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Controversies in Science/Was there a mitochondrial Eve?/A Critique Of

(Review Paper) Cited in

Points Made
- Evidence for African origin is ambiguous. Inappropriate tests were used to confirm, and when appropriate tests were done, results could not provide significant support for African origin.

- The time of origin was originally calculated without incorporating key sources of error, so there is much uncertainty about when this "Eve" is said to have lived. Despite this uncertainty, it has been found that she was very likely much older than 200 000 years.

- The out-of-Africa replacement theory is rejected by cladogram data and nuclear DNA patterns. The data and patterns are consistent with population-expansion events and gene flow.

Methods
Cladograms were examined and compared to the potential roots of humanity. No significantly conclusive results were found, so we do not have certainty that there was one location from which humans originated. The time to coalescence is seen as a variable as there is much potential for deviation.

Results
The article asserts that by one millions years ago humans were widespread in the Old World and had genetic contact with each other for much of the one million year time period. Concluding humans with regional subdivisions are from one evolutionary lineage, and can be traced using mtDNA. Cann, Stoneking, and Wilson, in their 1987 efforts to find the maxium parsimony tree failed because only one data order was tried. .


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Controversies in Science/Was there a mitochondrial Eve?/A critique of Evidence for mitochondrial DNA recombination in a human population of island Melanesia

(Review Paper) Cited in

Points Made
452 people from the island of Nguna were identified as having separate mtDNA lineages. Their DNA type was found to be a recombination of mtDNA types of individuals in other remote areas which is highly unlikely. The asexual and cloned nature of mtDNA inheritance has resulted in mutations in offspring due to no diversions from common ancestors. .

Mutations were proven through phylogenetic analysis of multiple mtDNA lineages.

Methods
There were three different methods that were used – DNA analysis, Mitochondrial DNA analysis, and phylogenetic analysis of the DNA sequences. There were samples taken from 452 people from 16 different areas. The mitochondrial DNA analysis detected the 9-bp deletion by means of PCR analysis. The phylogenetic analysis is a substation model used to look at the nucleotide frequencies. There was a bootstrap analysis used to see the similarities in the haplotypes combinations I and II, I and III, and also on the combination II and III.

DNA differences cannot be used alone in determining the differences between species as a whole. Our species is not diverse enough to compare the DNA alone. The DNA is basically the same from person to person, with only a few differences. We must consider other factors as well like bone characteristics, etc. Bmina836 (talk) 02:33, 28 March 2012 (UTC)

Results
The results indicate there human population in the area sampled derives from three main principal mtDNA Lineages.

The 452 individual DNA sequences fell into 175 different mtDNA types which were then divided into three major groups called haplogroups I, II and III. In haplogroup I, 25% of the total sample was found to belong to the 175 mtDNA types. The first lineage had the previously described Polynesian motive and was found mainly in Trobriand Islands near the east coast of PNG, in the Roro and was also present in different in islands of the Vanuatu archipelago. Haplogroup II was found to be the largest group and most diverse of all samples (48%). It was mainly found in PNG highlands and Vanuatu, but the PNG had the most diversity of haplotype. Haplogroup III was the third mtDNA lineage found and was the smallest one of the three. It is found in the highlands of PNG and in Vanuatu.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Talk:Controversies in Science/Was there a mitochondrial Eve?/A Critique of Simulating the mitochondrial dna inheritance

It appears that our academic source backs up this statement as well - in fact ours states that divergence occurred approximately 156 000 years ago. Dlodh574 (talk) 01:59, 28 March 2012 (UTC)

Monte Carlo techniques are manmade and not relevant142.109.6.1 02:01, 19 September 2012 (UTC).Areil314 (talk) 02:02, 19 September 2012 (UTC)


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Talk:Controversies in Science/Was there a mitochondrial Eve?

To support this study that we all originated from Africa we tested 30 DNA polymorphisms from 14 populations worldwide and they all traced back to the same date around the time that it is believed modern humans spread from Africa; thus, favoring the replacement model. Shayd948 (talk) 02:12, 28 March 2012 (UTC)


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Controversies in Science/Was there a mitochondrial Eve?/A critique of Recent African origin of modern humans revealed by complete sequences of hominoid mitochondrial DNAs.

(Review Paper) Cited in

Points Made
There is evidence to support the existence of mitochondrial eve because comparison of synonymous changes between African, Japanese, European, orangutan, chimpanzee and gorillas mitochondrial DNA points to shallow human ancestry, and specifically that modern humans originate from Africa.

Modern Humans came from Africa and traveled to Eurasia essentially replacing Homo Erectus and the Neanderthals with gene exchange. Neanderthal mDNA is outside the variance seen in the Human DNA and thus it can be determined that the two groups separated 660,000 years ago, give or take 140,000 years.

Methods
Published in 1997 was an analysis of mtDNA extracted from Neanderthals. The first Neanderthal fossil was found in Neander Valley, Germany. Part of the fossil was then ground up to extract mtDNA for further comparison to modern human mtDNA.

The upside to mtDNA is that it can be cleanly traced from the mother's genetics which go straight to the child unlike male genetics which can get twisted and confused. They took DNA samples from 3 different nationalities (Africans, Japanese and Europeans) and compared their DNA with those of chimpanzees, gorillas and Pygmies which are three main Africans apes. They compared the DNA differences.

Results
As you can see in the table above, the lowest number of synonymous changes was shown in the RNA genes between European-Japanese, European-African and African-Japanese. The minimal differences suggest that the last common ancestor for humans lived about 143,000 ± 18,000 years ago. The mtDNA compared that displayed the highest levels of synonymous differences between human and non-human were between European, African and Japanese, and orangutan mtDNA. All data considered, the results support the theory of recent African origin or modern humans, because of a mitochondrial Eve. The results also suggest that human’s closest living relative is the chimpanzee.

If human's closest living relatives are chimpanzee's it not a stretch to think that if neanderthal's were living they would be an even close relative and therefore humans may indeed be older than 143,000 ± 18,000 years ago.Jhill171 (talk) 02:42, 28 March 2012 (UTC)
 * It is said that 99% of our DNA is linked to chimpanzees and orangutans therefore looking back in anthropology in evolution we encounter a phase where we were a direct link to of monkeys to humans, therefore like stated above a relative to human can be much older than what science can study. Because we are at the moment limited on the scientific curve of study to be able to find the answers to the questions we are asking. Computers and our research is still fairly limited to how we can apply it.Sambo1428 (talk)


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Talk:Controversies in Science/Was there a mitochondrial Eve?/A critique of Absence of Regional Affinities of Neandertal DNA With Living Humans Does Not Reject Multiregional Evolution

How would Relethford explain how the Neandertal mDNA remained isolated if they were recorded across different regions of the planet?

