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4.9E: Exercises - Mathematics

4.9E: Exercises - Mathematics



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Solve Rational Equations

In the following exercises, solve each rational equation.

1. (dfrac{1}{a}+dfrac{2}{5}=dfrac{1}{2})

Answer

(a=10)

2. (dfrac{6}{3}-dfrac{2}{d}=dfrac{4}{9})

3. (dfrac{4}{5}+dfrac{1}{4}=dfrac{2}{v})

Answer

(v=dfrac{40}{21})

4. (dfrac{3}{8}+dfrac{2}{y}=dfrac{1}{4})

5. (1-dfrac{2}{m}=dfrac{8}{m^{2}})

Answer

(m=-2,; m=4)

6. (1+dfrac{4}{n}=dfrac{21}{n^{2}})

7. (1+dfrac{9}{p}=dfrac{-20}{p^{2}})

Answer

(p=-5, ; p=-4)

8. (1-dfrac{7}{q}=dfrac{-6}{q^{2}})

9. (dfrac{5}{3 v-2}=dfrac{7}{4 v})

Answer

(v=14)

10. (dfrac{8}{2 w+1}=dfrac{3}{w})

11. (dfrac{3}{x+4}+dfrac{7}{x-4}=dfrac{8}{x^{2}-16})

Answer

(x=-dfrac{4}{5})

12. (dfrac{5}{y-9}+dfrac{1}{y+9}=dfrac{18}{y^{2}-81})

13. (dfrac{8}{z-10}-dfrac{7}{z+10}=dfrac{5}{z^{2}-100})

Answer

(z=-145)

14. (dfrac{9}{a+11}-dfrac{6}{a-11}=dfrac{6}{a^{2}-121})

15. (dfrac{-10}{q-2}-dfrac{7}{q+4}=1)

Answer

(q=-18, ; q=-1)

16. (dfrac{2}{s+7}-dfrac{3}{s-3}=1)

17. (dfrac{v-10}{v^{2}-5 v+4}=dfrac{3}{v-1}-dfrac{6}{v-4})

Answer

no solution

18. (dfrac{w+8}{w^{2}-11 w+28}=dfrac{5}{w-7}+dfrac{2}{w-4})

19. (dfrac{x-10}{x^{2}+8 x+12}=dfrac{3}{x+2}+dfrac{4}{x+6})

Answer

no solution

20. (dfrac{y-5}{y^{2}-4 y-5}=dfrac{1}{y+1}+dfrac{1}{y-5})

21. (dfrac{b+3}{3 b}+dfrac{b}{24}=dfrac{1}{b})

Answer

(b=-8)

22. (dfrac{c+3}{12 c}+dfrac{c}{36}=dfrac{1}{4 c})

23. (dfrac{d}{d+3}=dfrac{18}{d^{2}-9}+4)

Answer

(d=2)

24. (dfrac{m}{m+5}=dfrac{50}{m^{2}-25}+6)

25. (dfrac{n}{n+2}-3=dfrac{8}{n^{2}-4})

Answer

(m=1)

26. (dfrac{p}{p+7}-8=dfrac{98}{p^{2}-49})

27. (dfrac{q}{3 q-9}-dfrac{3}{4 q+12}=dfrac{7 q^{2}+6 q+63}{24 q^{2}-216})

Answer

no solution

28. (dfrac{r}{3 r-15}-dfrac{1}{4 r+20}=dfrac{3 r^{2}+17 r+40}{12 r^{2}-300})

29. (dfrac{s}{2 s+6}-dfrac{2}{5 s+5}=dfrac{5 s^{2}-3 s-7}{10 s^{2}+40 s+30})

Answer

(s=dfrac{5}{4})

30. (dfrac{t}{6 t-12}-dfrac{5}{2 t+10}=dfrac{t^{2}-23 t+70}{12 t^{2}+36 t-120})

31. (dfrac{2}{x^{2}+2 x-8}-dfrac{1}{x^{2}+9 x+20}=dfrac{4}{x^{2}+3 x-10})

Answer

(x=-dfrac{4}{3})

