Question 27

What is the formula of an integral created using the trapezium rule?

Answer 27

0.5h * (f(x_0) + 2 f(x_1) + 2 f(x_2) + … + f(x_n))

Question 28

What is the formula of an integral created using Simpsons rule?

Answer 28

i.e. h/3 * (f(a) + 4 * sum^(n/4)_(i=1) f(a + (2i-1)h) + 2 * sum^(n/2 -1)_(i=1) f(a +2ih) + f(b).
h/3 * (f(x_0) + 4 f(x_1) + 2 f(x_2) + 4 f(x_3) + … f(x_n))

Question 29

How to compute a Gaussian integral using a given formula?

Answer 29

1. Create a small formula to rewrite a and b into a and b given in the integral e.g. (y = 1/10 x -1), rewrite this to x = ..
2. Fill in the formula in the integral with f(x)
3. Calculate using the given formula.

Question 30

What is the algorithm for the bisection method?

Answer 30

1. Get a, and b
2. Find c = (a+b)/2
3. Check sign(a) == sign(c), if so then c = a_(n+1)
   else c = b_(n+1)
4. Repeat until interval is small enough or f(c) < error_2

Question 31

What are advantages and disadvantages of the False Position method?

Answer 31

+ Convergence guaranteed
+ Faster convergence then bisection
- More calculations per iteration
- No calculation for the speed of the convergence

Question 32

What are advantages and disadvantages of the Newtons Method for root finding?

Answer 32

• does not always work
• no interval given
  + fast convergence
  + generalization in higher dimensions
  + has error estimation
  + just 1 initial estimate needed

Question 33

What is the Secant method?

Answer 33

Same method as Newtons method, however this uses an estimation for the derivative.
x_(n+1) = x_n - (x_n - x_(n-1))/(f(x_n) - f(x_(n-1))

Question 34

How to get a Polynomial using the Vandermonde Matrix?

Answer 34

Create the following system of equations:
p_0 + p_1 x + p_2 x^2 + p_3 x^3 + … = y
for every data point.

Question 35

How to get the number of Intervals for an integral within an error?

Answer 35

1. Use the formula given by the test to get maximum error for one.
2. Apply this for all.
3. Rewrite this to n > …

Question 36

What does it mean for an integration rule to be more efficient?

Answer 36

For the same partitions, the error of rule a is smaller then rule b. Or for a certain error you need fewer intervals.

Question 37

What is the algorithm of the False Position method?

Answer 37

1. Get a and b
2. Draw a line between trough (a, f(a)) and (b, f(b))
3. Crosses x-axis at c in (a, b)
4. Check sign of f(c), then choose new a and b
5. Repeat until interval is small enough, or f(c) < error_2

Question 38

What is the formula of the Gaussian rule?

Answer 38

G_n(f) = A_0 f(x_0) + A_1 f(x_1) + … + A_n f(x_n)

Question 39

How many degrees of polynomial is a Gaussian integral accurate?

Answer 39

The number of nodes + number of weights - 2, is the number of unknown variables. This -1.