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# Longer DNA exhibits greater potential for cell-free gene expression

## Abstract

Cell-free gene expression systems have been valuable tools for understanding how transcription/translation can be regulated in living cells. Many studies have investigated the determining factors that affect gene expression. Here we report the effect of the length of linearized reporter DNAs encoding the firefly luciferase gene so as to exclude the influence of supercoiling. It is found that longer DNA molecules exhibit significantly greater potency in gene expression; for example, the expression level for DNA with 25.7 kbp is 1000-times higher than that for DNA of 1.7 kbp. AFM observation of the DNA conformation indicates that longer DNA takes shrunken conformation with a higher segment density in the reaction mixture for gene expression, in contrast to the stiff conformation of shorter DNA. We propose an underlying mechanism for the favorable effect of longer DNA on gene expression in terms of the enhancement of access of RNA polymerase to the shrunken conformation. It is expected that the enhancement of gene expression efficiency with a shrunken DNA conformation would also be a rather general mechanism in living cellular environments.

## Introduction

Gene expression in living cells is strictly self-regulated to ensure that the correct amounts of proteins are made at the most appropriate timing and location for maintaining cellular homeostasis. Gene regulation can occur at any point in gene expression, from the start of the transcription phase to the translation phase. Due to the complexity of gene regulation, unveiling the complete mechanism of gene regulation has been a long-standing quest. To elucidate the complex mechanism of gene regulation in living cellular systems, it is necessary to closely investigate each phase of gene expression and also to shed light on the cooperative effects between transcription and translation. One of the best methods for such an investigation may be the use of a coupled transcription/translation system which allows artificial control of the experimental conditions in vitro1,2,3. The methodology of cell-free gene expression developed during the last couple of decades in molecular biology is expected to provide a useful experimental system for the purpose to gain deeper insights into the coupled transcription/translation dynamics1,2,3. In this regard, cell-free gene expression systems are one of the most widely used techniques in molecular biology1,2,3. The efficiency of cell-free gene expression is known to be affected by various factors such as vector type4,5,6,7, DNA topology4,5,6,7, and reaction components8,9,10,11, as well as the template DNA concentration12,13,14. It is noted that there exists a longstanding controversy regarding the topological effect of supercoiling, or linking number, in circular DNAs on their genomic activity4,5,6,7. Although these factors have been studied extensively, little is known about how the length of the DNA template influences gene expression. The aforementioned complex factors have made it difficult to investigate the impact of the DNA length on gene regulation, which has been an unexplored topic. In the present study, we evaluated the efficiency of protein translation, focusing on the effect of DNA templates of different lengths on cell-free protein synthesis using a luciferase assay. To exclude the influence of supercoiling, we used linear DNA molecules (PCR amplicons or linearized by restriction enzyme) as a template.

## Results

We compared the levels of gene expression of reporter DNAs encoding the firefly luciferase gene (luc-gene) with different lengths of 1717 bp, 4331 bp, and 25,690 bp; hereafter we may call as luc1.7k, luc4.3k and luc25.7k, respectively (Fig. 1a–c and Fig. S1 in the Supplementary Information). We used three different conditions: the same luc-gene concentration (Fig. 1d), the same nucleotide-unit concentration (Fig. 1e), and a combination of both (the same luc-gene and nucleotide-unit concentration adjusted by the addition of noncoding DNA, nc2.6k) (Fig. 1f). The expression level from luc25.7k is 1000-times higher than that with luc1.7k, indicating that the protein yield per target gene is increased by 1000-times of magnitude for a longer template, luc25.7k (Fig. 1d). Under the condition of the same nucleotide-unit concentration, where the concentrations of the targets are inversely proportional to the length of DNA molecules (Fig. 1e), the expression level from luc25.7k is highest despite the fact that luc25.7k has the fewest targets. Interestingly, in a comparison of luc1.7k and luc4.3k, luc4.3k induced a 7-times (Fig. 1d) and 2-times (Fig. 1e) increase in the gene expression level, respectively. These findings indicate that deletion of a 2.6-kb noncoding region results in a lower protein yield, i.e., the magnitude of gene expression increases when the luciferase-coding region is connected to the noncoding region. To validate this result, the level of gene expression was compared under the condition that both the amount of luc-gene and the nucleotide concentration were unified by the addition of extra nc2.6k to the reaction buffer (Fig. 1f). It is shown that the addition of nc2.6k to luc1.7k and luc4.3k caused an order of magnitude greater protein yields (Fig. 1d,f), suggesting the intrinsic effect of nucleotide concentration on gene expression efficiency. Nevertheless, it is clear that the expression efficiency per target is enhanced significantly when the length of the template DNA molecule is longer (Fig. 1f).

The above results clearly indicate that the level of gene expression markedly increased in response to an increase in the length of templates, in addition to the effect of nucleotide concentration in the reaction buffer solution. It is revealed that the potency of expression is in the order of luc25.7k > luc4.3k > luc1.7k under all conditions tested here (Fig. 1). It depends not on the number of targets, but rather on the length of the template DNA. These results also suggest that the noncoding region in reporter DNAs is involved in the regulation of gene expression.

The design of the linear DNAs shown in Fig. 1 may not be convincing enough to investigate the influence of DNA length on gene expression as the non-specific sequences of both up and down streams of the T7 RNA promoter are different in each gene. To address the concern in experimental design, we next designed an in vitro translation assay with four kinds of reporter genes of different nucleotide lengths. The four reporter DNAs encoding firefly luciferase with a T7 promoter were obtained from an identical 25.7-kbp plasmid DNA cleaved with restriction enzyme Sal I, Afl II, Aat II, or ApaL I, respectively (Fig. 2a). Sal I yields a single 25.7-kbp linear reporter gene, and the other enzymes generate reporter genes of different lengths with non-coding DNA fragments (Fig. 2b). The resulting DNA mixtures with reporter genes of different lengths (ApaLI 2.8k, AatII 3.3k, AflII 11.2k and luc25.7k) were used to evaluate their translational efficiency as DNA templates for continuous coupled transcription/translation. As expected, reporter gene expression increased in a nucleotide-length-dependent manner on the DNA templates while the amount and composition of nucleotides remained identical in each reaction (Fig. 2c,d). These results strongly suggest that DNA length positively affects gene activities in a cell-free gene expression system.

To determine the effect of reporter gene size on gene expression activity, single-molecule observations of each linear reporter gene were performed by AFM. The AFM images of each DNA in 10 mM Tris–HCl buffer (pH7.5) revealed a linearized structure, indicating that DNA molecules tend to exhibit a more winding structure as their length increases (Fig. 3a). Notably, in a 0.05% reaction mixture for in vitro gene expression, luc25.7k formed a shrunken structure with a higher density of DNA segments than that of luc4.3k and luc1.7k (Fig. 3b and Fig. S2 in the Supplementary Information). Based on these observations, we consider that the sliding and hopping behaviors of RNA polymerase, which have been studied extensively, is enhanced with longer DNA. In addition, proteins coexisting in the gene expression reaction buffer tend to localize onto the shrunken conformation for the larger DNA molecules, suggesting that the effective affinity of RNA polymerase for DNA is enhanced in longer DNA (see the schematic representations on the lower pictures in Fig. 3b). In the following discussion, we may argue the mechanism on the promotion of gene expression in relation to the physico-chemical characteristics of longer DNA molecules.

Cells regulate gene expression to maintain cellular functions. Gene expression starts with transcription and is continuously followed by translation events. Hence, the initiation of transcription is an efficient point for controlling protein production. In the present study, we demonstrated that DNA length is a crucial factor for gene expression by using a coupled transcription/translation system. The positive effects of DNA length on gene expression efficiency can mainly be explained by increased transcription activity, while mRNA stability, nucleases’ activity, and template degradation should also be taken into account. In fact, it appeared that mRNA was slightly less stable than ribosomal RNA when the stability of mRNA in reticulocyte cell lysates was examined by RT-qPCR (Table S1 in the Supplementary Information). Meanwhile, it is noted that the lengths and stabilities of transcripts from each reporter gene in Fig. 2 are the same as those of nucleotide from the T7 promoter to its terminator. It is thus regarded that the translation phase is not involved in controlling gene expression in the present experiments in a cell-free system.