32. (dfrac{5}{x^{2}+4 x+3}+dfrac{2}{x^{2}+x-6}=dfrac{3}{x^{2}-x-2})

33. (dfrac{3}{x^{2}-5 x-6}+dfrac{3}{x^{2}-7 x+6}=dfrac{6}{x^{2}-1})

Answer

no solution

34. (dfrac{2}{x^{2}+2 x-3}+dfrac{3}{x^{2}+4 x+3}=dfrac{6}{x^{2}-1})

Solve Rational Equations that Involve Functions

35. For rational function, (f(x)=dfrac{x-2}{x^{2}+6 x+8}):

  1. Find the domain of the function
  2. Solve (f(x)=5)
  3. Find the points on the graph at this function value
Answer
  1. The domain is all real numbers except (x eq-2) and (x eq-4)
  2. (x=-3, x=-dfrac{14}{5})
  3. ((-3,5),left(-dfrac{14}{5}, 5 ight))

36. For rational function, (f(x)=dfrac{x+1}{x^{2}-2 x-3}):

  1. Find the domain of the function
  2. Solve (f(x)=1)
  3. Find the points on the graph at this function value

37. For rational function, (f(x)=dfrac{2-x}{x^{2}-7 x+10}):

  1. Find the domain of the function
  2. Solve (f(x)=2)
  3. Find the points on the graph at this function value
Answer
  1. The domain is all real numbers except (x eq 2) and (x eq 5)
  2. (x=dfrac{9}{2})
  3. (left(dfrac{9}{2}, 2 ight))

38. For rational function, (f(x)=dfrac{5-x}{x^{2}+5 x+6}):

  1. Find the domain of the function
  2. Solve (f(x)=3)
  3. Find the points on the graph at this function value

Solve a Rational Equation for a Specific Variable

In the following exercises, solve:

39. (dfrac{c}{r}=2 pi ext { for } r)

Answer

(r=dfrac{C}{2 pi})

40. (dfrac{I}{r}=P ext { for } r)

41. (dfrac{v+3}{w-1}=dfrac{1}{2} ext { for } w)

Answer

(w=2 v+7)

42. (dfrac{x+5}{2-y}=dfrac{4}{3} ext { for } y)

43. (a=dfrac{b+3}{c-2} ext { for } c)

Answer

(c=dfrac{b+3+2 a}{a})

44. (m=dfrac{n}{2-n} ext { for } n)

45. (dfrac{1}{p}+dfrac{2}{q}=4 ext { for } p)

Answer

(p=dfrac{q}{4 q-2})

46. (dfrac{3}{s}+dfrac{1}{t}=2 ext { for } s)

47. (dfrac{2}{v}+dfrac{1}{5}=dfrac{3}{w} ext { for } w)

Answer

(w=dfrac{15 v}{10+v})

48. (dfrac{6}{x}+dfrac{2}{3}=dfrac{1}{y} ext { for } y)

49. (dfrac{m+3}{n-2}=dfrac{4}{5} ext { for } n)

Answer

(n=dfrac{5 m+23}{4})

50. (r=dfrac{s}{3-t} ext { for } t)

51. (dfrac{E}{e}=m^{2} ext { for } c)

Answer

(c=dfrac{E}{m^{2}})

52. (dfrac{R}{T}=W ext { for } T)

53. (dfrac{3}{x}-dfrac{5}{y}=dfrac{1}{4} ext { for } y)

Answer

(y=dfrac{20 x}{12-x})

54. (c=dfrac{2}{a}+dfrac{b}{5} ext { for } a)

Writing Exercises

55. Your class mate is having trouble in this section. Write down the steps you would use to explain how to solve a rational equation.

Answer

Answers will vary.

56. Alek thinks the equation (dfrac{y}{y+6}=dfrac{72}{y^{2}-36}+4) has two solutions, (y=-6) and (y=4). Explain why Alek is wrong.


Hinfsyn

z represents the error outputs to be kept small.

y represents the measurement outputs provided to the controller.

nmeas and ncont are the number of signals in y and u, respectively. y and u are the last outputs and inputs of P , respectively. hinfsyn returns a controller K that stabilizes P and has the same number of states. The closed-loop system CL = lft(P,K) achieves the performance level gamma , which is the H norm of CL (see hinfnorm ).