Due to the discovery of the Lake Mungo 3 specimen (one of the Neanderthal skeletal findings), there is two methods of approaching this argument. The Mungo 3 specimen displays an mtDNA sequence that degrades with time, which could potentially have resulted in the extinction of certain species based upon their mtDNA sequence. This implies that there is potential that Neanderthal involvement in humanity could have been more significant and has simply decayed over the generations. The other argument presents that while Neanderthal and Human varied more so than any of the human or chimpanzee subspecies, they are closer tied to humans than 66% of chimpanzee subspecies. This identifies that the variances within one race is more drastic than the mtDNA variances between humans and Neanderthals.Nitoryu (talk) 02:05, 19 September 2012 (UTC)

I think the holes in the mitochondrial eve theory is what make the multiregional theory more probable.Mcder (talk) 03:09, 19 September 2012 (UTC)

I debate the fact that there is common mitochondrial DNA between humans and Neanderthals. In articles I found there was no proof of similarities. The only proof to defend against Mitochondrial Eve is bone structure; which in my option has not been completely proven.Asmit (talk) 03:10, 21 September 2012 (UTC)


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Talk:A critique of The Global Warming Debate: A Review of the State of Science

I agree that population directly affects the temperature of the climate because as the population increases the CO2 emissions are likely to increase as well. These emissions are absorbed by the land and the ocean which affects the climate and the atmosphere. Shayd948 (talk)

 * Although humans industrialization must have an impact on the environment, this evidence does not seem evident to humans causing global warming. These are just two cities who happen to be industrialized, and two cities who are not. It doesn't seem to take into account that the climate fluctuates naturally. To find it completely relevant, I think there would need to be a larger sample size. Cthom4751 (talk) 17:23, 13 April 2012 (UTC)

 * There are also other factors that play into an increase in surface temperatures – C02 emissions - greenhouse emissions and global warming. For example how close the highly populated cities are to climates that will absorb the C02 emissions. If a city is closer to the ocean or a densely forested area, the more C02 emissions are going to be absorbed. The further away a city is from these sources, the less likely the same amount of C02 emissions are going to be absorbed.Jrach339 (talk) 00:48, 4 April 2012 (UTC)

 * Hingane's study does not appear to address the issue of global warming, but merely argues that a larger occupied surface area will have a higher thermal signature (no doubt due to highly reflective construction materials and reduced vegetation). The study of four cities is also not comprehensive enough to draw any reliable conclusions because it lacks any statistical reliability in terms of the contributions toward global warming. This is an interesting study, but it does not appear relevant to the debate at hand.Ixby 02:39, 15 February 2012 (UTC)

This article states that the higher the population is, the higher the surface temperature is, but that only concludes that the two are relative to each other. This article does not address the surface area covered by this population. Not all industrialized cities have a high number of people in a concentrated area, and some highly populated cities are not always concentrated in one massive area. Take for example cities that have many bordering hamlets or suburb areas. It is more about the more concentrated areas with more people, the higher the C02 emissions are, therefore the bigger the increase in the greenhouse effect, ultimately leading to a rise in temperatures.


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Talk:Controversies in Science/Was there a mitochondrial Eve?/A critique of Evidence for mitochondrial DNA recombination in a human population of island Melanesia

Lfamc (talk) 02:26, 28 March 2012 (UTC)It seems that the results of this study show that the genes found in the Melanesian populations are from three different lineages. However, the results of this study do not state or prove that these lineages are different from the mtDNA found in Africa. Therefore, what is the link between the difference of lineages and the non existence of mitochondrial eve.


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Genomics/Recent African origin of modern humans

(Review Paper) Cited in Genomics refutes an exclusively African origin of humans.

Results
Based on this simulation the nulear genome disproves the “Recent Africa Origin” theory because it shows how mtDNA can occur in a multiregional world. This simulation concludes that 80% of homo sapiens' nuclear loci (taken from all around the world) have DNA from non-African archaic humans.


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Controversies in Science/Was there a mitochondrial Eve?/Genomics refutes an exclusively African origin of humans.

(Review Paper) Cited in

Points Made
The Monte Carlo simulation shows that the current data has been misinterpreted because the sample size is so small. Mitochondrial DNA can be interpreted and limited just to Africa, however, based under the assumption that there are multi-regions the Monte Carlo simulation shows that this theory is flawed

Methods
The method used is called the Monte Carlo model which is a computer simulation dispersing individuals. The simulation showed individuals moving at the pace of a parent and child walking together which worked out to 320 km2/generation. There were a total of 101 locations Locations are spaced 115 km apart and distance was approximately 11,500 km, roughly the distance from Ethiopia to China. Each location comprises 100 individuals which simulated the population size to 10,100 individuals. This is the common estimate of the long-term effective population size of humans.

All individuals are represented by a genome. Individuals are divided into three characterized groups such as "modern", "archaic" and "hypbrid." The group they are placed in depends on whether or not they have each of or a combination of alleles in the C genotypal loci.

Results
Based on this simulation the nulear genome disproves the “Recent Africa Origin” theory because it shows how mtDNA can occur in a multiregional world. This simulation concludes that 80% of homo sapiens nuclear loci (taken from all around the world) have DNA from non-African archaic humans.


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Talk:Controversies in Science/Was there a mitochondrial Eve?/A Critique of A complete Neanderthal Mitochondrial genome sequence determined by high-throughput sequencing: A Review

This point does not support mitochondria eve theory because they point out that there were differences found in the mitochondria DNA from Neandertals to contemporary groups. If we had mixex, our mitochondria DNA would have some similarity to it.Arodr451 (talk) 02:51, 28 March 2012 (UTC)


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Talk:Controversies in Science/Was there a mitochondrial Eve?/a critique of Human demographic history: refining the recent African origin model

There can only be a small amount of difference between two species in order for them to be able to produce fertile offspring. For example, a horse and a donkey can be bred to produce a mule, but mules are infertile. Therefore, limited human diversity is only proof that humans are able to reproduce, not that we cannot have descended from slightly different species thousands of years ago. PolkaDots576 (talk) 02:54, 28 March 2012 (UTC)


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University of Florida/Egm4313/s12.team11.R5

Intermediate Engineering Analysis Section 7566 Team 11 Due date: March 30, 2012.

Problem Statement
Given: Find $$ R_{c} \!$$ for the following series: 1. $$ r(x) = \sum_{k=0}^\infty (k+1) k x^k \!$$ 2. $$ r(x) = \sum_{k=0}^\infty \frac{(-1)^k}{\gamma^k} x^{2k} \!$$ Find $$ R_{c} \!$$ for the Taylor series of 3. $$ sin(x) \!$$ at  $$ x = 0 \!$$

4. $$ log(1+x) \!$$ at  $$ x=0 \!$$

5. $$ log(1+x)\!$$ at  $$ x=1 \!$$

Solution
The radius of convergence $$ R_{c} \!$$ is defined as $$ R_{c} = [\lim_{k \to \infty} |\frac{d_{k+1}}{d_{k}}|]^{-1} \!$$ 1. $$ R_{c} = [\lim_{k \to \infty} |\frac{(k+2)(k+1)}{(k+1)(k)}|]^{-1} \!$$ $$ R_{c} = [\lim_{k \to \infty} |\frac{(k+2)}{(k)}|]^{-1} \!$$ $$ R_{c} = [\lim_{k \to \infty} |\frac{k(1+\frac{2}{k})}{k}|]^{-1} \!$$ $$ R_{c} = [\lim_{k \to \infty} |1+\frac{2}{k}|]^{-1} \!$$ $$ R_{c} = [1+0]^{-1} \!$$ $$ R_{c} = 1 \!$$ 2. $$ R_{c} = [\lim_{k \to \infty} |\frac{\frac{(-1)^{k+1}}{\gamma^{k+1}}}{\frac{(-1)^{k}}{\gamma^{k}}}|]^{-1} \!$$ $$ R_{c} = [\lim_{k \to \infty} |\frac{-1}{\gamma}|]^{-1} \!$$ $$ R_{c} = [\lim_{k \to \infty} \frac{1}{\gamma}]^{-1} \!$$ $$ R_{c} = [\frac{1}{\gamma}]^{-1} \!$$ $$ R_{c} = \gamma \!$$