## Discussion

A plausible mechanism for the promoting effect of DNA length on transcriptional efficiency may be represented by the difference in the conformation of DNA molecules. It is getting clearer that the higher-order structure of the template DNA molecule plays an important role as a controlling factor on the activity of gene expression. To date, not a number of studies have reported that transcriptional condensates with genomic DNA serve as regulatory hubs in gene expression15,16,17,18,19,20,21,22,23,24. As for the characteristic property of the high-order structure on DNA, it has been established that giant DNA above the size of several tens kbp undergoes a large discrete transition from an elongate coil state into a tightly compact state, accompanied by the increase of concentration of various polycationic species such as polyamines25,26. It is known that transcriptional activity is completely inhibited in the tightly compact state27,28. Recently, it has been found that gene expression is promoted at the lower concentrations of polyamines than the concentration that causes compaction9. That is, maximum expression is generated in the shrunken state where the density of DNA segments is higher than in the elongated coil state by avoiding tight packing. Inspection for the actual structure of the shrunken conformation has indicated enhancement of parallel orientation between DNA segments in a shrunken state of long DNA9,10,29. In other words, the basic property of the alignment of DNA segments resembles that of liquid crystals, where rod-like elements prefer to be arranged in parallel under weak repulsive interactions9.

If we consider the above-mentioned effect of the shrunken conformation in the reaction mixture on gene expression, we can expect that formation of the shrunken conformation with a preference for parallel alignment accelerates gene expression and that such a promoting effect would be caused by the greater ability of RNA polymerase to access shrunken DNA. We denote the length of template DNA as L, corresponding to the ‘contour length’ in the definition of polymer physics30. For simplicity, if we assume that the average distance between DNA segments is nearly constant, the volume, v, of a DNA molecule in the shrunken state would exhibit a scaling relationship with L as $$v \sim L$$, which is different from a polymer chain in a good solvent, $$v \sim {L}^\frac{9}{5}$$, and an ideal chain, $$v \sim {L}^\frac{3}{2}$$30. The preferential binding of RNA polymerase together with the enhanced location of NTPs would be effective in the inner volume of the shrunken state. Thus, we may consider that the effective volume of the active region of translation is given by subtracting the infectivity on the surface region: $${v}_{E} \sim (L-{\alpha L}^\frac{2}{3})$$, where we used an approximation of scaling relationships; $$v\sim L,\left[\mathrm{Surface Area}\right]\sim {L}^\frac{2}{3}$$, and $$\alpha$$ is a positive constant. Now, we may evaluate the relative activity of gene expression as in the case of Fig. 1e. By taking the variables n (number density of DNA) and C (density of DNA in base pair units), the relationship is obtained as in $$nL=C.$$ From these relationships, the total active volume of DNA molecules is deduced to be $${V}_{E }\sim n{v}_{E} \sim \frac{C}{L}\left(L- \alpha {L}^\frac{2}{3}\right)=C(1-\alpha {L}^{-\frac{1}{3}})$$. Thus, the value $${L}^{-\frac{1}{3}}$$ corresponds to the degree of the decrease in gene expression, which is 2.5, 2.0 and 1.0 for template DNAs of luc1.7k, luc4.3k and luc25.7k, respectively. By taking the gene expression of luc1.7k as a control, the increase in activity is estimated from the calculations of “2.5–2.0 = 0.5” and “2.5–1.0 = 1.5”, i.e., 0.5 and 1.5, for luc4.3k and luc25.7k, respectively. These expected values of the increase in genetic activities correspond to the observed experimental trend, despite the simplicity of the theoretical argument.

In the present article, we found that longer DNA enhances the efficiency of gene expression. It is known that DNA molecules much larger than the persistence length (150–200 bp) behave as semiflexible polymer chains31,32,33. Here, it has become clear that such relatively long DNA molecules exhibit a shrunken conformation in reaction buffer, which increases the opportunity for encountering RNA polymerase and NTPs due to an increase in their effective concentration around DNA. Although the discovery of the size effect of DNA on gene expression is still in a preliminary stage, it would be promising to extend this insight by shedding light on the actual mechanisms of gene regulation in living cells through extensive studies for longer DNA molecules with the sizes of Mbp–Gbp. It would also be interesting to examine the effect of DNA length on the activity of other enzymes, such as restriction enzymes, DNA repair enzymes, DNA polymerase, etc.

## Methods

### Preparation of linear reporter genes by PCR

A 4.3-kbp plasmid DNA (Luciferase T7 Control DNA: 4331 bp) was purchased from Promega (Madison, WI, USA). A 1.7-kbp linear reporter DNA, a 4.3-kbp linear reporter DNA and a 2.6-kbp linear non-coding DNA were amplified by PCR using the following primers: for the 1.7-kbp linear DNA (forward 5′-TAATACGACTCACTATAGGGAGACC-3′; reverse 5′-CAATTTGGACTTTCCGCCCTTC-3′); for the 4.3-kbp linear DNA (forward 5′-TAATACGACTCACTATAGGGAGACC-3′; reverse 5′-ATTTCGATAAGCCAGCTGC-3′); and for the 2.6-kbp non-coding DNA (forward 5′-CTGTATTCAGCGATGACGAAATTC-3′; reverse 5′-ATTTCGATAAGCCAGCTGC-3′). 1 ng of the 4.3-kbp parental plasmid was used as a template for PCR. PCR was performed with KOD FX neo (Toyobo, Tokyo, Japan) using a Thermal Cycler Dice (TP600) (Takara Bio Inc, Shiga, Japan), according to the manufacturer’s protocol. The PCR products were precipitated with 0.3 M sodium acetate/ethanol, washed with 70% ethanol, and dried. The DNA pellets were dissolved in 50 μL of TE buffer (pH 8.0) and stored at − 20 °C until use. Each template DNA was analyzed using a DNA ladder marker (1 kbp DNA Ladder One, Nacalai Tesque, Kyoto, Japan) by standard 1% agarose gel electrophoresis (Fig. S1b,d in the Supplementary Information).

### Preparation of linear reporter genes by restriction enzyme treatment

A parental 25.7-kbp plasmid DNA was purchased from Kazusa Genome Technologies (Chiba, Japan). The whole sequence of the plasmid is shown in Supplementary Sequence 1. Restriction enzymes Sal I, Afl II, Aat II and ApaL I were purchased from Takara Bio Inc. 2 μg of the 25.7-kbp plasmid was cleaved with 30 U of each restriction enzyme in 100 μL of T buffer (Takara Bio Inc) containing 0.1% BSA. The reaction mixtures were incubated for 120 min at 37 °C. After enzymatic digestion, the DNA mixtures were precipitated with 0.3 M sodium acetate/ethanol, washed with 70% ethanol, and dried. The DNA pellets were dissolved in 50 μL of TE buffer (pH 8.0) and stored at − 20 °C until use. Each digested DNA was analyzed using a DNA ladder marker (1 kbp DNA Ladder One, Nacalai Tesque, Kyoto, Japan) by standard 1% agarose gel electrophoresis (Fig. S1c,d in the Supplementary Information).

### AFM observation

For AFM imaging using an SPM-9700 (Shimadzu, Kyoto, Japan) with a High-Throughput Scanner (Shimadzu, Kyoto, Japan), 0.3 μM DNA was dissolved in 10 mM Tris–HCl buffer (pH 7.5) including 10 μM spermidine or in 0.05% TnT Quick Master Mix (in vitro gene expressions reaction mixture). The DNA solutions were transferred onto a freshly cleaved mica surface and then incubated for 10–30 min at room temperature (24 °C). Subsequently, the samples were rinsed with ultra-pure water, dried with nitrogen gas and imaged by AFM. All measurements were performed in air using the tapping mode. The cantilever, OMCL-AC200TS-C3 (Olympus, Tokyo, Japan), was 200 μm long with a spring constant of 9–20 N/m. The scanning rate was 2 Hz and images were captured using the height mode in a 512 × 512 pixel format. The obtained images were plane-fitted and flattened by the computer program supplied with the imaging module.

### Luciferase reporter assay

Cell-free in vitro transcription/translation of Luc-gene was performed with a TnT T7 Quick Coupled Transcription/Translation System (Promega) according to the manufacturer's instructions and previous reports9,10,11. Reaction mixtures containing the reporter DNAs were incubated for 90 min at 30 °C on a Dry Thermo Unit (TAITEC, Saitama, Japan). The expression levels of luciferase were evaluated following the addition of luciferin as a luciferase substrate (Luciferase Assay Reagent, Promega) by detecting the emission intensity at around 565 nm using a luminometer (MICROTEC Co., Chiba, Japan).

## Data availability

The data analyzed in this paper are available from the corresponding author on reasonable request.

## References

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Katzen, F., Chang, G. & Kudlicki, W. The past, present and future of cell-free protein synthesis. Trends Biotechnol. 23, 150–156 (2005).

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Ahn, J.-H. et al. Cell-free synthesis of recombinant proteins from PCR-amplified genes at a comparable productivity to that of plasmid-based reactions. Biochem. Biophys. Res. Commun. 338, 1346–1352 (2005).

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Moore, S. J. et al. Rapid acquisition and model-based analysis of cell-free transcription–translation reactions from nonmodel bacteria. Proc. Natl. Acad. Sci. USA 115, E4340–E4349 (2018).

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Marshall, R. & Noireaux, V. Quantitative modeling of transcription and translation of an all-E. coli cell-free system. Sci. Rep. 9, 11980 (2019).