[ K , CL , gamma ] = hinfsyn( P , nmeas , ncont , gamTry ) calculates a controller for the target performance level gamTry . Specifying gamTry can be useful when the optimal controller performance is better than you need for your application. In that case, a less-than-optimal controller can have smaller gains and be more numerically well-conditioned. If gamTry is not achievable, hinfsyn returns [] for K and CL , and Inf for gamma .

[ K , CL , gamma ] = hinfsyn( P , nmeas , ncont , gamRange ) searches the range gamRange for the best achievable performance. Specify the range with a vector of the form [gmin,gmax] . Limiting the search range can speed up computation by reducing the number of iterations performed by hinfsyn to test different performance levels.

[ K , CL , gamma ] = hinfsyn( ___ , opts ) specifies additional computation options. To create opts , use hinfsynOptions . Specify opts after all other input arguments.

[ K , CL , gamma , info ] = hinfsyn( ___ ) returns a structure containing additional information about the H synthesis computation. You can use this argument with any of the previous syntaxes.


Use Excel for Two-way ANOVA

This exercise is concerned with analysis of variance (ANOVA) in Chapter 10. In particular with the situation when you have two predictor variables, that is two-way ANOVA.

Using Excel for two-way ANOVA (analysis of variance)

Introduction

Excel can carry out the necessary calculations to conduct ANOVA and has several useful functions that can help you:

  • FDIST – calculates an exact p-value for a given F-value (and degrees of freedom).
  • VAR – computes the variance of a series of values.
  • COUNT – counts the number of items, useful for degrees of freedom.
  • AVERAGE – calculates the mean.

However, Excel is most suitable for one-way ANOVA, where you have a single predictor variable. When you have two predictor variables two-way ANOVA is possible, but can be tricky to arrange.

In order to carry out the calculations you need to have your data arranged in a particular layout, let’s call it sample layout or “on the ground” layout. This is not generally a good layout to record your results but it is the only way you can proceed sensibly using Excel. In this exercise you’ll see how to set out your data and have a go at the necessary calculations to perform a two-way ANOVA.

The data for this exercise are available as a file: Two Way online.xlsx. There are two worksheets, one with the bare data and one completed version so you can check your work.

You can use the Analysis ToolPak to carry out the computations for you but you’ll still need to arrange the data in a particular manner. The Analysis ToolPak is not “active” by default so you may need to go to the options/settings and look for Add-Ins.

The exercise data

The data you’ll use for this exercise are in the file Two Way online.xlsx and are presented in the following table:

Exercise data for two-way anova.

Water vulgaris sativa
Lo 9 7
Lo 11 6
Lo 6 5
Mid 14 14
Mid 17 17
Mid 19 15
Hi 28 44
Hi 31 38
Hi 32 37

These data represent the growth of two different plant species under three different watering regimes. The first column shows the levels of the Water variable, this is one of the predictor variables and you can see that there are three levels: Lo, Mid and Hi. The next two columns show the growth results for two plant species, labelled vulgaris and sativa. These two columns form the second predictor variable (we’ll call that Plant, which seems suitable).

This layout is the only way that Excel can deal with the values but it is not necessarily the most useful general layout for your data. The scientific recording layout is more powerful and flexible.

The other thing to note is that there are an equal number of items (replicates) in each “block”. Here there are only 3 observations per block. This replication balance is important the more unbalanced your situation is the more “unreliable” the result is. In fact, if you use the Analysis ToolPak for 2-way ANOVA you must have a completely balanced dataset (or the routine refuses to run).

Calculate Column Sums of Squares

Start by opening the example datafile: Two Way online.xlsx. Make sure you go to the Data worksheet (the worksheet Completed is there for you to check your results). The data are set out like the previous table, in sample layout. You will need to calculate the various sums of squares and to help you the worksheet has some extra areas highlighted for you.

Start by calculating the column SS, that is the sums of squares for the Plant predictor.

