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## Introduction

With the advent of resting-state fMRI, there has been an explosion of interest in characterizing intrinsic neural networks, as well as describing their various contributions to human cognition and behavior. Of particular interest to researchers has been the default mode network (DMN), a functionally-related system of regions which show greater metabolic activity at rest compared to task1. Regions of the DMN include posterior cingulate, medial-prefrontal, hippocampal, and lateral temporal areas, and are often defined as regions showing strong functional connectivity with the posterior cingulate during rest2. While the exact function of the DMN has proven elusive, it is involved in a wide variety of cognitions, such as autobiographical memory, spontaneous thought, and the integration of social information3,4,5,6. Furthermore, disruption of DMN connectivity is linked to psychopathological states such as schizophrenia, ADHD, conduct disorder, and depression7,8,9,10.

One key form of task engagement is feedback learning, during which individuals must maintain mental representations of the task structure and reinforcement history in order to guide future behavior18,19,20. Processes which impact feedback learning are important during adolescence, as teens are particularly sensitive to performance-relevant cues in their environment, showing both developmental19,21 and inter-individual20,22,23 differences in sensitivity to positive and negative feedback information. Given the role of DMN in disrupting attentional and engagement processes14,15, failure to suppress DMN regions during task should be related to decreased feedback learning, as adolescents disengage (as a result of decreased attention and/or motivation) from task information. This disengagement could have particularly important consequences for adolescents, as deficits in the ability to learn from feedback can have negative impacts on adolescent behavior19,20.

Given the transient nature of DMN suppression during task, one key unanswered question remains when considering potential mechanisms involved in down-regulating (or suppressing) the default mode network during task engagement: Does the DMN need to show reduced (i.e., a decrease in the absolute level of activation), or more stable (i.e., reduced fluctuations in activation) activity to achieve suppression? This question distinguishes between the mean level of neural activation (or deactivation as the case may be) and the stability over time of DMN activity as the mechanism by which the brain reduces the influence of functional networks on behavior. Under the first hypothesis, we would expect that individuals who show greater task engagement (indexed by increased feedback learning) would show the greatest DMN deactivation during task. Alternatively, feedback learning may be reflected in more stable DMN activity over time, as suppression of the DMN causes the network to become less responsive to changing task dynamics. Importantly, these two explanations may not be mutually exclusive, as stable and strong deactivation might co-occur.

In the current study, we tested the hypothesis that stability in task-negative regions of the DMN contributes to feedback learning (i.e., participants’ ability to extract and respond to information cues from the task environment), over and above absolute level of deactivation, against the alternative hypothesis that suppression of the DMN is primarily achieved through reductions in univariate activation (i.e., deactivation). To do so, adolescents completed a risky decision-making task, the Balloon Analog Risk Task (BART) during functional magnetic resonance imaging (fMRI). Using an ROI-based approach for both traditional univariate and model-based network analyses, we extracted parameter estimates of DMN deactivation and stability for each individual. We then entered these two types of parameters as simultaneous predictors of adolescents’ engagement on the task. While previous work has focused on the absolute deactivation of DMN, we hypothesized that DMN stability would be a more powerful predictor of feedback learning than absolute level of deactivation in these regions. However, we further predicted that stability and level of deactivation would be positively related, with stronger deactivation being associated with greater DMN stability, suggesting a possible reconciliation of this hypothesis with previous conceptualizations of the DMN during task.