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Golkaram, M., Hellander, S., Drawert, B. & Petzold, L. R. Macromolecular crowding regulates the gene expression profile by limiting diffusion. PLoS Comp. Biol. 12, e1005122 (2016).

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## Acknowledgements

This work was supported in part by JSPS KAKENHI (JP20H02859 to S.S. and JP20H01877 to K.Y.) and ISHIZUE 2019 of Kyoto University Research Development Program (to S.S.).

## Author information

Authors

### Contributions

T.N., Y.Y., K.Y. and S.S. conceptualized the study. T.N., Y.Y. and S.S. designed and performed in vitro coupled transcription/translation experiments. T.N. performed the AFM imaging. T.N., Y.Y., K.Y. and S.S. interpreted the data and wrote the manuscript. S.S. supervised all aspects of this work.

### Corresponding author

Correspondence to Shin-ichi Sato.

## Ethics declarations

### Competing interests

The authors declare no competing interests.

### Publisher's note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Rights and permissions

Reprints and Permissions

Nishio, T., Yoshikawa, Y., Yoshikawa, K. et al. Longer DNA exhibits greater potential for cell-free gene expression. Sci Rep 11, 11739 (2021). https://doi.org/10.1038/s41598-021-91243-x

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text:
# Need Formula Concerning RC Timing

#### Mathew Lewis McGill

Joined Nov 5, 2015
4
I have a circuit that generates a pulse width. What I need is a formula to determine the time it will take for the output to switch given a voltage at the noninverting input to the comparator.

#### Papabravo

Joined Feb 24, 2006
12,562
Let's see a +5V source
R=1.62 MΩ
C=10nF

The response of an RC circuit is

$$v=5(1-e^{\frac{-t}{RC}})$$
You should be able to answer your own question with a bit of algebra

#### Mathew Lewis McGill

Joined Nov 5, 2015
4
Yippy, a quiz.
OK, was you hoping that I'd show you? Cause I found that equation; It's all over the place.
I was kind'a hoping to get an equation to solve "t"

#### atferrari

Joined Jan 6, 2004
3,489
Yippy, a quiz.
OK, was you hoping that I'd show you? Cause I found that equation; It's all over the place.
I was kind'a hoping to get an equation to solve "t"
Not even that "bit of algebra" available Mathew?

#### Papabravo

Joined Feb 24, 2006
12,562
Let's see a +5V source
R=1.62 MΩ
C=10nF

The response of an RC circuit is

$$v=5(1-e^{\frac{-t}{RC}})$$
You should be able to answer your own question with a bit of algebra
$$\frac{v}{5} = 1-e^{\frac{-t}{RC}} \frac{v}{5}-1=-e^{\frac{-t}{RC}} 1-\frac{v}{5}=e^{\frac{-t}{RC}}$$
Take the natural logarithm of both sides
$$ln(1-\frac{v}{5})=ln(e^{\frac{-t}{RC}})$$
Can you take it from there?

#### Mathew Lewis McGill

Joined Nov 5, 2015
4
for a voltage of 125mv, the delay is 410.2ms

You're good with algebra, but seriously struggle with providing a simple straight answer.

By the way, I plugged the original equation in Excel and massaged the numbers.

If it's not too painful, I'd appreciate that equation now.

Thanks

#### Papabravo

Joined Feb 24, 2006
12,562
$$\frac{v}{5} = 1-e^{\frac{-t}{RC}} \frac{v}{5}-1=-e^{\frac{-t}{RC}} 1-\frac{v}{5}=e^{\frac{-t}{RC}}$$
Take the natural logarithm of both sides
$$ln(1-\frac{v}{5})=ln(e^{\frac{-t}{RC}})$$
Can you take it from there?
Guess not.
$$ln(1-\frac{v}{5})=\frac{-t}{RC} -RCln(1-\frac{v}{5})=t$$
Are you really sure you could not have done that?
Remember ln is the natural logarithm, or log to the base e (≈2.718281828...), and log is logarithm to the base 10. Seriously struggle with providing a straight answer? If I show you, but you don't understand how it is done, you'll just keep having to ask other people to solve your problems for you. Struggle a bit on you own and you'll remember how to do it next time it comes up. We're here to help not spoon feed answers to you.

Also -- try not to bite the hand that feeds you next time. Petulance is unbecoming.

Last edited:

#### Mathew Lewis McGill

Joined Nov 5, 2015
4
Was just looking for the equation and didn't want to have to spell out my limitations. Sorry for coming off badly.

Thank you.

#### WBahn

Joined Mar 31, 2012
24,854
Congratulations on tricking someone into doing your homework for you. Doesn't happen too often around here, but it does happen from time to time.

#### Papabravo

Joined Feb 24, 2006
12,562
Doesn't matter. He only has enough fish for a day -- not enough to feed himself long enough to get a degree.

#### WBahn

Joined Mar 31, 2012
24,854
Doesn't matter. He only has enough fish for a day -- not enough to feed himself long enough to get a degree.
Too true, my friend, all too true.

format: plain


text:
# Multiclass classification with softmax

I have 5 classes. I’m using a softmax function and getting 5 probabilities in each row that add up to 1 in total.
To calculate the loss do I just pick the column with the highest probability, assign it to the column number and then calculate loss. For eg: if column 1 has highest probability- that row is classified as Class 1 and let’s say the true output is Class 0. So the MSE loss is (1-0)^2 = 1. Or is there any better method?

Hi jpj!

Yes, you should use CrossEntropyLoss as your loss criterions (and not
use `Softmax` because `CrossEntropyLoss` expects logits rather than
probabilities as its predicted `input`).

Best.

K. Frank

1 Like


text:
# Evaluate sqrt(7x)(sqrtz-7sqrt7)?

• Evaluate sqrt(7x)(sqrtz-7sqrt7)?

Positive: 52 %
Get the answer to Integral of sqrt(7*x^9) with the Cymath math problem solver ... Evaluate. Expand. Factor. Graph. Integrate. Prime Factorization.
Positive: 49 %

### More resources

... {x \rightarrow 2}\frac{\sqrt{3 - x ... How do I evaluate the following ... Go to How would I evaluate $\lim_{x \rightarrow 2} \frac{\sqrt{6 ... Read more Positive: 52 % Evaluating limit \lim_{x\to\infty}(x^2-\sqrt ... so that y= x^2 \sqrt{1+7/x^2 + 1/x^4}. ... Evaluate \lim_{x \to \infty} \sqrt{x} ... Read more Positive: 47 % \int\limits_0^1\int\limits_0^1 xy\sqrt{x^2+y^2}\,dydx=\int\limits_0^1 x\,dx\frac12\int\limits_0^1 2y ... Evaluate \int_0^1 \int_\sqrt{y}^1 \int_0^{x^2+y ... Read more Positive: 33 % ... {\tan x}\, dx$ using only the basic techniques like substitution? Update ... Is it possible to evaluate [math]\int\sqrt{\sec ... \cos 7x}{ 1+2 ...


text:
# Math Help - Finding angles

1. ## Finding angles

Find all the angles between $0^\circ$ and $360^\circ$ which satisfy the equation $sec(2x+30^\circ)=sec(2x-60^\circ)$

2. Originally Posted by Punch
Find all the angles between $0^\circ$ and $360^\circ$ which satisfy the equation $sec(2x+30^\circ)=sec(2x-60^\circ)$
$\sec{(2x + 30^\circ)} = \sec{(2x - 60^\circ)}$

$\cos{(2x + 30^\circ)} = \cos{(2x - 60^\circ)}$.

Use the identity $\cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$

So $\cos{2x}\cos{30^\circ} - \sin{2x}\sin{30^\circ} = \cos{2x}\cos{60^\circ} + \sin{2x}\sin{60^\circ}$

$\frac{\sqrt{3}}{2}\cos{2x} - \frac{1}{2}\sin{2x} = \frac{1}{2}\cos{2x} + \frac{\sqrt{3}}{2}\sin{2x}$

$\frac{\sqrt{3} - 1}{2}\cos{2x} = \frac{\sqrt{3} + 1}{2}\sin{2x}$

$\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \tan{2x}$

$\frac{(\sqrt{3} - 1)^2}{3 - 1} = \tan{2x}$

$\frac{3 - 2\sqrt{3} + 1}{2} = \tan{2x}$

$\frac{4 - 2\sqrt{3}}{2} = \tan{2x}$

$2 - \sqrt{3} = \tan{2x}$

This might seem a bit off track, but if you're good with your half angle identities, you will know that

$\tan{\frac{\theta}{2}} = \frac{1 - \cos{\theta}}{\sin{\theta}}$.

Using this identity, what does $\tan{15^\circ}$ equal? (It will help you solve your equation...)