In the formula x represents each column, T is the overall data and n is the number of samples.

  1. Click in cell A12 and type a label, Col SS, for the sums of squares of the columns (the Plant predictor variable).
  2. In Cell B12 type a formula to calculate the SS for the vulgaris column: =(AVERAGE(B2:B10)-AVERAGE(B2:C10))^2*COUNT(B2:B10). You should get 7.11.
  3. In C12 type a similar formula to get the SS for the sativa column: =(AVERAGE(C2:C10)-AVERAGE(B2:C10))^2*COUNT(C2:C10). You should get 7.11.

You should now have the two sums of squares for the columns (the Plant predictor).

Calculate Row Sums of Squares

The row SS are calculated in a similar manner to the col SS.

  1. In cell E3 type a formula to compute the row SS for the Water block Lo: =(AVERAGE(B2:C4)-AVERAGE(B2:C10))^2*COUNT(B2:C4). You should get 880.07.
  2. In E6 compute row SS for the Mid block: =(AVERAGE(B5:C7)-AVERAGE(B2:C10))^2*COUNT(B5:C7). You should get 71.19.
  3. In E9 compute row SS for the Hi block: =(AVERAGE(B8:C10)-AVERAGE(B2:C10))^2*COUNT(B8:C10). You should get 1451.85.
  4. In E12 calculate the overall row SS: =SUM(E2:E10). You should get 2401.11.
  5. In F12 type a label, Row SS, to remind you what this value is.

You’ve now got the row and column SS.

Calculate Error Sums of Squares (within groups SS)

The next step is to determine the error sums of squares. You can get this by multiplying the block variance by the number of replicates for each group minus 1. This is essentially a tinkering of the formula for variance:

  1. In H3 type a formula to calculate the SS for the Lo Water and vulgaris Plant block: =VAR(B2:B4)*2. You should get 12.67.
  2. Repeat the previous step for the rest of the blocks. You’ll need to highlight the appropriate values for each block. Note that the *2 part is the same for all blocks, as there are three replicates for each block.
  3. In H12 type a formula to calculate the overall error SS: =SUM(H2:I10). You should get 69.33.
  4. In I13 type a label, Error SS, to remind you what the value represents.

You now have the error term for the ANOVA. This is an important value, as you’ll need it to calculate the final F-values.

Calculate Interaction Sums of Squares

The final SS component is that for the interactions between the two variables (Water and Plant). To do this you use the means of the various groups and the replication like so:

The formula looks horrendouns but in reality it is more tedious than really hard. The first mean is the mean of a single block. The next Xa, is essentially the column mean. The Xb mean is the “row” mean. The final mean (double overbar) is the overall mean. The n is the number of replicates in each block.

  1. In K3 type a formula for the interaction SS for the fisrt block: =(AVERAGE(B2:B4)-AVERAGE(B2:B10)-AVERAGE(B2:C4)+AVERAGE(B2:C10))^2*COUNT(B2:B4). You should get 14.81.
  2. In K6: =(AVERAGE(B5:B7)-AVERAGE(B2:B10)-AVERAGE(B5:C7)+AVERAGE(B2:C10))^2*COUNT(B5:B7). Gives 7.26.
  3. In K9: =(AVERAGE(B8:B10)-AVERAGE(B2:B10)-AVERAGE(B8:C10)+AVERAGE(B2:C10))^2*COUNT(B8:B10). Gives 42.81.
  4. In L3: =(AVERAGE(C2:C4)-AVERAGE(C2:C10)-AVERAGE(B2:C4)+AVERAGE(B2:C10))^2*COUNT(C2:C4). Gives 14.81.
  5. In L6: =(AVERAGE(C5:C7)-AVERAGE(C2:C10)-AVERAGE(B5:C7)+AVERAGE(B2:C10))^2*COUNT(C5:C7). Gives 7.26.
  6. In L9: =(AVERAGE(C8:C10)-AVERAGE(C2:C10)-AVERAGE(B8:C10)+AVERAGE(B2:C10))^2*COUNT(C8:C10). Gives 42.81.
  7. In K12 type a formula to get the total interaction SS: =SUM(K2:L10). You should get 129.78.
  8. In L12 type a label, Interact SS, to remind you what the value represents.

