Old Exams
We provide these exams _as is_, that is we will not explain the steps in the solutions, but you can use these to discuss with your fellow students and train.
| Exam | Problems | Solutions | Remarks |
|---|---|---|---|
| Autumn 2024 4N/4D | pdf (incl. solutions) | ||
| Summer 2024 4N/D | pdf (incl. solutions) | Correct tableau exercise 4a) Solution in Task 5b) should be (l_1 = \frac{-1}{12}…\) and \(l_4 = \frac{1}{12}...\) Solution of problem 9 should be 1.3297 |
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| Spring 2024 – 4N | pdf (incl. solutions) | ||
| Autumn 2023 – 4N/D | pdf (incl. solutions) | ||
| Summer 2023 – 4N/D | pdf (incl. solutions) | ||
| Spring 2023 – 4N | pdf (incl. solutions) | ||
| Summer 2022 – 4N/D | pdf (incl. solutions) | In Task 6b the second to last step; the first + is a minus, which in effect makes the first term in the last line vanish. | |
| Autumn 2022 – 4N/D | Task 3 should be \(b_n=\frac{4-6(-1)^n}{n\pi}\). In the solution for task 4 in the point \(k<0:\) in the two-oint-list close to the end in each case it should be \(G(t)=Ce^{4kt}\) instead of \(G(t)=Ce^{-4kt}\). Also \(n=0\) is allowed. In task 10 (a) the central difference scheme has \(x_i^2\) instead of \(x_i\), so we get \(\frac{u(x_i+h)-u(x_i)+u(x_i-h)}{h^2}+\mathscr O(h^2)+x_i^2u(x_i)=\sin(\pi x_i),\ i=1,\dots,N-1\), and the last equations for the finite difference scheme should contain \(U_{i-1}-(2-h^2x_i^2)U_i+U_{i+1}=h^2\sin(\pi x_i),\ i=1,\dots,N-1\) |
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| Spring 2022 - 4N | pdf (incl. Solutions) | In problem 1, the second to last line, when collecting the coefficients for \(x^3\) instead of \(\frac{25}{8}\) it should read \(0\) and for $x$ the $\frac{2}{3}$ has a sign error. | |
| Autumn 2021 - 4N | |||
| Autumn 2020 - 4N/D | This was a home exam. | ||
| Summer 2020 - 4N/D | pdf TMA4125_exam_q_7.py | This was a home exam. In the solution problem 3, a factor \(\sqrt{2\pi}\) is missing in the convolution theorem. The final answer should be \(\sqrt{a\pi} e^{-\frac{x^2}{4a}}\). |
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| Spring 2020 - 4N | This was a home exam. There is a sign error in the solution of problem 2b. At the end, it should be \((1-(-1)^n)\) instead of \(1+(-1)^n\). |
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| Autumn 2019 - 4N/D | |||
| Summer 2019 - 4N/D | There is a sign error in the solution of problem 2b. The last line should (still) have a minus upfront, cf. also Kreyszig, p. 527 (Sec. 11.9, Example 3). The solution of problem 8 is wrong. What is implemented there is an explicit Euler method, not an implicit one. |
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| Spring 2019 - 4N | |||
| Autumn 2018 - 4N/D | pdf(norsk) , pdf(english) | Here is a more detailed solution of Problem 2a. | |
| Spring 2018 - 4N | |||
| Autumn 2017 - 4D | |||
| Summer 2017 - 4N/D | In the solution of problem 2b), the sign of k in equation (1) and (2) should be the opposite for the rest of the solution to make sense. In the solution of problem 5, the y_2 should use 0.995, and not 1 in their computation. | ||