3. Originally Posted by Punch
Find all the angles between $0^\circ$ and $360^\circ$ which satisfy the equation $sec(2x+30^\circ)=sec(2x-60^\circ)$
HI

$\cos (2x-60)=\cos (2x+30)$

$2x-60=2n(180) \pm (2x+30)$

$2x-60=2n(180) -2x-30$

$2n(180)=4x-30$

$180n=2x-15$ , where n=0,1,2,3

4. Originally Posted by Prove It
$\sec{(2x + 30^\circ)} = \sec{(2x - 60^\circ)}$

$\cos{(2x + 30^\circ)} = \cos{(2x - 60^\circ)}$.

Use the identity $\cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$

So $\cos{2x}\cos{30^\circ} - \sin{2x}\sin{30^\circ} = \cos{2x}\cos{60^\circ} + \sin{2x}\sin{60^\circ}$

$\frac{\sqrt{3}}{2}\cos{2x} - \frac{1}{2}\sin{2x} = \frac{1}{2}\cos{2x} + \frac{\sqrt{3}}{2}\sin{2x}$

$\frac{\sqrt{3} - 1}{2}\cos{2x} = \frac{\sqrt{3} + 1}{2}\sin{2x}$

$\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \tan{2x}$

$\frac{(\sqrt{3} - 1)^2}{3 - 1} = \tan{2x}$

$\frac{3 - 2\sqrt{3} + 1}{2} = \tan{2x}$

$\frac{4 - 2\sqrt{3}}{2} = \tan{2x}$

$2 - \sqrt{3} = \tan{2x}$

This might seem a bit off track, but if you're good with your half angle identities, you will know that

$\tan{\frac{\theta}{2}} = \frac{1 - \cos{\theta}}{\sin{\theta}}$.

Using this identity, what does $\tan{15^\circ}$ equal? (It will help you solve your equation...)
I can't continue

5. Originally Posted by Punch
I can't continue
I told you how to continue.

Use the final identity I gave you to evaluate $\tan{15^\circ}$.

$\tan{15^\circ} = \tan{\frac{30^\circ}{2}}$.

6. Originally Posted by Prove It
I told you how to continue.

Use the final identity I gave you to evaluate $\tan{15^\circ}$.

$\tan{15^\circ} = \tan{\frac{30^\circ}{2}}$.

may i ask where did u get the 15 from?

7. Experience.

Try it.

8. Originally Posted by Prove It
Experience.

Try it.
I know you are an expert lol but i need to know the way in solving the problems and not by luck... i can't have u sitting beside me in the exam right

9. or do you tangent inverse to find what 2x is equals to? or using special angle

1. Stop arguing with the person who is trying to help you, instead do as you've been instructed so that you learn something and hopefully will be able to remember this for an exam.

2. Use the half-angle identities for sine, cosine and tangent to find the exact surd answers when your angle is $15^\circ$. You can also find exact values when your angle is $22.5^{\circ}$, or any successive half angle.

11. Originally Posted by Punch
or do you tangent inverse to find what 2x is equals to? or using special angle
If you thought more about post #2 your life would be less stressful.

12. Sorry if my words offended you but i truly don't understand and am not arguing with u, i know u r trying to help me and i am truly thankful to u however i really do not understand where the 15 came from...

13. If you work out what tan15 is equal to, it will make sense where the 15 comes from and why it is being used.

alternatively
2 - root3 = tan2x
2 - root3 = (2tanx)/(1 - tan^2x)

simplify to find tanx

14. Oh so it's similar to using a special angle... $2 - \sqrt{3} = \tan{15}$
then use double angle formula to solve for x

Don't mind me asking but why is this type of question coming up when I don't learn anything about the special angles of tan15 and such? I have only heard of tan 30, tan 60, tan 90 and so on..

15. Originally Posted by Punch
Oh so it's similar to using a special angle... $2 - \sqrt{3} = \tan{15}$
then use double angle formula to solve for x

Don't mind me asking but why is this type of question coming up when I don't learn anything about the special angles of tan15 and such? I have only heard of tan 30, tan 60, tan 90 and so on..
hi again punch ,

nowdays questions don't confine to only whatever you have learnt so be prepared . It's good that you learnt the extras . However , in this case , tan 15 is related to whatever you have learnt , the special angle and double angle formula .

$\tan 30=\frac{1}{\sqrt{3}}$

$\tan 2(15)=\frac{1}{\sqrt{3}}$

$\frac{2\tan 15}{1-\tan^2 15}=\frac{1}{\sqrt{3}}$

$1-\tan^2 15=2\sqrt{3}\tan 15$

$\tan^2 15+2\sqrt{3}\tan 15 -1=0$

$
\tan 15=\frac{-2\sqrt{3}\pm \sqrt{12-4(-1)}}{2}
$

$=\pm 2-\sqrt{3}$

$-2-\sqrt{3}$ is not possible because tan 15 is in the first quadrant , everything here will be positive .

therefore , $\tan 15=2-\sqrt{3}$

Page 1 of 2 12 Last


text:
## Zentralblatt MATH

Publications of (and about) Paul Erdös

Zbl.No:  383.30001
Autor:  Erdös, Paul; Hwang, J.S.
Title:  On a geometric property of Lemniscates. (In English)
Source:  Aequationes Math. 17, 344-347 (1978).
Review:  Motivated by a property of polynomials of a complex variable, the authors prove the theorem below and discuses related open questions. Theorem. Let pn(w,wk) = prodk = 1n|w-wk|  (w,wk in R3) and E(pn) = {w: pn(w,wk) \leq 1}. If pn(w,wk) and pn^*(w,wk^*) are such that E(pn)\subseteq E(pn^*) and if all the zeros wk of pn lie on the same plane, then pn(w,wk)\equiv pn^*(w,wk^*). Moreover, the hypothesis E(pn)\subseteq E(pn{^*}) is not sufficient to deduce pn = pn^*. [For further properties of products pn(w,wk), see J.B.Diaz and D.B.Schaffer, Appl. Anal. 6, 109-117 (1977; Zbl 346.30003).]
Reviewer:  A.Giroux
Classif.:  * 30C10 Polynomials (one complex variable)

© European Mathematical Society & FIZ Karlsruhe & Springer-Verlag


text:
# Chapter 1 - Section 1.8 - Absolute Value Equations and Inequalities - 1.8 Exercises: 54

no solution

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|-5x+7|-4\lt-6 ,$ use the properties of inequality to isolate the absolute value expression. Then analyze the resulting inequality. $\bf{\text{Solution Details:}}$ Using the properties of inequality, the statement above is equivalent to \begin{array}{l}\require{cancel} |-5x+7|\lt-6+4 \\\\ |-5x+7|\lt-2 .\end{array} The absolute value of $x,$ written as $|x|,$ is the distance of $x$ from $0,$ and hence is always a nonnegative number. In the same way, for any $x$, the expression at the left side is always a nonnegative number. This will never be $\text{ less than }$ the expression at the right. Hence, there is $\text{ no solution .}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
Post Categories:   DEFAULT

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text:
# Revision history [back]

The struct DMatch tells for a descriptor (feature) which descriptor (feature) from the train set is more similar. So there are a queryIndex a TrainIndex (which features decide the matcher that is more similar) and a distance. The distance represents how far is one feature from other (in some metric NORM_L1, NORM_HAMMING etc).

So you can determine if a matche is correct if you set a threshold for the distance. For example with SURF features you can match with knnMatch the 2 nearest features, and then you get a Matrix of Matches and you can take as good matches all that its distance are the half than the second I mean:

   matcher->knnMatch(desc1, trainDesc, matches, 2, Mat(), false );
for(int i=0; i<matches.size(); i++){
if(matches[i][0].distance > matches[i][1].distance * 0.6){
goodMatches.push_back(i);
}
}


text:
# Calculating Vmax (or Kcat) given specific activity of an enzyme over length of time incubated with an activator

Is it possible to figure out the $V_\mathrm{max}$ (or $k_\mathrm{cat}$) and/or $K_\mathrm m$ from a plot of enzyme activity expressed as specific activity over time (i.e. $\mathrm{U/mg}$ on the $Y$ and time on the $x$ axis). I've figured out how to express specific activity as turnover but I'm not sure this helps with calculating $V_\mathrm{max}$ or $k_\mathrm{cat}$.

The reaction: p38 phosphorylates MSK1 with Michaelis–Menten kinetics

Experimental conditions: $1\ \mathrm{U/mL}$ of p38 was added at time $t=0\ \mathrm s$ to a solution containing MSK1. They don't say how much MSK1 but express its activity as $\mathrm{U/mg}$ (which is analogous to $\mathrm{(\mu mol/min)/mL}$, otherwise known as specific activity)

The Data:

Time         MSK1 activity
(s)          ((µmol/min)/mg)
__________________________________________________
0               0
300             175
600             350
1200             450
1800             475
2400             500

Calculated turnover:

Time         MSK1 turnover
(s)          (1/(nmol s))
__________________________________________________
18000          300
36000          600
72000         1200
108000         1800
144000         2400
0            0

In actual fact, I though $k_{\rm cat}$ was the turnover and that this was a constant, specific for each enzyme. This data however suggests that as the reaction progresses more MSK1 is produced the turnover number increases, which makes sense and therefore calls into question the two definitions of $k_{\rm cat}$.