Now you have all the sums of squares values you need to complete the ANOVA.

Compute total SS and degrees of freedom

The total sums of squares can be calculated by adding the component SS together. On the other hand, it would be good to check your maths by calculating it from the total variance and df.

You’ll also need the degrees of freedom for the various components before you can construct the final ANOVA table.

  1. In A13 type a label, Total SS, for the overall sums of squares.
  2. In B13 type a formula to calculate the total SS: =VAR(B2:C10)*(COUNT(B2:C10)-1). You should get 2616.44.
  3. In A14 type a label, df, for the degrees of freedom.
  4. In B14 type a formula for the column df: =COUNT(B12:C12)-1. The result should be 1.
  5. In E14 type a formula for row df: =COUNT(E2:E10)-1. You should get 2.
  6. In H14 work out the error df: =C18*C19. You should get 2.
  7. In K14 work out the interaction df: =COUNT(B2:C10)-COUNT(K2:L10). You should get 12.

Now you have everything except the total df, which you can place in the final ANOVA table shortly.

Make the final ANOVA table

You can now construct the final ANOVA table and compute the F-values and significance of the various components. You want your table to end up looking like the following:

Completed anova table for two-way analysis of variance.

ANOVA
Source of Variation SS df MS F P-value F crit
Water 2403.1 2 1201.56 207.96 4.9E-10 3.89
Plant 14.22 1 14.22 2.46 0.1426 4.75
Water*Plant 129.78 2 64.89 11.23 0.0018 3.89
Error 69.33 12 5.78
Total 2616.44 17

  1. Start by typing a label, ANOVA, into cell A16.
  2. Type the rest of the labels for the ANOVA table as shown above.
  3. In the SS column you can place the sums of squares results, which you’ve already computed. So in B18: =E12. In B19: =SUM(B12:C12). and so on.
  4. The total SS can be =B13 or a sum of the individual SS components.
  5. In the df column you can place the values, which are already computed. The total SS in C22 needs to be determined: =COUNT(B2:C10)-1. You should get 17.
  6. The MS are worked out by dividing the SS by the df. For e.g. in D18: =B18/C18.
  7. Determine an F-value by dividing the MS for a row by the Error MS. So in E18: =D18/D21. Gives 207.96.
  8. Work out an exact p-value for each of your F-values using the FDIST function. You need your F-value, the df for that row and the error df. In F18 type: =FDIST(E18,C18,C21). The result should be 4.9E-10, which is highly significant.
  9. In F19: =FDIST(E19,C19,C21).
  10. In F20: =FDIST(E20,C20,C21).
  11. You can use the FINV function to get a critical value for F. You’ll need a level of significance (0.05), the df for that row and the error df. In G18 type: =FINV(0.05,C18,C21). The result is 3.89.
  12. In G19: =FINV(0.05,C19,C21).
  13. In G20: =FINV(0.05,C20,C21).

Now you have the final ANOVA table completed. You can see that the interaction term is highly significant (p = 0.0018). The Water treatment is also significant but the Plant variable is not (p = 0.14).

Graphing the result

You should plot your result as a chart of some kind. There are several ways you might proceed. Chapter 6 gives details about constructing charts in Excel and in R for various scenarios.

One option would be to make a bar chart, showing the mean for each block, and with the blocks grouped by watering treatment or by plant species. You should give an impression of the variability using error bars. The following column chart was produced using 95% confidence intervals:

Visualisation of two-way anova result. Bars show sample means and error bars are 95% CI.