However, in this problem, the series $$ x \!$$ term is $$ x^{2k} \!$$ not $$ x^k \!$$, as is the general form. Therefore, this implies: $$ |x^2| = \gamma \!$$ $$ |x| = \sqrt{\gamma} \!$$ $$ R_{c} = \sqrt{\gamma} \!$$ 3. The Taylor series for $$ sin(x) \!$$ is expressed as $$ sin(x) = \sum_{k=0}^\infty \frac{(-1)^{k}}{(2k+1)!}x^{2k+1} \!$$ $$\lim_{k \to \infty} \frac{\frac{(-1)^{k+1}}{(2(k+1)+1)!}x^{2(k+1)+1}}{\frac{(-1)^{k}}{(2k+1)!}x^{2k+1}} \!$$ $$\lim_{k \to \infty} \frac{\frac{1}{(2k+3)!}x^{2k+3}}{\frac{1}{(2k+1)!}x^{2k+1}} \!$$ $$\lim_{k \to \infty} \frac{\frac{1}{(2k+3)}x^{3}}{x} \!$$ $$\lim_{k \to \infty} \frac{1}{(2k+3)} x^{2} \!$$ Therefore: $$R_{c} = [\lim_{k \to \infty} \frac{1}{(2k+3)} ]^{-1}\!$$ $$R_{c} = \infty \!$$ 4. The Taylor series for $$ log(1+x) \!$$ at $$ x = 0 \!$$ is expressed as $$ log(1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^{k} \!$$ $$R_{c} = [\lim_{k \to \infty} |\frac{\frac{(-1)^{k+2}}{k+1}}{\frac{(-1)^{k+1}}{k}} |]^{-1} \!$$ $$R_{c} = [\lim_{k \to \infty} |\frac{\frac{(-1)}{k+1}}{\frac{(1)}{k}} |]^{-1} \!$$ $$R_{c} = [\lim_{k \to \infty} |\frac{\frac{(-1)}{k(1+\frac{1}{k})}}{\frac{1}{k}} |]^{-1} \!$$ $$R_{c} = [\lim_{k \to \infty} |\frac{(-1)}{(1+\frac{1}{k})}|]^{-1} \!$$ $$R_{c} = [1/1]^{-1} \!$$ $$R_{c} = 1 \!$$

5. The Taylor series for $$ log(1+x) \!$$ at $$ x = 1 \!$$ is expressed as $$ log(1+x) = log(2) + \sum_{k=1}^\infty \frac{(-1)^{k+1}}{2^{k} k}(x-1)^{k} \!$$ $$\lim_{k \to \infty} \frac{\frac{(-1)^{k+2}}{2^{k+1} (k+1)}(x-1)^{k+1}}{\frac{(-1)^{k+1}}{2^{k} k}(x-1)^{k}} \!$$ $$\lim_{k \to \infty} \frac{\frac{(-1)^{2}}{2(k+1)}(x-1)}{\frac{1}{k}} \!$$ $$\lim_{k \to \infty} \frac{\frac{1}{2k(1+\frac{1}{k})}(x-1)}{\frac{1}{k}} \!$$ $$\lim_{k \to \infty} \frac{\frac{1}{2(1+\frac{1}{k})}(x-1)}{1} \!$$ $$\frac{1}{2}(x-1) \!$$ For convergence: $$ \frac{1}{2}(x-1) < 1 \!$$ $$ x-1 < 2 \!$$ $$ x < 3 \!$$ Therefore, $$R_{c} = 3 \!$$

R5.2
Solved by: Andrea Vargas

Problem Statement
Part 1:Determine whether the following are linearly independent using the Wronskian

Part 2: Determine whether the following are linearly independent using the Gramian

Part 1
Using the Wronskian we check for linear independence.

We know from (1) and (2) in 7-35 that if $$ W=\det\begin{bmatrix} f & g \\ f' & g' \end{bmatrix} =fg'-gf'\neq 0\!$$ Then the functions are linearly independent.

Part 1.1
$$ f(x)=x^2 \!$$ $$ g(x)=x^4 \!$$

Taking the derivatives of each function:

$$ f'(x)=2x \!$$ $$ g'(x)=4x^3\!$$

$$ W=\begin{bmatrix} x^2 & x^4 \\ 2x & 4x^3 \end{bmatrix} = 4x^5-2x^5=2x^5\neq 0\!$$

$$ \therefore \!$$ f(x) and g(x) are linearly independent

Part 1.2
$$ f(x)=\cos(x) \!$$ $$ g(x)=\sin(3x) \!$$

Taking the derivatives of each function:

$$ f'(x)=-\sin(x) \!$$ $$ g'(x)=3\cos(3x)\!$$

$$ W=\begin{bmatrix} \cos(x) & \sin(3x) \\ -\sin(x) & 3\cos(3x) \end{bmatrix} = 3\cos(x)\cos(3x)+\sin(3x)\sin(x)=\neq 0\!$$

$$ \therefore \!$$ f(x) and g(x) are linearly independent

Part 2
Using the Gramian we check for linear independence. We know from the notes in (1) 7-34 that: $$\langle f,g \rangle :=\int_{a}^{b}=f(x)g(x)\!$$

and that the Gramian is defined as: $$ \Gamma (f,g)=det\begin{bmatrix} \langle f,f \rangle & \langle f,g \rangle \\ \langle g,f \rangle & \langle g,g \rangle \end{bmatrix}\!$$

Then f,g are linearly independent if $$\Gamma(f,g)\neq 0 \!$$

Part 2.1
$$ f(x)=x^2 \!$$ $$ g(x)=x^4 \!$$

Taking scalar products: $$\langle f,f \rangle:=\int_{-1}^{1}=f(x)f(x)=x^2*x^2=\int_{-1}^{1}=x^4=\left [ \frac{x^5}{5} \right ]_{-1}^{1}=\frac{2}{5}\!$$ $$\langle f,g \rangle=\langle g,f \rangle:=\int_{-1}^{1}=f(x)g(x)=x^2*x^4=\int_{-1}^{1}=x^6=\left [ \frac{x^7}{7} \right ]_{-1}^{1}=\frac{2}{7}\!$$ $$\langle g,g \rangle:=\int_{-1}^{1}=g(x)g(x)=x^4*x^4=\int_{-1}^{1}=x^8=\left [ \frac{x^9}{9} \right ]_{-1}^{1}=\frac{2}{9}\!$$

$$ \Gamma (f,g)=\det \begin{bmatrix} \frac{2}{5} & \frac{2}{7} \\ & \\ \frac{2}{7} & \frac{2}{9} \end{bmatrix}=0.08888-0.0816326531\neq 0\!$$ $$ \therefore \!$$ f(x) and g(x) are linearly independent