## 1 Answer

The enzyme catalysis reaction following MM kinetics can be depicted as:

$$\Large\ce{E + S<=>[k_f][k_r]ES->[k_{cat}]E + P }$$

Note that $V_{max} = k_{cat}.[E_0]$ i.e. conversion rate constant (turnover number) times the total amount of enzyme.

You should note that the enzyme here is p38 and substrate is MSK1 (which in turn is an enzyme). When MSK1 gets phosphorylated, it assumes an active form. MSK1 activity (and specific activity) would reflect the concentration of phosphorylated MSK1 (MSK1-P) i.e. the product. Note that the $k^{^{MSK1-P}}_{cat}$ is its intrinsic property and it does not change; only the $[E^{^{MSK1-P}}_0]$ changes.

I hope your doubt is clear.


text: I have some old bills that I would like the value of... one being "series of 1928b"silver certificate $1bill.. treasurer is woods.. can't read the other name... I also have series 1957, series 1957a, 1957b... series 1935d, 1935e... and series 1976$2 bills... all are in perfect condition!


text:
## Reverend Bayes, meet Countess Lovelace: Probabilistic Programming for Machine Learning

Andy Gordon's talk, Reverend Bayes, meet Countess Lovelace: Probabilistic Programming for Machine Learning, is online.

We propose a marriage of probabilistic functional programming with Bayesian reasoning. Infer.NET Fun turns the simple succinct syntax of F# into an executable modeling language â€“ you can code up the conditional probability distributions of Bayes' rule using F# array comprehensions with constraints. Write your model in F#. Run it directly to synthesize test datasets and to debug models. Or compile it with Infer.NET for efficient statistical inference. Hence, efficient algorithms for a range of regression, classification, and specialist learning tasks derive by probabilistic functional programming.

## Comment viewing options

### The recent developments

The recent developments around non-deterministic and probabilistic programming (aka weighted non-determinism) are very exciting. There are a lot of different implementation techniques with different strengths and weaknesses, like importance sampling, Gibbs sampling, belief propagation, etc. On the other hand, techniques for SAT solving like unit propagation, nogood learning and symmetry reduction have not yet been applied to probabilistic programming (as far as I know). Also, frequentist inference techniques like maximum likelihood & Bayesian loss function minimization can probably be done in the continuous case with automatic differentiation. Perhaps the bag of implementation techniques will continue to expand, or perhaps they will be unified into a single technique that combines the different strengths. In any case this direction of research is very exciting because the way you specify a generative model is often even higher level than the English prose you find in machine learning papers, yet you can automatically and generically do inference with them instead of having to hire a statistician and a computer scientist to develop inference algorithms. If the inference algorithms become good enough then that could transform how business decision making and experimental science are done.

### The elephant in the Bayesian living room

Almost nobody knows how to handle queries like

P[X | Y = Z^2]

generally. Getting probabilistic and logical inference together could start to address this.

It's not like Bayesians don't know it's a problem. It shows up every time they want to observe the output of a deterministic function of random variables. But they can't say what those kinds of queries mean, because Bayes' law for densities can't express it. They just model around it for now.

FWIW, I'm working on this, and having some initial success with an arrow type whose inhabitants calculate conservative approximations of images and preimages. It's a fairly direct translation of the measure-theoretic approach.

### Are you sure it's an

Are you sure it's an elephant? According to various Bayesian algorithms, it might be a pillar, a rope, a tree branch, a hand fan, a wall, or a solid pipe.

### Time to become a

Time to become a Contributing Editor?

### Contributing Editor

You should just force that privilege on people you deem worthy.

### I just force their hand...

I just force their hand...

### Texts?

Are there any comprehensive texts people here would recommend on probabilistic programming as a general topic? Is this something that's sitting in the primary literature at the moment?


text:
### Network

G

#### Geronimo Stempovski

Jan 1, 1970
0
Unfortunately there is an equation I don't quite understand. Maybe someone
is in the mood for discussing and explaining the correctness of the equation
to me.
The formula I am talking about is (1) in the paper [
http://www.ee.ucla.edu/faculty/papers/yang-ckk_ieeeTransCircSystems2_nov2006.pdf ]

For high-common mode signaling (which standard would that be, anyway? TTL?
CMOS? SSTL?) it is assumed
P = V*Vswing/Z0 = V*Vrx/Z0*H(f)

For low-common mode signaling (LVDS? CML? LVPECL?) it states
P = Vswing^2/2*Z0 = Vrx^2/2*Z0*H(f)^2

What I don't understand is the factor 2 (2*Z0) in the calculation of the
low-common mode signaling. Furthermore I'm not sure if the H(f)^2 is
correct.

Any help is highly appreciated! Thanks a lot in advance!

Regards, Gero

O

#### operator jay

Jan 1, 1970
0
Geronimo Stempovski said:
dissipation.
Unfortunately there is an equation I don't quite understand. Maybe
someone is in the mood for discussing and explaining the correctness
of the equation to me.
The formula I am talking about is (1) in the paper [
http://www.ee.ucla.edu/faculty/papers/yang-ckk_ieeeTransCircSystems2_nov2006.pdf ]

For high-common mode signaling (which standard would that be,
anyway? TTL? CMOS? SSTL?) it is assumed
P = V*Vswing/Z0 = V*Vrx/Z0*H(f)

For low-common mode signaling (LVDS? CML? LVPECL?) it states
P = Vswing^2/2*Z0 = Vrx^2/2*Z0*H(f)^2

What I don't understand is the factor 2 (2*Z0) in the calculation of
the low-common mode signaling. Furthermore I'm not sure if the
H(f)^2 is correct.

Any help is highly appreciated! Thanks a lot in advance!

Regards, Gero

I don't know the topic or the equation in question, however,

P = Vpeak/sqrt(2) * Vpeak/sqrt(2) / R = Vpeak^2 / 2 * R
would give power for a simple AC voltage applied to an R. Here the
factor of 2 is 'changing' the voltages to their RMS values. Could it
be similar in the equation you are looking at?

G

#### Geronimo Stempovski

Jan 1, 1970
0
operator jay said:
factor of 2 is 'changing' the voltages to their RMS values. Could it be
similar in the equation you are looking at?

Thank you, Jay. Unforutnately not. Could I mail you the paper and you have a

N

#### Noway2

Jan 1, 1970
0
Geronimo said:
Unfortunately there is an equation I don't quite understand. Maybe someone
is in the mood for discussing and explaining the correctness of the equation
to me.
The formula I am talking about is (1) in the paper [
http://www.ee.ucla.edu/faculty/papers/yang-ckk_ieeeTransCircSystems2_nov2006.pdf ]

For high-common mode signaling (which standard would that be, anyway? TTL?
CMOS? SSTL?) it is assumed
P = V*Vswing/Z0 = V*Vrx/Z0*H(f)

For low-common mode signaling (LVDS? CML? LVPECL?) it states
P = Vswing^2/2*Z0 = Vrx^2/2*Z0*H(f)^2

What I don't understand is the factor 2 (2*Z0) in the calculation of the
low-common mode signaling. Furthermore I'm not sure if the H(f)^2 is
correct.

Any help is highly appreciated! Thanks a lot in advance!

Regards, Gero

NOTE: Cross posting linker were removed and this response was posted
only to sci.electronics.design

I am not familiar with the terms high-common mode and low-common mode,
but the methods you refer to in parentheses seem to suggest single sided
transmission with a common ground return versus differential.

If this is the case, then the 2X factor could be due to that you have
differential signaling and hence two conductors with currents of equal
magnitude and opposite direction flowing in both conductors?

J

#### Jim Granville

Jan 1, 1970
0
Geronimo said:
Unfortunately there is an equation I don't quite understand. Maybe someone
is in the mood for discussing and explaining the correctness of the equation
to me.
The formula I am talking about is (1) in the paper [
http://www.ee.ucla.edu/faculty/papers/yang-ckk_ieeeTransCircSystems2_nov2006.pdf ]

For high-common mode signaling (which standard would that be, anyway? TTL?
CMOS? SSTL?) it is assumed
P = V*Vswing/Z0 = V*Vrx/Z0*H(f)

For low-common mode signaling (LVDS? CML? LVPECL?) it states
P = Vswing^2/2*Z0 = Vrx^2/2*Z0*H(f)^2

What I don't understand is the factor 2 (2*Z0) in the calculation of the
low-common mode signaling. Furthermore I'm not sure if the H(f)^2 is
correct.