The critical value for t can be determined using the TINV function and the degrees of freedom: =(TINV(0.05,16)). Note that in this case df = 16 because we are splitting the blocks by plant (there are 9-1 + 9-1 = 16 degrees of freedom). The size of the error bars is then:


Breaking $u$-Substitution

While learning $u$-substitution, I noticed a potential glitch in its application to definite integrals. If $u$ is defined as a non-injective function of the original integrand variable, then given only the $u$ definition along with the $u$-substituted definite integral, it would be impossible to recover the bounds of the original definite integral. For example, given $int_4^9e^u du, ext < where >u = x^2, ext < thus >du = 2x dx$ it would be impossible to determine whether the original integral had bounds $-2$ to $-3$, $-2$ to $3$, $2$ to$-3$, or $2$ to $3$. While the u-substitution definite integral algorithm can be completed without the need to inverse-substitute the bounds at the end, and while at least in this problem, the answer turns out the same regardless of which pair of original bounds was present, the geometric picture for each case is different each definite integral corresponds to the area under a different part of the curve. Seemingly, the reason the signs of the original bounds are unimportant in this case is that the original integrand, $<2xe>^<(x^2)>$, is an odd function, so adding, i.e., the area between $x = -2$ and $x = 2$ to the area between $x = 2$ and $x = 3$ won't change the result.

So a good candidate for a case where the information loss due to a u-substitution would matter might involve

1) A non-injective $u$ definition, and

2) A non-odd original integrand, that doesn't become odd when shifted in any direction by a constant, and

3) An integrand that is locally linear everywhere

This is actually trickier to find than one might expect. The integrand $<2xe>^<(x^2)>$ won't work because it's odd, while the integrand $3x^2e^<(x^3)>$ won't work because $u = x^3$ is injective. $4x^3e^<(x^4)>$ is odd, $u = x^5$ is injective, and so on. The absolute value, ramp, and digital root functions aren't odd, but they aren't locally linear everywhere, so they can't really be integrated. The most promising integrand I found is $cos(x)e^$. Integrating it from $ to $-3pi$ on Desmos yields a different answer than integrating from $ to $-2pi$, but the difference is approximately $10^<-15>$, so it could simply be a rounding anomaly.

Has anyone found an integrand and corresponding $u$ definition that fit this bill? If not, do we know for certain that the information lost in a non-injective u-substitution is never important?


3 Answers 3

You need to be careful with your limits of integration. In general, if you're doing a $u$ -substitution to solve a definite integral, there are two potential ways to deal with the limits of integration:

(1) Adjust them when you make your substitution.

(2) Don't worry about them initially. Get an antiderivative in terms of $u$ , then switch back to the original variable and apply Fundamental Theorem of Calculus.

For an example, let's consider the simple one $int_2^3xe^,dx$ , with the substitution $u=x^2$ . In either case, we have $frac=2x$ , so $du=2x,dx$ , and so $x,dx=frac12,du.$ (Note that I didn't solve for $dx$ , here, but rather for $x,dx$ , which are the extraneous $x$ -terms when we make the substitution $e^mapsto e^u$ in the original integrand.)

Using method (1), $x=2$ implies $u=2^2=4$ , and $x=3$ implies $u=3^2=9$ . Hence, $int_2^3xe^,dx=frac12int_4^9e^u,du=frac12left[e^u ight]_^9=frac12(e^9-e^4).$

Using method (2), $int xe^,dx=frac12int e^u,du=frac12e^u=frac12e^.$ (Since we're about to apply the FTC, we don't need an integration constant, here.) Thus, by FTC, we have $int xe^,dx=left[frac12e^ ight]_^3=frac12(e^9-e^4).$

Pick whichever method works for you, and stick with it. (I recommend the first one, personally.) You seem to be favoring the latter. However, you must be cautious with the notation. $int_0^e^xsin(e^x-1),dx eqint_0^sin u,du$ , for example. As a potential alternative, you could write $int_0^e^xsin(e^x-1),dx=int_^sin u,du,$ to make clear that the latter limits of integration are on the variable $x$ .


Related Resources

The various resources listed below are aligned to the same standard, (6EE04) taken from the CCSM (Common Core Standards For Mathematics) as the Expressions and equations Worksheet shown above.

Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y stands for. Reason about and solve one-variable equations and inequalities.