Part 2.2
$$ f(x)=\cos(x) \!$$ $$ g(x)=\sin(3x) \!$$

Taking scalar products: $$\langle f,f \rangle:=\int_{-1}^{1}=f(x)f(x)=\int_{-1}^{1}=\cos^2(x)dx\!$$ We can use the trig identity for power reduction $$\cos^2(x)=\frac{1}{2}\cos(2x)+\frac{1}{2}\!$$ Then we have, $$\int_{-1}^{1}\frac{1}{2}\cos(2x)+\frac{1}{2}dx=\left [ \frac{1}{4}\sin(2x)+\frac{1}{2}x \right ]_{-1}^{1}\!$$ $$=0.7273243567-(-0.7273243567)=1.454648713\!$$

$$\langle f,g \rangle=\langle g,f \rangle:=\int_{-1}^{1}f(x)g(x)dx\!$$ $$=\int_{-1}^{1}\sin(3x)\cos(x)dx=\left [ \frac{1}{8}(-2\cos(2x)-\cos(4x)) \right ]_{-1}^{1}\!$$ $$=0.185742-0.185742=0\!$$

$$\langle g,g \rangle:=\int_{-1}^{1}g(x)g(x)dx=\int_{-1}^{1}\sin(3x)\sin(3x)dx\!$$ $$=\frac{1}{2}\int_{-1}^{1}1-\cos(6x)dx=\left [ \frac{1}{2}(x-\frac{1}{6}\sin(6x)) \right ]_{-1}^{1}\!$$ $$=0.523285+0.523285=1.04656925\!$$

$$ \Gamma (f,g)=\det \begin{bmatrix} 1.454648713 & 0 \\ 0 & 1.04656925 \end{bmatrix}=1.522390612\neq 0\!$$ $$ \therefore \!$$ f(x) and g(x) are linearly independent

Conclusion
By both methods (the Wronskian and the Gramian) we obtain the same results.

Problem Statement
Verify using the Gramian that the following two vectors are linearly independent.

$$\mathbf{b_1}=2\mathbf{e_1}+7\mathbf{e_2}\!$$ $$\mathbf{b_1}=1.5\mathbf{e_1}+3\mathbf{e_2}\!$$

Solution
$$ \Gamma (f,g)=det\begin{bmatrix} \langle \mathbf{b1},\mathbf{b1} \rangle & \langle \mathbf{b1},\mathbf{b2} \rangle \\ \langle \mathbf{b2},\mathbf{b1} \rangle & \langle \mathbf{b2},\mathbf{b2} \rangle \end{bmatrix}\!$$

We know from (3) 8-9 that: $$\langle \mathbf{b1},\mathbf{b2} \rangle= \mathbf{b1}\cdot \mathbf{b2} \!$$

We obtain, $$\langle \mathbf{b1},\mathbf{b1} \rangle= (2)(2)+(7)(7)=4+49=53\!$$ $$\langle \mathbf{b1},\mathbf{b2} \rangle= \langle \mathbf{b2},\mathbf{b1} \rangle=(1.5)(2)+(7)(3)=3+21=24\!$$ $$\langle \mathbf{b2},\mathbf{b2} \rangle= (1.5)(1.5)+(3)(3)=2.25+9=11.25\!$$

Then, $$ \Gamma (f,g)=det\begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix}=596.25-576=20.25\neq0\!$$ $$ \therefore \!$$ b_1 and b_2 are linearly independent

Problem Statement
Show that $$ y_p(x)=\sum_{i=0}^{n}y_{p,i}(x)\!$$ is indeed the overall particular solution of the L2-ODE-VC $$ y''+p(x)y'+q(x)y=r(x) \!$$ with the excitation $$ r(x)=r_1(x)+r_2(x)+...+r_n(x)=\sum_{i=0}^{n}r_i(x)\!$$.

Discuss the choice of $$ y_p(x)\!$$ in the above table e.g., for $$ r(x)=k \cos\omega x\!$$ why would you need to have both $$ \cos \omega x \!$$ and $$\sin \omega x \!$$ in $$ y_p(x) \!$$ ?

Solution
Because the ODE is a linear equation in y and its derivatives with respect to x, the superposition principle can be applied:

$$r_1(x) \!$$ is a specific excitation with known form of $$ y_{p1}(x)\!$$ and $$r_2(x) \!$$ is a specific excitation with known form of $$ y_{p2}(x)\!$$

$$+\begin{cases} & \text{ } y_{p1}''+p(x)y_{p1}'+q(x)y_{p1}=r_1(x) \\ & \text{ } y_{p2}''+p(x)y_{p2}'+q(x)y_{p2}=r_2(x) \end{cases}\!$$

becomes

$$(y_{p1}+y_{p2})''+p(x)(y_{p1}+y_{p2})'+q(x)(y_{p1}+y_{p2})=r_1(x)+r_2(x)\!$$

proving that

$$ y_p(x)=\sum_{i=0}^{n}y_{p,i}(x)\!$$ is indeed the overall particular solution of the L2-ODE-VC $$ y''+p(x)y'+q(x)y=r(x) \!$$ with the excitation $$ r(x)=\sum_{i=0}^{n}r_i(x)\!$$

According to Fourier Theorem periodic functions can be represented as infinite series in terms of cosines and sines:

$$f(x)=a_0+\sum_{n=1}^{\infty}[a_n\cos\omega x+b_n\sin\omega x] \!$$

where the coefficients $$ a_0, a_n , b_n \!$$ are the Fourier coefficients calculated using Euler formulas.

So even though the system is being excited by functions like $$ r(x)=k\cos \omega x \!$$ the particular solution would still include both $$ \sin \omega x \!$$ and $$ \cos \omega x \!$$ in $$ y_p(x) \!$$ because the excitation is a periodic function that can be represented as the Fourier infinite series in terms of both $$ \sin \omega x \!$$ and $$ \cos \omega x \!$$ times the Fourier coefficients

Problem Statement
Show that $$cos(7x)$$ and $$sin(7x)$$ are linearly independant using the Wronskian and the Gramain (integrate over 1 period)

Solution
$$f=cos(7x),g=sin(7x)$$ One period of $$ 7x=\pi/7$$ Wronskian of f and g $$ W(f,g)=det\begin{bmatrix} f & g\\ f' & g' \end{bmatrix}$$

Plugging in values for $$f,f',g,g';$$ $$ W(f,g)=det\begin{bmatrix} cos(7x) & sin(7x)\\ -sin(7x) & cos(7x) \end{bmatrix}$$ $$=7cos^2(7x)+7sin^2(7x)$$ $$=7[cos^2(7x)+sin^2(7x)]$$ $$=7[1]$$  They are linearly Independant using the Wronskian. $$ = \int_{a}^{b}f(x)g(x)dx$$ $$\Gamma(f,g)=det\begin{bmatrix}  & \\  &  \end{bmatrix}$$ $$\int_{0}^{\pi/7}cos^2(7x)dx=\pi/14$$ $$\int_{0}^{\pi/7}sin^2(7x)dx=\pi/14$$ $$\int_{0}^{\pi/7}cos(7x)*sin(7x)dx=0$$ $$\Gamma(f,g)=det\begin{bmatrix} \pi/14 & 0\\ 0 & \pi/14 \end{bmatrix}$$ $$\Gamma(f,g)=\pi^2/49$$ They are linearly Independent using the Gramain.