| Spring 2017 - 4N | In task 1 the second sum of \(g(x)\) should be from \(k=0\). There is a minus sign typo in the Euler identity for the sin function in the solution of 3c. In the solution of task 8 on page 7 in the first two big formulas each the second \(u_{m-1}^{n+1}\) has to be replaced by \(u_{m+1}^{n+1}\). |
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| Autumn 2016 - 4N | |||
| Autumn 2016 - 4D | |||
| Summer 2016 - 4N | |||
| Summer 2016 - 4D | |||
| Spring 2016 - 4N | |||
| Autumn 2015 - 4N | |||
| Autumn 2015 - 4D | |||
| Summer 2015 - 4D | Oppgave 6 ble glemt i LF. feil i LF i oppgave 4) bruk av formelark gir k_1=0 og y_1=y_0+(1/2)k_2 = 1,00346… |
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| Autumn 2014 - 4N | There is an error in problem 5a: The interpolation points xi should be -2,-1,0,1,2 instead of -2,1,0,1,2. |
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| Høsten 2014 - 4D | |||
| Sommeren 2014 - 4D | |||
| Autumn 2013 - 4N | |||
| Høsten 2013 - 4D | |||
| Sommeren 2013 - 4D | |||
| Autumn 2012 - 4N | pdf (incl. solutions) | ||
| Høsten 2012 - 4D | |||
| Sommeren 2012 - 4D | \(x_2^{(1)} = -3{,}0420\) og ikke \(x_2^{(1)} = 3{,}0420\) slik det står i løsningsforslaget. På tredje linje i løsningen til opppgave 6 skal siste ledd \(U_{i, j-1}\) være \(U_{i-1, j}\). Sjette linje mangler også en slutt-parentes. |
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| Høsten 2011 - 4D | En fortegnsfeil forekommer i starten av løsningen til oppgave 3a. | ||
| Sommeren 2011 - 4D | En regnefeil forekommer på tredje siste linje av oppgave 4a: \(p(3/4)\) er ikke \(3/16\), men \(9/16\). | ||
| Spring 2011 - 4N | In problem 4b, the ODE should have been y''-6y'+9y=0. | ||
| Autumn 2010 - 4N | |||
| Høsten 2010 - 4D | I den Lagrange-interpolasjonsbaserte løsningen av oppgave 1 hadde det opprinnelig sneket seg inn en \(3\) for mye på nest siste linje. Dette er nå rettet opp slik at det som står der nå, er korrekt. | ||
| Sommeren 2010 - 4D | |||
| Spring 2010 - 4N | |||
| Autumn 2009 - 4N | |||
| Høsten 2009 - 4D | |||
| Sommeren 2009 - 4D | |||
| Spring 2009 - 4N | |||
| Autumn 2008 - 4N | |||
| Høsten 2008 - 4D | |||
| Sommeren 2008 - 4D | |||
| Spring 2008 - 4N | |||
| Autumn 2007 - 4N | In problem 1, the integration should be with respect to u and not t (in other words: replace the dt in the integral by du). In the solution of problem 2b, the time dependent part should be e((1-(n/2)^2)t instead of e1-(n/2)^2 t. In problem 4b, the approximation for u' '(0.1) is 4.545 and not 5.03 as claimed in the solutions. |
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| Høsten 2007 - 4D | |||
| Sommeren 2007 - 4D | |||
| Autumn 2006 - 4N | |||
| Spring 2006 - 4N | There is an error in problem 2a: The function f is equal to 1 for π/2 < x < π, not for π/2 < x < 1. |
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| Høsten 2006 - 4D | I første setning av oppgave 3c forekommer en fortegnsfeil. | ||
| Sommeren 2006 - 4D | I oppgave 6 bruker LF feil nedre randbetingelse, \(\cos\pi x/2\). | ||
| Høsten 2005 - 4D | |||
| Sommeren 2005 - 4D | |||
| Høsten 2004 - 4D | |||
| Sommeren 2004 - 4D | Skrivefeil i løsning på oppgave 3b: Her menes \(7/y\), ikke \(7/u\). Dette er en eksamen hvor LF bare er en fasit. |
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| Høsten 2003 - 4D | |||
You can find more past papers at: https://wiki.math.ntnu.no/tma4130/2011h/gamle_eksamensoppgaver