Worksheet

Simplifying Expressions

Similar to the above listing, the resources below are aligned to related standards in the Common Core For Mathematics that together support the following learning outcome:

Apply and extend previous understandings of arithmetic to algebraic expressions


This integral is more complicated than it looks and only requires comfort with hyperbolic trigonometric definitions. My initial instinct is to look at the expression inside the $cosh$ to see if I can simplify it. In fact we have by the definition of the hyperbolic trigonometric functions,

Now using log laws we have that,

We note at this point that,

There is a lot in common between the above two equations which motivates us to press forward, we now use the following identity which we can prove just with the definitions of the hyperbolic trig functions,

$cosh(x+y) = cosh(x) cosh(y) + sinh(x) sinh(y)$

We use to simplify the $cosh$ in the integral, we find that,

$cosh(3x + ( anh^<-1>(3x) - anh^<-1>(x))) = cosh(3x) cosh( anh^<-1>(3x) - anh^<-1>(x)) + sinh(3x) sinh( anh^<-1>(3x) - anh^<-1>(x)) $

Using the following definitions of the hyperbolic functions,

We find that using $(*)$ , leaving the details to you,

Putting this all together,

Finally the integral simplifies to,

$I = 2 int_<-1/3>^ <1/3>(1-3x^2)cosh(3x) mathrmx + 2 int_<-1/3>^ <1/3>2x sinh(3x) mathrmx $

Which is a lot easier to compute with IBP and I will leave that to you to complete, this does indeed give the correct answer.


Splitting the app into model, view, and controller files

In our previous two Flask apps we have had the view displayed for the user in a separate template file, and the computations as always in compute.py , but everything else was placed in one file controller.py . For illustration of the MVC concept we may split the controller.py into two files: model.py and controller.py . The view is in templates/view.html . These new files are located in a directory hw3_flask The contents in the files reflect the splitting introduced in the original scientific hello world program in the section Application of the MVC pattern. The model.py file now consists of the input form class:

The file templates/view.html is as before, while controller.py contains

The statements are indentical to those in the hw2 app, only the organization of the statement in files differ.


Abstract

In this work, the characteristic-wise alternative formulation of the seventh and ninth orders conservative weighted essentially non-oscillatory (AWENO) finite difference schemes is derived. The polynomial reconstruction procedure is applied to the conservative variables rather than the flux function of the classical WENO scheme. The numerical flux contains a low order term and high order derivative terms. The low order term can use arbitrary monotone fluxes that can enhance the resolution and reduce numerical dissipation of the fine scale structures while capturing shocks essentially non-oscillatory. The high order derivative terms are approximated by the central finite difference schemes. The improved performance in terms of accuracy, essentially non-oscillatory shock capturing and resolution for the complex shocked flow with fine scale structures in the classical one- and two-dimensional problems is demonstrated. However, the inclusion of the high order derivative terms is prone to generate Gibbs oscillations around a strong discontinuity and might result in a negative density and/or pressure. Therefore, a positivity-preserving limiter [Hu et al. J. Comput. Phys. 242 (2013)] is adopted to ensure the positive density and pressure in the shocked flows with extreme conditions, such as Mach 2000 jet flow problem.


Abstract

This paper is dedicated to numerical computation of higher order derivatives in Simulink. In this paper, a new module has been implemented to achieve this purpose within the Simulink-based Infinity Computer solution, recently introduced by the authors. This module offers several blocks to calculate higher order derivatives of a function given by the arithmetic operations and elementary functions. Traditionally, this can be done in Simulink using finite differences only, for which it is well-known that they can be characterized by instability and low accuracy. Moreover, the proposed module allows to calculate higher order Lie derivatives embedded in the numerical solution to Ordinary Differential Equations (ODEs). Traditionally, Simulink does not offer any practical solution for this case without using difficult external libraries and methodologies, which are domain-specific, not general-purpose and have their own limitations. The proposed differentiation module bridges this gap, is simple and does not require any additional knowledge or skills except basic knowledge of the Simulink programming language. Finally, the block for constructing the Taylor expansion of the differentiated function is also proposed, adding so another efficient numerical method for solving ODEs and for polynomial approximation of the functions. Numerical experiments on several classes of test problems confirm advantages of the proposed solution.


Watch the video: LECTURE. EXERCISE 1-6. STD 12 DR. G M SUTARIYA (August 2022).