Problem Statement
Find 2 equations for the 2 unknowns M,N and solve for M,N.

Solution
$$ y_p(x)=Mcos7x+Nsin7x$$ $$y'_p(x)=-M7sin7x+N7cos7x$$ $$y''_p(x)=-M7^2cos7x-N7^2sin7x$$ Plugging these values into the equation given ($$y''-3y'-10y=3cos7x$$) yields; $$-M7^2cos7x-N7^2sin7x-3(-M7sin7x+N7cos7x)-10(Mcos7x+Nsin7x)=3cos7x$$ Simplifying and the equating the coefficients relating sin and cos results in; $$-59M-21N=3$$ $$-59N+21M=0$$ Solving for M and N results in; $$M=-177/3922, N=-63/3922$$

Problem Statement
Find the overall solution $$ y(x)$$ that corresponds to the initial conditions $$y(0)=1, y'(0)=0$$. Plot over three periods.

Solution
From before, one period $$=\pi/7$$ so therefore, three periods is $$3\pi/7.$$ Using the roots given in the notes $$\lambda_1=-2,\lambda_2=5$$, the homogenous solution becomes; $$y_h(x)=c_1e^{-2x}+c_2e^{5x}$$ Using initial condtion $$y(0)=1$$; $$1=c_1+c_2$$ $$y'_h(x)=-2c_1e^{-2x}+5c_2e^{5x}$$ with $$y'(0)=0$$ $$0=-2c_1+5c_2$$ Solving for the constants; $$c_1=5/7,c_2=2/7$$ $$y_h(x)=5/7e^{-2x}+2/7e^{5x}$$ Using the $$ y_p(x)$$ found in the last part; $$y=y_h+y_p$$ $$y=5/7e^{-2x}+2/7e^{5x}-177/3922cos7x-63/3922sin7x$$

R5.6
solved by Luca Imponenti

Problem Statement
Complete the solution to the following problem

 * $$y''+4y'+13y=2e^{-2x}cos(3x)\!$$

where

$$y_h=e^{-2x}[Acos(3x)+Bsin(3x)]\!$$

and

$$y_p=xe^{-2x}[Mcos(3x)+Nsin(3x)]\!$$

Find the overall solution $$y(x)\!$$ corresponds to the initial condition:
 * $$y(0)=1 \, \ y'(0)=0\!$$

Plot the solution over 3 periods.

Particular Solution
Taking the derivatives of the particular solution $$y_p(x)\!$$

$$y_p=xe^{-2x}[Mcos(3x)+Nsin(3x)]\!$$

$$y'_p=e^{-2x}[sin(3x)(N-2Nx-3Mx)+cos(3x)(3Nx+M-2Mx)]\!$$

$$y''_p=e^{-2x}[sin(3x)(12Mx-6M-5Nx-4N)+cos(3x)(6N-5Mx-4M-12Nx)]\!$$

Plugging these into the ODE yields

$$sin(3x)(6M-12Mx+5Nx+4N)+cos(3x)(5Mx+4M+12Nx-6N)+\!$$ $$4[sin(3x)(N-2Nx-3Mx)+cos(3x)(3Nx+M-2Mx)]+$$ $$13x[Mcos(3x)+Nsin(3x)]=2cos(3x)\!$$

Equating like terms allows us to solve for M and N

$$sin(3x)[(12Mx-6M-5Nx-4N)+4(N-2Nx-3Mx)+13Nx]=0\!$$

$$cos(3x)[(6N-5Mx-4M-12Nx)+4(3Nx+M-2Mx)+13Mx]=2cos(3x)\!$$

$$-6M=0\!$$

$$6N=2\!$$

$$M=0 \, \ N=\frac{1}{3}\!$$

So the particular solution is

$$y_p=\frac{1}{3}xe^{-2x}sin(3x)\!$$

Overall Solution
The overall solution in the sum of the homogeneous and particular solutions

$$y(x)=y_h(x)+y_p(x)\!$$

$$y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]+\frac{1}{3}xe^{-2x}sin(3x)\!$$

To find A and B we apply the initial conditions

$$y(0)=1 \, \ y'(0)=0\!$$

$$y(0)=1=A\!$$

Taking the derivative

$$y'(x)=\frac{d}{dx}[e^{-2x}[cos(3x)+Bsin(3x)]+\frac{1}{3}xe^{-2x}sin(3x)]\!$$

$$y'(x)=e^{-2x}[(3B+x-2)cos(3x)-(2B+\frac{2}{3}x+\frac{8}{3})sin(3x)]\!$$

$$y'(0)=0=3B-2\!$$

$$B=\frac{2}{3}\!$$

Giving us the overall solution

$$y(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)+\frac{1}{3}xsin(3x)]\!$$

Plot
The period for $$cos(3x) \, \ sin(3x)\!$$ is $$\frac{2\pi}{3}$$

Plotting the solution $$y(x)\!$$ over 3 periods yields

R5.7
Solved by Daniel Suh

Problem Statement
$$v=4e_{1}+2e_{2}=c_{1}b_{1}+c{2}b_{2}\!$$

$$b_{1}=2e_{1}+7e_{2}\!$$

$$b_{2}=1.5e_{1}+3e_{2}\!$$ 1. Find the components $$c_{1},c_{2}\!$$ using the Gram matrix. 2. Verify the result by using $$b_{1}\!$$ and $$b_{2}\!$$, and rely on the non-zero determinant matrix of $$b_{1}\!$$ and $$b_{2}\!$$ relative to the bases of $$e_{1}\!$$ and $$e_{2}\!$$.