Any help is highly appreciated! Thanks a lot in advance!

That does seem mangled.
When in doubt, check the dimensions of the answer ?

Normally where frequency is used in power calcs, it is of the form
of power dissipation capacitance : W = Fo * Cp * Vcc^2

-jg

T

#### Terry Given

Jan 1, 1970
0
Jim said:
Geronimo said:
dissipation.
Unfortunately there is an equation I don't quite understand. Maybe
someone is in the mood for discussing and explaining the correctness
of the equation to me.
The formula I am talking about is (1) in the paper [
http://www.ee.ucla.edu/faculty/papers/yang-ckk_ieeeTransCircSystems2_nov2006.pdf
]

For high-common mode signaling (which standard would that be, anyway?
TTL? CMOS? SSTL?) it is assumed
P = V*Vswing/Z0 = V*Vrx/Z0*H(f)

For low-common mode signaling (LVDS? CML? LVPECL?) it states
P = Vswing^2/2*Z0 = Vrx^2/2*Z0*H(f)^2

What I don't understand is the factor 2 (2*Z0) in the calculation of
the low-common mode signaling. Furthermore I'm not sure if the H(f)^2
is correct.

Any help is highly appreciated! Thanks a lot in advance!

That does seem mangled.
When in doubt, check the dimensions of the answer ?

Normally where frequency is used in power calcs, it is of the form
of power dissipation capacitance : W = Fo * Cp * Vcc^2

-jg

it looks like its analagous to an RMS calculation, but made on an
unknown dataset of a particular (hardware) flavour. its a mere proof by
blatant assertion, so it'd take a fair bit of digging or maths to figure
out where it came from.

ya gotta love professional comics.

Cheers
Terry

F

#### Fred Bloggs

Jan 1, 1970
0
Unfortunately there is an equation I don't quite understand. Maybe someone
is in the mood for discussing and explaining the correctness of the equation
to me.
The formula I am talking about is (1) in the paper [
http://www.ee.ucla.edu/faculty/papers/yang-ckk_ieeeTransCircSystems2_nov2006.pdf ]

For high-common mode signaling (which standard would that be, anyway? TTL?
CMOS? SSTL?) it is assumed
P = V*Vswing/Z0 = V*Vrx/Z0*H(f)

For low-common mode signaling (LVDS? CML? LVPECL?) it states
P = Vswing^2/2*Z0 = Vrx^2/2*Z0*H(f)^2

What I don't understand is the factor 2 (2*Z0) in the calculation of the
low-common mode signaling. Furthermore I'm not sure if the H(f)^2 is
correct.

The background presentation in the paper is somewhat abbreviated because
the authors are implicitly assuming the reader is participating in an
ongoing discussion of fundamental limitations of I/O throughput due to
channel, architectural, and power efficiency considerations. Both power
equations are of the form V*I where V is DC power supply for the
individual channel circuit driver and I is the current switched into the
line termination *at the destination* , making this is a calculation of
power supplied by the DC supply that is dissipated in the termination at
the destination. In the case of HCM the channel driver is powered by
Vdd, making V=Vdd, and the current switched from it into the line is
I=Vswing/Zo, where it is assumed the line is terminated in a matched
resistive impedance Zo, so that P=Vdd*Vswing/Zo. In the case of LCM, the
channel driver is powered by a DC supply of magnitude Vswing and the
driver is a series terminated circuit, by the MOSFET triode region
channel resistance, putting Vswing/2 on the Zo termination at the
destination. Therefore, for LCM, V=Vswing and I=(Vswing/2)/Zo. This may
seem inconsistent with the HCM equation but it is not when you
understand that the assumption in both cases is that the receiver
equalization makes I*Zo=Vrx/H(f). This confusion is just a scale factor
anyway and does not affect the calculation in section II for minimum
energy per bit versus data rate gamma*f over channel H(f).

A

#### [email protected]

Jan 1, 1970
0
This is a nice explanation but still a bit puzzling. Especially
confusing is the fact that
Vswing has a different definition in the cases of LCM and HCM. Also,
according to the paper, the signaling power is that dissipated by the
termination and the lossy transmission line. If Vswing stands for the
voltage measured at the output pin of the transmitter and the line is
lossless, wouldn't the power always be Vswing^2/Z0 ?

Also, in the case of HCM, Vdd is not really the voltage across the
channel' so the product V*I is not really the power dissipated by the
channel+termination. If I have two equal resistances in a series and
put V volts across both resistors and try to measure the power
dissipated by the bottom one, it will not be V*V(across R)/R=V^2/2R
but rather V^2/4R. Why is this not an accurate model of the HCM
transmitter? Or is the load resistor not considered part of the
driver? Or (more likely) they just use their expression because it has
the dimension of power and is related to the actual power dissipated
at the receiving end?

Finally, their analysis does not really care about the 2 in the
denominator (or any other constants for that matter) but does care
about the power of H(f) there. I can belive the fact that LCM is more
sensitive to the frequency response since the HCM transmitter is a
current source and therefore in V*I, the I part does not depend on
H(f) (as noone looses current along the way in a transmission line).
Is this the idea of their analysis (greatly simplified, of course)?

In any case, thanks for the explanation. I would appreciate any
clarification on the questions above.
destination. Therefore, for LCM, V=Vswing and I=(Vswing/2)/Zo. This may
seem inconsistent with the HCM equation but it is not when you
understand that the assumption in both cases is that the receiver
equalization makes I*Zo=Vrx/H(f).

This part is somewhat unclear to me. Regardless of what happens at the
receiver end, wouldn't it always be that
I*Z0=Vout and Vrx=H(f)*Vout so I*Z0=Vrx/H(f) (provided the load looks
resistive and that's essentially what equalization will achieve)?
Essentially, equalization will just make the reactive component of the
load invisible' to the reciever but this is already assumed in all
the models above, so how does this relate to the factor of 2 in
equation (2) in the paper?
This confusion is just a scale factor
anyway and does not affect the calculation in section II for minimum
energy per bit versus data rate gamma*f over channel H(f).

Agreed.

Thanks again

Alex

F

#### Fred Bloggs

Jan 1, 1970
0
This is a nice explanation but still a bit puzzling. Especially
confusing is the fact that
Vswing has a different definition in the cases of LCM and HCM.

The definition is the same and the difference between HCM and LCM is
that HCM powers the channel drivers from the logic level power supply,
Vdd, and LCM powers its drivers from a low voltage power supply on the
order of Vswing.
Also,
according to the paper, the signaling power is that dissipated by the
termination and the lossy transmission line. If Vswing stands for the
voltage measured at the output pin of the transmitter and the line is
lossless, wouldn't the power always be Vswing^2/Z0 ?

Lossless or not, the transmitter puts Vswing across the line at its
output, so the power into the line will be Vswing^2/Zo is correct.
Also, in the case of HCM, Vdd is not really the voltage across the
channel' so the product V*I is not really the power dissipated by the
channel+termination.

V*I is the generic calculation of the power *supplied* by the circuit DC
power source. If the line draws current Vswing/Zo from the driver, then
the DC supply, Vdd, must provide Vswing/Zo amperes to the driver, making
the power provided by the DC supply Vdd*Vswing/Zo. The paper is
interested in computing the power supply requirements as well as the
driver circuit internal dissipation requirements as a function of data rate.

If I have two equal resistances in a series and
put V volts across both resistors and try to measure the power
dissipated by the bottom one, it will not be V*V(across R)/R=V^2/2R
but rather V^2/4R. Why is this not an accurate model of the HCM
transmitter? Or is the load resistor not considered part of the
driver? Or (more likely) they just use their expression because it has
the dimension of power and is related to the actual power dissipated
at the receiving end?

They break the total power drawn from the DC power supply into two
components: the power dissipated by the line and its termination; and
the power dissipated by the driver electronics and whatever internal
source termination it uses. The receiver Vrx, the line loss H(f), and
the characteristic impedance Zo, are all that are necessary to compute
the power required by the line and its termination at a particular
Nyquist frequency f. The driver internal dissipation is not so
straightforward and must be computed on a case by case basis as a
function of CMOS process and circuit topology.
Finally, their analysis does not really care about the 2 in the
denominator (or any other constants for that matter) but does care
about the power of H(f) there. I can belive the fact that LCM is more
sensitive to the frequency response since the HCM transmitter is a
current source and therefore in V*I, the I part does not depend on
H(f) (as noone looses current along the way in a transmission line).
Is this the idea of their analysis (greatly simplified, of course)?