Gram Matrix
$$ T(b_{1},b_{2})=\begin{bmatrix}  &  \\  &  \end{bmatrix}$$ $$ = \!$$ Thus,

$$ =  = <(2)(2) + (7)(7)> = 53\!$$

$$ =  = <(1.5)(1.5) + (3)(3)> = 11.25\!$$

$$ = <b_{2}, b_{1}> = <b_{1}\cdot b_{2}> = <(2)(1.5) + (7)(3)> = 24\!$$

$$ T(b_{1},b_{2})=\begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix}$$

$$\Gamma = det[T] = (53)(11.25)-(24)(24) = 20.25\!$$

Defining c
Define: $$c =\begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix}$$ $$d = \begin{bmatrix} <b_{1}, v> \\ <b_{2}, v> \end{bmatrix}= \begin{bmatrix} d_{1} \\ d_{2} \end{bmatrix}\!$$

If $$\Gamma \neq 0 \!$$, then $$\Gamma^{-1}\!$$ exists

$$c = \Gamma^{-1}d\!$$

Finding c
$$\Gamma = 20.25 \neq 0 \!$$ thus, $$\Gamma ^{-1}\!$$exists

$$d =\begin{bmatrix} d_{1} \\ d_{2} \end{bmatrix} = \begin{bmatrix} (2)(4)+(7)(2) \\ (1.5)(4)+(3)(2) \end{bmatrix} = \begin{bmatrix} 22 \\ 12 \end{bmatrix}\!$$

$$\begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix} = \begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix}^{-1}\begin{bmatrix} 22 \\ 12 \end{bmatrix}\!$$

$$\begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix} = \begin{bmatrix} -2 \\ 5.33 \end{bmatrix}\!$$

Part 2 Solution
$$v = 4e_{1} + 2e_{2} \equiv c_{1}b_{1} + c_{2}b_{2}\!$$

$$c_{1}b_{1} + c_{2}b_{2} = (-2)(2e_{1}+7e_{2}) + (5.33)(1.5e_{1} + 3e_{2})\!$$

$$c_{1}b_{1} + c_{2}b_{2} = -4e_{1} - 14e_{2} + 8e_{1} + 16e_{2}\!$$

$$c_{1}b_{1} + c_{2}b_{2} = 4e_{1} + 2e_{2}\!$$

$$v = 4e_{1} + 2e_{2} \equiv c_{1}b_{1} + c_{2}b_{2}\!$$

$$\therefore \!$$ solution is correct

Problem Statement
Find the integral $$ \int x^n log(1+x)dx \!$$ for $$ n=0 \!$$ and $$ n=1 \!$$ Using integration by parts, and then with the help of of General Binomial Theorem $$ (x+y)^n = \sum^n_{k=0} \binom{n}{k} x^{n-k} y^k \!$$

Solution
For $$ n=0 \!$$: $$ \int x^0 log(1+x)dx = \int log(1+x)dx \!$$ For substitution by parts, $$ u = log(1+x), du=\frac{1}{1+x}, dv=dx, v=x \!$$ $$\int log(1+x)dx = x log(1+x) - \int \frac{x}{1+x}dx \!$$ $$ \int log(1+x)dx = x log(1+x) - \int (1 - \frac{1}{1+x})dx \!$$ $$ \int log(1+x)dx = x log(1+x) - x + log(1+x) + C\!$$ Therefore: $$ \int log(1+x)dx = (x+1) log(1+x) - x + C \!$$ Using the General Binomial Theorem: $$ (x+y)^0 = \sum^0_{k=0} \binom{0}{k} x^{0-k} y^k = 1 \!$$ Therefore: $$ \int (1) log(1+x)dx = \int log(1+x)dx \!$$ Which we have previously found that answer as: $$ \int log(1+x)dx = (x+1) log(1+x) - x + C \!$$ For $$ n=1 \!$$: $$ \int x^1 log(1+x)dx = \int x log(1+x)dx \!$$ Initially we use the following substitutions: $$ t = 1+x, x = t-1, dt = dx \!$$ $$\int x log(1+x)dx = \int (t-1) log(t)dt = \int (t log(t) - \log(t))dt \!$$ First let us consider the first term: $$ \int t log(t) dt \!$$ Next, we use the integration by parts: $$ u = \log {t}, du = \frac{1}{t} dt, dv = t dt, v = \frac{1}{2}t^2 \!$$ $$ \int t log(t) dt = \frac{1}{2}t^2 log(t) - \int \frac{1}{2} t^2 (\frac{1}{t} dt) \!$$ $$ \int t log(t) dt = \frac{1}{2}t^2 log(t) - \int \frac{1}{2} t dt) \!$$ $$ \int t log(t) dt = \frac{1}{2}t^2 log(t) - \frac{1}{4} t^2 \!$$  Next let us consider the second term: $$ \int log(t) dt \!$$ Again, we will use integration by parts: $$ u = \log {t}, du = \frac{1}{t} dt, dv = dt, v = t \!$$ $$ \int t log(t) dt = t log(t) - \int t (\frac{1}{t} dt) \!$$  $$ \int t log(t) dt = t log(t) - \int dt \!$$  $$ \int t log(t) dt = t log(t) - t \!$$  Therefore: $$ \int (t log(t) - \log(t))dt = \frac{1}{2}t^2 log(t) - \frac{1}{4} t^2 - (t log(t) - t) \!$$  $$ \int (t log(t) - \log(t))dt = \frac{1}{2}t^2 log(t) - \frac{1}{4} t^2 - t log(t) + t \!$$  Re-substituting for t: $$ \int x log(1+x)dx = \frac{1}{2}(1+x)^2 log(1+x) - \frac{1}{4} (1+x)^2 - (1+x) log(1+x) + (1+x) + C\!$$  $$ \int x log(1+x)dx = (1+x)( \frac{1}{2}(1+x) log(1+x) - \frac{1}{4} (1+x)- log(1+x) + 1 ) + C\!$$  $$ \int x log(1+x)dx = (1+x)( \frac{1}{2} x log(1+x) - \frac{1}{2} log(1+x) - \frac{1}{4} x + \frac{3}{4}) + C\!$$ Therefore: $$ \int x log(1+x)dx = (1+x)(\frac{1}{2} x log(1+x) - \frac{1}{2} log(1+x) - \frac{1}{4} x + \frac{3}{4} ) + C\!$$ Using the General Binomial Theorem for the integral with t substitution $$ \int (t-1) log(t)dt \!$$: $$ (x+y)^1 = (t+(-1))^1 = (t+(-1)) = \sum^1_{k=0} \binom{1}{k} x^{1-k} y^k = \sum^1_{k=0} \binom{1}{k} t^{1-k} (-1)^k = t-1 \!$$ Therefore: $$ \int (t-1) log(t)dt = \int x log(1+x)dx \!$$ Which we have previously found that answer as: $$ \int x log(1+x)dx = (1+x)(\frac{1}{2} x log(1+x) - \frac{1}{2} log(1+x) - \frac{1}{4} x + \frac{3}{4} ) + C\!$$

R5.9
Solved by: Gonzalo Perez

Problem Statement
Consider the L2-ODE-CC (5) p.7b-7 with $$ log(1+x) \! $$ as excitation:

$$ y''-3y'+2y=r(x) \! $$  (5) p.7b-7

$$ r(x)=log(1+x) \! $$    (1) p.7c-28

and the initial conditions

$$ y(\frac{-3}{4})=1, y'(\frac{-3}{4})=0 \! $$.

Part A
Project the excitation $$ r(x) \! $$ on the polynomial basis

$$ {b_j (x) = x^j, j=0, 1, ..., n} \! $$   (1)

i.e., find $$ d_j \! $$ such that

$$ r(x) \approx r_n (x) = \sum_{j=0}^{n} d_j x^j \! $$   (2)

for x in $$ [\frac{-3}{4}, 3] \! $$, and for n = 3, 6, 9.

Plot $$ r(x) \! $$ and $$ r_n (x) \! $$ to show uniform approximation and convergence.