Yes, it is all very straightforward. Once again:

for the HCM case:

the receiver requires Vrx at its input,

this means the transmitter must place Vrx/H(f) at its side of the line,

Vrx/H(f) across the line input requires (Vrx/H(f))/Zo amperes,

the (Vrx/H(f))/Zo amperes supplied by the driver, into the line,
ultimately is supplied by the DC power supply Vdd to the driver,

so the DC power supply Vdd is providing Vdd*(Vrx/H(f))/Zo watts to make
just the line portion of the bit transmission happen.

The LCM case is similar.

In any case, thanks for the explanation. I would appreciate any
clarification on the questions above.

This part is somewhat unclear to me. Regardless of what happens at the
receiver end, wouldn't it always be that
I*Z0=Vout and Vrx=H(f)*Vout so I*Z0=Vrx/H(f) (provided the load looks
resistive and that's essentially what equalization will achieve)?
Essentially, equalization will just make the reactive component of the
load invisible' to the reciever but this is already assumed in all
the models above, so how does this relate to the factor of 2 in
equation (2) in the paper?

First of all, the I at the receiver end is not the same as the I at the
transmitter end. A transmission line does not obey the laws of lumped
element circuit analysis because it is entails energy transmission by
wave propagation. In both LCM and HCM, you have (Vrx/H(f))/Zo amperes
must be injected into the line at the transmitter side. One difference
is that for HCM, Vswing=Vrx/H(f), and for LCM, Vswing/2=Vrx/H(f). The
second difference is that for HCM, the DC voltage supply for the circuit
is Vdd, and for LCM, DC voltage supply for the circuit is Vswing. The
resulting power required of the DC supply for HCM and LCM are then
Vdd*Vrx/(H(f)*Zo), and Vswing*Vrx/(H(f)*Zo), respectively. But because
Vswing/2=Vrx/H(f) in the LCM case, Vswing=2*Vrx/H(f), making the power
in terms of Vrx, 2*(Vrx/H(f))^2/Zo watts, and not (Vrx/H(f))^2/(2*Zo) as
they state in the paper.

A

#### [email protected]

Jan 1, 1970
0
Fred,

V*I is the generic calculation of the power *supplied* by the circuit DC
power source. If the line draws current Vswing/Zo from the driver, then
the DC supply, Vdd, must provide Vswing/Zo amperes to the driver, making
the power provided by the DC supply Vdd*Vswing/Zo. The paper is
interested in computing the power supply requirements as well as the
driver circuit internal dissipation requirements as a function of data rate.

The only remark I have is that this is not the power that goes into
signaling: some of it is spent on driving circuitry in the driver, say
the load resistor in the HCM driver but I guess it is a power metric
as this is the minimal power the power supply must provide for driving
the line.
First of all, the I at the receiver end is not the same as the I at the
transmitter end. A transmission line does not obey the laws of lumped
element circuit analysis because it is entails energy transmission by
wave propagation.

This is certainly true, however, conservation of charge will guarantee
that the integral of I is the same at both ends. This, with
transmitter pre-emphasis (that will take care of phase disparity) is
enough to make the power computation valid. Now, that I have reread
it, what you said makes perfect sense, except that you used

In any case, thanks.

Alex

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• anonymous
Please help! :( The coordinates of the midpoints of the sides of a triangle are A(0, 1), B(4, 0), and C(2, –2). Find the coordinates of the vertices of the triangle. Hint: You should plot out this problem using graph paper! A. (–2, 1); (2, 3); (6, –2) B. (–2, –1); (–2, 3); (5, –3) C. (–2, –2); (0, 3); (0, –1) D. (–2, –1); (2, 3); (6, –3)
Mathematics

Looking for something else?

Not the answer you are looking for? Search for more explanations.


text:
# Homework Help: Need help on Ohm's Law

1. Feb 12, 2005

### maxpayne_lhp

Well, I'm quite idle in Lunar New year's holidays so I decide to do some exercises on Physics, I got stuck when solving the problem, then I tried with the suggestion on my text book, but still, I don't understand the problem, please help me out! btw, sorry, something has gotten on LaTex or I'm too bad at it, so I just place the raw syntax, sorry!
Well, we have the figure:
http://img.photobucket.com/albums/v649/maxpayne_lhp2/5.jpg
And also, we have:
$$\emf=48V, r(Internam resistance)=0\omega$$
$$R_1=2\onega$$
$$R_2=8\onega\]$$
$$R_3=6\onega$$
$$R_4=16\onega$$
Well, they told me to find out the value of $$U_{MN}$$ (Well, in my native scientific system taugh in school,$$U_{MN}$$stands for the potential diferece between M and N.
Well, obviously, I found that
$$U_{MN} = U_{MA}+U_{AN}$$
$$=U_{AN}-U_{AM}$$
Then, I tended to account for the Intensity of the current thru AB via the formula:
$$I=\frac{\emf}{R_{AB}}$$
Then, I found out that I=8 (A)
But soon, I saw something wrong and looked up in the suggestion corner, they told me:
As r=0. we have
$$U_{MN}=\emf.$$ Let the current flows thru AMB and ANB from A to B, then:
$$U_{AN}=\frac{U_{AB}}{R_2+R_4} . R_2 = \frac{\emf}{R_2+R_4} . R_2= 16V$$
Do the same thing with $$U_{AM}$$
Though, I don't really understand what they mean, so please help me out! What did I do wrong?
Thanks for your time and help!

Last edited: Feb 12, 2005
2. Feb 12, 2005

### Andrew Mason

There is no resistance between M and N so there is no potential difference. M and N are at the same potential.

To find the potential difference across R1 and R2, work out resistance for R1 and R2 in parallel: $1/R_{1,2}=1/R_1 + 1/R_2$ and for R3 and R4 in parallel: $1/R_{3,4}=1/R_3 + 1/R_4$. Add $R_{total}=R_{1,2}+R_{3,4}$. Then work out the total current using $I = V/R_{total}$. That will allow you to work out the potential from A to M and from M to B.

AM

3. Feb 12, 2005

### maxpayne_lhp

Ohh, sorry, I'm so clumsy, I didn't mean to draw a straight line connecting M and N, there was nothing connecting them at all. yes, they meant to cross the resistances.
And that's helped. Thanks!
BTW, I got difficulty when I need to go down for the next line, what's the syntax in LaTex for it? If you don't have time, please send me a link so that I can refer. Thanks alot!

Last edited: Feb 12, 2005
4. Feb 12, 2005

### xanthym

Then the currents I_13 thru R_1 & R_3 and I_24 thru R_2 & R_4 are given by:

$$:(1): \ \ \ \ I_{13} = \frac {E} {R_1 + R_3}$$

$$:(2): \ \ \ \ I_{24} = \frac {E} {R_2 + R_4}$$

from which you can determine all required voltage drops. The overall equivalent resistance R will then be:

$$:(3): \ \ \ \ R = \frac {E} {I_{13} + I_{24}} = \frac {1} { \frac {1} {R_1 + R_3} + \frac {1} {R_2 + R_4} }$$

~~

Last edited: Feb 12, 2005
5. Feb 12, 2005

### maxpayne_lhp

Oh thanks, I got it, it's $$U_{MN}=4V$$
Thanks alot!

6. Feb 12, 2005

### maxpayne_lhp

Excuse me, still, I don't know what
Thanks!
btw, if we figure the another way, as thru R3 and R4, is it the same?
Thanks

7. Feb 12, 2005

### xanthym

The equivalent resistance R is the value of a SINGLE resistor that can REPLACE ALL the resistors R_1, R_2, R_3, and R_4 between your battery. In other words, this resistor R will draw the same current that ALL your other resistors draw TOGETHER when connected to a battery. It's useful to know R in terms of the OVERALL operation of ALL the resistors. However, you still need to use the individual resistor values to determine voltage drops between the individual resistors inside your circuit -- just like you were doing.

8. Feb 12, 2005

### maxpayne_lhp

Well, okay, Thanks!
Thanks for all of your time and help!


text:
# Far point: Difference between revisions

In visual perception, the far point is the point at which an object must be placed along the optical axis of the eye for its image to be focused on the retina when the eye is not accommodating. It is sometimes described as the farthest point from the eye at which images are clear.

For an unaccommodated emmetropic eye, the far point is at infinity, but for the sake of practicality, infinity is considered to be 6 m because the accommodation change from 6 m to infinity is negligible.

For an unaccommodated myopic eye, the far point is closer than 6 m. It depends upon the refractive error of the person's eye .

For an unaccommodated hypermetropic eye, incident light must be converged before entering the eye so as to focus on the retina. In this case (the hypermetropic eye) the focus point is behind the retina in virtual space, rather than on the retina screen.