Note that:

$$ \left \langle x^i,r \right \rangle = \int_{a}^{b} x^i log(1+x) dx \! $$    (3)

Solution
To solve this problem, it is important to know that the scalar product is defined as the following:

$$ \left \langle b_0, b_0 \right \rangle = \int x^0\cdot x^0 dx \! $$.

Therefore, it follows that:

$$ \left \langle b_i,b_j \right \rangle = \int x^i\cdot x^j dx \! $$, where $$ b_i (x)=x^i \! $$ and $$ b_j (x) = x^j \! $$.

We know that if $$ b_1, b_2 \! $$ are linearly independent, then by theorem on p.7c-37, the matrix is solvable.

According to this and (3)p.8-14:

If $$ \Gamma \neq 0 \Rightarrow \Gamma^{-1} \! $$ exists $$ \Rightarrow c= \Gamma^{-1}d \! $$.      (3)p.8-14

Now let's define the Gram matrix $$ \Gamma \! $$ as a function of $$ b_i \! $$:

$$ \Gamma (b_i)=\begin{bmatrix} \left \langle b_0,b_0 \right \rangle & \left \langle b_0,b_1 \right \rangle & ... & \left \langle b_0,b_n \right \rangle\\ ... & ... & ... & ...\\ ... & ... & ... & ...\\ \left \langle b_n,b_0 \right \rangle & \left \langle b_n,b_1 \right \rangle & ... & \left \langle b_n,b_n \right \rangle \end{bmatrix} \! $$      (1)p.8-13

Defining the "d" matrix as was done in (3)p.8-13, we get:

$$ d = \begin{Bmatrix} \left \langle b_0,r \right \rangle\\ \left \langle b_1,r \right \rangle\\ ...\\

\left \langle b_n,r \right \rangle\end{Bmatrix} \! $$.       (3)p.8-13

And according to (1)p.8-15: $$ r_n(x) = \sum_{0}^{n}c_i x^i \! $$         (1)p.8-15

Now, we can find the values to compare $$ r_n \! $$ to $$ y \! $$.

Using Matlab, this is the code that was used to produce the results:

The Matlab code above produced the following graph:

Where $$ r_n(x) \! $$ is represented by the dashed line and the approximation,$$ y(x) \! $$, is represented by the red line. This code can work for all n values.

Part B
In a seperate series of plots, compare the approximation of the function $$ \log(x+1) \!$$ by Taylor series expansion about $$ x=0 \!$$.

Where: $$ f(x)= \sum_{n=0}^\infty \frac{f^{(n)}(\hat x)}{n!}(x- \hat x)^n \!$$

Solution
For n=1:

$$ \log(x+1) = \frac{x}{\log(10)} \! $$

For n=2:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)} \! $$

For n=3:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)} \! $$

For n=4:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)} \!$$

For n=5:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)} \! $$

For n=6:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)} \! $$

For n=7:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)} \!$$

For n=8:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)} \! $$

For n=9:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)} \! $$

For n=10:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})} \! $$

For n=11:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})} \!$$

For n=12:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$ $$ -\frac{x^{12}}{\log(10^{12})} \! $$

For n=13:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$ $$ -\frac{x^{12}}{\log(10^{12})}+\frac{x^{13}}{\log(10^{13})} \! $$

For n=14:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$ $$ -\frac{x^{12}}{\log(10^{12})}+\frac{x^{13}}{\log(10^{13})}-\frac{x^{14}}{\log(10^{14})} \! $$

For n=15:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$ $$ -\frac{x^{12}}{\log(10^{12})}+\frac{x^{13}}{\log(10^{13})}-\frac{x^{14}}{\log(10^{14})}+\frac{x^{15}}{\log(10^{15})} \! $$

For n=16:

$$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$ $$ -\frac{x^{12}}{\log(10^{12})}+\frac{x^{13}}{\log(10^{13})}-\frac{x^{14}}{\log(10^{14})}+\frac{x^{15}}{\log(10^{15})}-\frac{x^{16}}{\log(10^{16})} \!$$

Using Matlab to plot the graph:

Part 2
Find $$ y_n (x) \! $$ such that:

$$ y_n'' + ay_n'+by_n = r_n (x) \! $$   (1) p.7c-27

with the same initial conditions as in (2) p.7c-28.

Plot $$ y_n (x) \! $$ for n = 3, 6, 9, for x in $$ [\frac{-3}{4}, 3] \! $$.

In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.

Solution
First, we find the homogeneous solution to the ODE: The characteristic equation is: $$\lambda^2-3\lambda+2=0\!$$ $$(\lambda-2)(\lambda-1)=0\!$$ Then, $$\lambda=1,2\!$$ Therefore the homogeneous solution is: $$y_h=C_1e^(2x)+c_2e^x\!$$

Now to find the particulate solution

For n=3:

$$ r(x)=\sum_{0}^{n}-\frac{(-1)^nx^n}{n\ln (10)} \!$$

$$ r(x)=\frac{x}{\ln(10)}-\frac{x^2}{2\ln(10)}+\frac{x^3}{3\ln(10)} \!$$

We can then use a matrix to organize the known coefficients:

$$ \begin{bmatrix} 2 &-3  &2  &0  & 0\\ 0&  2  &-6  &6  & 0\\ 0&  0&  2  &-9  &12  \\ 0&  0&  0&  2  &-12\\ 0&  0&  0& 0 &  2  \\  \end{bmatrix} \begin{bmatrix} K_{0} \\ K_{1} \\ K_{2} \\ K_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{ln(10)} \\ \frac{1}{2ln(10)} \\ \frac{1}{3ln(10)} \end{bmatrix}\!$$ Then, using MATLAB and the backlash operator we can solve for these unknowns: Therefore $$ y_{p4}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4 \!$$ Superposing the homogeneous and particulate solution we get $$ y_{n}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4+C_1e^{2x}+C_2e^{x} \!$$ Differentiating: $$y'_{n}=3.7458+3.1486x+1.1943x^2+0.2172x^3+2C_1e^{2x}+C_2e^{x}\!$$ Evaluating at the initial conditions: $$y(-0.75)=0.9698261719+0.231301601C_!+0.4723665527C_2=1\!$$ $$y'(-0.75)=1.9645125+0.4462603203C_1+0.4723665527C_2=0\!$$ We obtain: $$ C_1=-4.46\!$$ $$ C_2=0.055 \!$$ Finally we have: $$ y_{n}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4-4.46e^{2x}+0.055e^{x} \!$$

For n=6:

$$ r(x)=\sum_{0}^{n}-\frac{(-1)^nx^n}{n\ln (10)} \!$$

$$ r(x)=\frac{x}{\ln(10)}-\frac{x^2}{2\ln(10)}+\frac{x^3}{3\ln(10)}-\frac{x^4}{4\ln(10)}+\frac{x^5}{5\ln(10)}-\frac{x^6}{6\ln(10)} \!$$

We can then use a matrix to organize the known coefficients:

$$ \begin{bmatrix} 2 &-3  &2  &0  & 0 &0 &0 &0\\ 0&  2  &-6  &6  & 0 &0 &0 &0\\ 0&  0&  2  &-9  &12  &0 &0 &0\\ 0&  0&  0&  2  &-12 &20 &0 &0 \\ 0&  0&  0& 0 &  2 &-15 &30 &0\\ 0 &0 &0 &0 &0 &2 &-18 &42\\ 0 &0 &0 &0 &0 &0  &2 &-21\\ 0 &0 &0 &0 &0 &0 &0 &2\\  \end{bmatrix} \begin{bmatrix} K_{0} \\ K_{1} \\ K_{2} \\ K_{3} \\ K_{4} \\ K_{5} \\ K_{6} \end{bmatrix} \begin{bmatrix} 0 \\ \frac{1}{ln(10)} \\ \frac{1}{2ln(10)} \\ \frac{1}{3ln(10)} \\ \frac{1}{4ln(10)} \\ \frac{1}{5ln(10)} \\ \frac{1}{6ln(10)} \end{bmatrix}\!$$ Then, using MATLAB and the backlash operator we can solve for these unknowns: Therefore $$ y_{p7}=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7 \!$$ Superposing the homogeneous and particulate solution we get $$ y_{n}=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7+C_1e^{2x}+c_2e^x \!$$ Differentiating: $$y'_{n}=375.3933+371.213x+181.644x^2+57.9784x^3+13.46x^4+2.1714x^5+0.214x^6+2C_1e^{2x}+C_2e^x\!$$ Evaluating at the initial conditions: $$y(-0.75)=178.816+0.2231301601C_!+0.4723665527C_2=1\!$$ $$y'(-0.75)=178.413+0.4462603203C_1+0.4723665527C_2\!$$ We obtain: $$ C_1=-2.6757\!$$ $$ C_2=-375.173 \!$$ Finally $$ y_{n}=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7+-2.6757e^{2x}-375.173e^x \!$$

For n=9:

$$ r(x)=\sum_{0}^{n}-\frac{(-1)^nx^n}{n\ln (10)} \!$$

$$ r(x)=\frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)} \! $$

We can then use a matrix to organize the known coefficients: $$ \begin{bmatrix} K_{0} \\ K_{1} \\ K_{2} \\ K_{3} \\ K_{4} \\ K_{5} \\ K_{6} \\ K_{7} \\ K_{8} \\ K_{9} \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{ln(10)} \\ \frac{1}{2ln(10)} \\ \frac{1}{3ln(10)} \\ \frac{1}{4ln(10)} \\ \frac{1}{5ln(10)} \\ \frac{1}{6ln(10)} \\ \frac{1}{7ln(10)} \\ \frac{1}{8ln(10)} \\ \frac{1}{9ln(10)} \end{bmatrix}\!$$ Then, using MATLAB and the backlash operator we can solve for these unknowns: Therefore $$ y_{p11}=1753158.594+1752673.419x+875851.535x^2+291627.134x^3+72745.1129x^4+14484.362x^5+2392.510x^6+335.632x^7+40.417x^8\!$$ $$+4.1499x^9+0.3474x^(10)+0.0197x^(11) \!$$ Superposing the homogeneous and particulate solution we get $$ y_{n}=1753158.594+1752673.419x+875851.535x^2+291627.134x^3+72745.1129x^4+14484.362x^5+2392.510x^6+335.632x^7+40.417x^8\!$$ $$+4.1499x^9+0.3474x^(10)+0.0197x^(11)+C_1e^{2x}+C_2e^(x) \!$$ Differentiating: $$y'_{n}=0.2167 x^10+3.474 x^9+37.3491 x^8+323.336 x^7+2349.42 x^6+14355.1 x^5+72421.8 x^4+290980. x^3+874881. x^2\!$$ $$+1.7517x10^6 x+1.75267x10^6+2C_1e^{2x}+C_2e^x\!$$ Evaluating at the initial conditions: $$y(-0.75)=828254+0.2231301601C_!+0.4723665527C_2=1\!$$ $$y'(-0.75)=828145+0.4462603203C_1+0.4723665527C_2=0\!$$ We obtain: $$ C_1=-484.022\!$$ $$ C_2=-1753750 \!$$ Finally $$ y_{n}=1753158.594+1752673.419x+875851.535x^2+291627.134x^3+72745.1129x^4+14484.362x^5+2392.510x^6+335.632x^7+40.417x^8\!$$ $$+4.1499x^9+0.3474x^(10)+0.0197x^(11)-484.022e^{2x}-1753750e^x \!$$

Here is the graph for this problem using Matlab:


text:
User:Egm4313.s12.team11.arrieta/Report 5

Problem R5.4
Solved by Francisco Arrieta

Problem Statement
Show that $$ y_p(x)=\sum_{i=0}^{n}y_{p,i}(x)\!$$ is indeed the overall particular solution of the L2-ODE-VC $$ y''+p(x)y'+q(x)y=r(x) \!$$ with the excitation $$ r(x)=r_1(x)+r_2(x)+...+r_n(x)=\sum_{i=0}^{n}r_i(x)\!$$.

Discuss the choice of $$ y_p(x)\!$$ in the above table e.g., for $$ r(x)=k \cos\omega x\!$$ why would you need to have both $$ \cos \omega x \!$$ and $$\sin \omega x \!$$ in $$ y_p(x) \!$$ ?

Solution
Because the ODE is a linear equation in y and its derivatives with respect to x, the superposition principle can be applied:

$$r_1(x) \!$$ is a specific excitation with known form of $$ y_{p1}(x)\!$$ and $$r_2(x) \!$$ is a specific excitation with known form of $$ y_{p2}(x)\!$$

$$+\begin{cases} & \text{ } y_{p1}''+p(x)y_{p1}'+q(x)y_{p1}=r_1(x) \\ & \text{ } y_{p2}''+p(x)y_{p2}'+q(x)y_{p2}=r_2(x) \end{cases}\!$$

becomes

$$(y_{p1}+y_{p2})''+p(x)(y_{p1}+y_{p2})'+q(x)(y_{p1}+y_{p2})=r_1(x)+r_2(x)\!$$

proving that

$$ y_p(x)=\sum_{i=0}^{n}y_{p,i}(x)\!$$ is indeed the overall particular solution of the L2-ODE-VC $$ y''+p(x)y'+q(x)y=r(x) \!$$ with the excitation $$ r(x)=\sum_{i=0}^{n}r_i(x)\!$$

According to Fourier Theorem periodic functions can be represented as infinite series in terms of cosines and sines:

$$f(x)=a_0+\sum_{n=1}^{\infty}[a_n\cos\omega x+b_n\sin\omega x] \!$$

where the coefficients $$ a_0, a_n , b_n \!$$ are the Fourier coefficients calculated using Euler formulas.

So even though the system is being excited by functions like $$ r(x)=k\cos \omega x \!$$ the particular solution would still include both $$ \sin \omega x \!$$ and $$ \cos \omega x \!$$ in $$ y_p(x) \!$$ because the excitation is a periodic function that can be represented as the Fourier infinite series in terms of both $$ \sin \omega x \!$$ and $$ \cos \omega x \!$$ times the Fourier coefficients