## Optical Calculation

Given the far point ${\displaystyle FP}$ (in meters) of a person, the optical power of a convex lens (i.e. converging lens) necessary for them to see distant object clearly can be calculated as below[1]:

1. ^ "Vision Correction | Physics". courses.lumenlearning.com. Retrieved 2019-12-05.


text:
# qwerkyapp.com

Home > Error Function > Error Function Complementary Approximation

# Error Function Complementary Approximation

## Contents

Should I ever use the pronoun "ci"? Springer-Verlag. The error and complementary error functions occur, for example, in solutions of the heat equation when boundary conditions are given by the Heaviside step function. This note presents a closed-form approximation for the error, complementary error, and scaled complementary error functions, with maximum relative errors within 0.8%. this content

Pathak, Evaluation of Blocking Probability for Downlink in Poisson Networks, IEEE Wireless Communications Letters, 2015, 4, 6, 625CrossRef3Renato Cicchetti, Antonio Faraone, Gianni Orlandi, Diego Caratelli, Real-Argument Incomplete Hankel Functions: Accurate and Mankar, Bibek R. and Stegun, I.A. (Eds.). "Repeated Integrals of the Error Function." §7.2 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. doi:10.3888/tmj.16–11.Schöpf, Supancic ^ E.

## Complementary Error Function Table

I have used this in my work and got incredible accuracy with just one term in the sum. It is defined as:[1][2] erf  ( x ) = 1 π ∫ − x x e − t 2 d t = 2 π ∫ 0 x e − t In order of increasing accuracy, they are: erf  ( x ) ≈ 1 − 1 ( 1 + a 1 x + a 2 x 2 + a 3 x

1. However, it can be extended to the disk |z| < 1 of the complex plane, using the Maclaurin series erf − 1  ( z ) = ∑ k = 0
2. Positive integer values of Im(f) are shown with thick blue lines.
3. If L is sufficiently far from the mean, i.e. μ − L ≥ σ ln  k {\displaystyle \mu -L\geq \sigma {\sqrt {\ln {k}}}} , then: Pr [ X ≤ L
5. Math.
6. The approximation is very useful in solving atmospheric science problems by providing analytical solutions.
7. R.
8. ISBN978-1-4020-6948-2. ^ Winitzki, Sergei (6 February 2008). "A handy approximation for the error function and its inverse" (PDF).
9. The closed solution can be incorporated into analytical solutions for a broad range of physical and engineering problems.An example of one such problem, and the driver for achieving the above expression,
10. Supancic, "On Bürmann's Theorem and Its Application to Problems of Linear and Nonlinear Heat Transfer and Diffusion," The Mathematica Journal, 2014.

New York: Dover, pp.299-300, 1972. Derivative and integral The derivative of the error function follows immediately from its definition: d d z erf  ( z ) = 2 π e − z 2 . {\displaystyle I consider $\tanh{(x)}$ to be the somewhat prettyier function, and so I wanted to find an approximation to $\text{erf}$ with "nice functions" by a short expression. Complementary Error Function In Matlab Another form of erfc  ( x ) {\displaystyle \operatorname  2 (x)} for non-negative x {\displaystyle x} is known as Craig's formula:[5] erfc  ( x | x ≥ 0

The denominator terms are sequence A007680 in the OEIS. Complementary Error Function Calculator Bonetto, Path suppression of strongly collapsing bubbles at finite and low Reynolds numbers, Physical Review E, 2013, 87, 6CrossRef Advertisement PDF PDF Info References Figures Close article support pane Browse Publications Both functions are overloaded to accept arguments of type float, double, and long double. recommended you read For previous versions or for complex arguments, SciPy includes implementations of erf, erfc, erfi, and related functions for complex arguments in scipy.special.[21] A complex-argument erf is also in the arbitrary-precision arithmetic

For large enough values of x, only the first few terms of this asymptotic expansion are needed to obtain a good approximation of erfc(x) (while for not too large values of Complementary Error Function Mathematica Spanier, J. Sometimes, when one wants to evaluate these functions as accurately as possible, rational Chebyshev approximations (Cody, 1969) can be used. New Exponential Bounds and Approximations for the Computation of Error Probability in Fading Channels.

## Complementary Error Function Calculator

The value of the constant and the way of adding the constant are restrained by the requirement of Equations (5) and (6). http://onlinelibrary.wiley.com/doi/10.1002/asl.154/pdf Contact the MathWorld Team © 1999-2016 Wolfram Research, Inc. | Terms of Use THINGS TO TRY: inverse erf inverse erfc erf current community blog chat Mathematics Mathematics Meta your communities Sign Complementary Error Function Table LCCN65-12253. Complementary Error Function Excel Mathematics and Computation 23: 631–637.CrossRef | Web of Science Times Cited: 70Cody WJ. 1990.

Interactive Entries>webMathematica Examples> History and Terminology>Wolfram Language Commands> Less... news M. Press, William H.; Teukolsky, Saul A.; Vetterling, William T.; Flannery, Brian P. (2007), "Section 6.2. Your cache administrator is webmaster. Inverse Complementary Error Function

Boston, MA: Academic Press, p.122, 1997. Wolfram Education Portal» Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. Cody's rational Chebyshev approximation algorithm.[20] Ruby: Provides Math.erf() and Math.erfc() for real arguments. have a peek at these guys I obtained that $k$ value by the requirement that "norm" given by $\int_0^\infty\text{erf}(x)-f(x)dx,$ i.e.

Indeed, Φ ( x ) = 1 2 π ∫ − ∞ x e − t 2 2 d t = 1 2 [ 1 + erf  ( x 2 Complementary Error Function Ti 89 What does Peter Dinklage eat on camera in Game of Thrones? Weisstein. "Bürmann's Theorem" from Wolfram MathWorld—A Wolfram Web Resource./ E.

## However, I found that for $k=\sqrt{\pi}\log{(2)}$ the situation gets "better".

For complex, the Faddeeva package provides a C++ complex implementation. ACM Transactions on Mathematical Software 16: 29–37.CrossRef | Web of Science Times Cited: 5Decker DL. 1975. H. Gamma Function Approximation This directly results from the fact that the integrand e − t 2 {\displaystyle e^{-t^ − 2}} is an even function.

Journal of Geophysical Research 107(D23):4698, DOI: 10.1029/2001JD001429.Wiley Online Library | Web of Science Times Cited: 40Lohmann U, Kärcher B, Hendricks J. 2004. I'm asking for the cases where we need to avoid letting the total area go over 1. –Mehrdad Jan 24 '14 at 5:06 3 Hello. The system returned: (22) Invalid argument The remote host or network may be down. http://qwerkyapp.com/error-function/error-function-erf-approximation.html Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply.

Not the answer you're looking for? Decker (1975) provides and cites approximations that are quick to compute, but all fail to give expressions in closed form. Given a value of a, the scaled complementary error function can be estimated by Equation (8), then the complementary error and error functions can be approximated in turn according to their How was photo data processed and transferred back to Earth from satellites in the pre-digital era?

For integer , (16) (17) (18) (19) (Abramowitz and Stegun 1972, p.299), where is a confluent hypergeometric function of the first kind and is a gamma function. One can calculate F(I) with a given λ by a calculator, even by hand.Having derived an analytical expression for the cumulative distribution function, the recurrence interval is then (11)This relatively simple The steps taken to find Equation (8) are: multiplying the numerator and denominator of Equation (7) by a, separating the denominator into , and adding a constant to the latter part. The purpose being for example, that if I see somewhere that for a computation I have to integrate erf, that I can think to myself "oh, yeah that's maybe complicated, but

For |z| < 1, we have erf  ( erf − 1  ( z ) ) = z {\displaystyle \operatorname ζ 2 \left(\operatorname ζ 1 ^{-1}(z)\right)=z} . These values are given in Table I, together with their accuracies when used in Equation (8) to calculate eerfc(x), and hence erf(x) and erfc(x). Referenced on Wolfram|Alpha: Erfc CITE THIS AS: Weisstein, Eric W. "Erfc." From MathWorld--A Wolfram Web Resource. Some authors discuss the more general functions:[citation needed] E n ( x ) = n ! π ∫ 0 x e − t n d t = n ! π ∑

MathCAD provides both erf(x) and erfc(x) for real arguments. Each line is for a single value of a. the approximation is compact/rememberable while the values are even better, from a numerical point of view. J.; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W., NIST Handbook of Mathematical Functions, Cambridge University Press, ISBN978-0521192255, MR2723248 External links MathWorld – Erf Authority control NDL: 00562553 Retrieved from

Find duplicates of a file by content Four line equality A positive integer gets reduced by 9 times when one of its digits is deleted.... And however you choose the integration bounds for an interval, the area difference is no more than $0.017$. However, for −1 < x < 1, there is a unique real number denoted erf − 1  ( x ) {\displaystyle \operatorname Γ 0 ^{-1}(x)} satisfying erf  ( erf A parameterization of cirrus cloud formation: Homogeneous freezing including effects of aerosol size.

Computers and Geosciences 32: 1007–1024.CrossRef | Web of Science Times Cited: 4 | ADSSwamee PK. 2002. is the double factorial: the product of all odd numbers up to (2n–1).