## Methods

Sixty-seven adolescent participants completed an fMRI scan. One participant was scanned using the wrong head coil, and another was excluded for excessive movement (>10% of slices with movement in excess of 2 mm), resulting in a final sample of 65 adolescents (Mage = 13.32, SD = 0.62, range = 12.42–14.83; 56 Caucasian, 2 African American, 7 mixed race/multiple responses). Participants were largely from high income households (1 $0–14,999; 3$15–29,999; 6 $30–44,999; 4$45–59,999; 8 $60–74,999; 9$75–89,999; 30 > $90,000; and 4 not reported), with highly educated parents (3 completed a high school diploma; 8 completed some college; 4 completed an associate degree; 21 completed a bachelor’s degree; 5 completed some graduate school; 18 completed a master’s degree; and 3 completed a professional degree). Written informed assent was obtained for all participants under the age of 18, and written informed consent was obtained from each participants’ parent and/or legal guardian. All methods were carried out in accordance with the relevant guidelines and regulations outlined by the Declaration of Helsinki and experimental protocols were approved by the University of Illinois, Urbana-Champaign Institutional Review Board. ### Risky Decision-Making Task Participants completed a version of the Balloon Analogue Risk Task (BART), a well-validated experimental paradigm24,25 that has been adapted for fMRI in developmental populations19,26. The BART measures participants’ willingness to engage in risky behavior in order to earn rewards, and is associated with real-life risk taking in adolescents20,27 and adults24,25. During the scan session, participants were presented with a sequence of 24 balloons that they could pump up to earn points. Each pump decision was associated with earning one point but also increased the risk that a balloon would explode. If participants pumped a balloon too many times, the balloon would explode and participants would lose all the points they had earned for that balloon. However, if participants chose to cash out before the balloon exploded, the points they earned would be added to the running total of points, which was presented on the screen as a points meter. Participants were instructed that their goal was to earn as many points as possible during the task. Each event (e.g., larger balloon following a pump, new balloon following cashed or explosion outcomes) was separated with a random jitter (500–4000 ms). Balloons exploded after 4 to 10 pumps, and the order of balloons was presented in a fixed order (after being pseudo-randomly ordered prior to data collection), although none of this information was made available to participants. The BART was self-paced and would not advance unless the participant made the choice to either pump or cash out. Participants were told that they could win a$10 gift card at the end of the neuroimaging session if they earned enough points during the task. The point threshold for winning this prize was intentionally left ambiguous so that participants were motivated to continue earning points throughout the task. In reality, all participants were given a \$10 gift card after completing the scan session.

To measure adolescents’ task engagement during the risky decision-making task, we calculated two indices of feedback learning. Specifically, we were interested in how adolescents used feedback information from previous trials in order to guide their current behavior, and to adapt that behavior when it results in maladaptive outcomes26. Previous research using the BART has shown that adolescence is a time of increased feedback learning (compared with childhood), and that individual differences in feedback learning predict differences in risk behavior19. In the current study, we were interested in two types of feedback learning to measure task engagement.

To obtain these indices, we took a multi-level modeling approach utilizing the SAS software package (SAS version 9.4; SAS Institute Inc., Durham, NC), in which trials (24 balloons) were nested within adolescents, and the level 1 outcome was the final number of pump decisions made on a given balloon. To obtain our learning indices, we modeled pump number at the trial level as dependent on (1) previous feedback and (2) the size of that feedback. Consistent with previous research27,28, we also controlled for the overall trial number and the outcome of the current balloon, resulting in the following Level 1 equation:

$$\begin{array}{c}Number\,of\,Pump{s}_{ij}={\gamma }_{0j}+{\gamma }_{1j}Trial\,Numbe{r}_{ij}+{\gamma }_{2j}Current\,Outcom{e}_{ij}\\ \,+\,{\gamma }_{3j}Previous\,Outcome\,Valenc{e}_{ij}+{\gamma }_{4j}Previous\,Outcome\,Valu{e}_{ij}+{\mu }_{0j}+{\varepsilon }_{ij}\end{array}$$

Total pumps on a particular balloon trial (i) for a given adolescent (j) was modeled as a function of the average number of pumps across the task (γ0j), the trial number (γ1j; range = 0–23), the outcome of the current trial (γ2j; coded Cash-Out = 0, Explosion = 1), the outcome of the previous trial (γ3j; coded Cash-Out = 0, Explosion = 0), and the size of the previous outcome (γ4j), in which we calculated a 75th percentile threshold for each participant’s pump behavior based on their individual data. For previous trial outcomes on balloons where adolescents pumped above this threshold, the predictor was coded as 1, as points earned or lost on these trials were high value for the individual, while all other previous trials were coded as 0, indicating lower value earnings or loss28. As our focus was on adolescents’ individual sensitivity to the valence and value of previous feedback, these parameters were allowed to vary randomly in our model in order to gain individual effect estimates. Nesting of trials within balloons was achieved by modeling a between-person random intercept (μ0j) assumed to be independent and identically distributed and follow a Normal distribution with a constant variance (i.e., $${u}_{0j} \sim {\rm{N}}[0,{\tau }_{00}]$$). Finally, the individual-level residuals errors (εij) were assumed to be independent and identically distributed, following a Normal distribution with a constant variance (i.e., $${\varepsilon }_{ij} \sim {\rm{N}}[0,{\sigma }^{2}]$$). In order to use these two metrics of task engagement, we extracted empirical Bayes estimates for each adolescent. Empirical Bayes estimates are optimally-weighted averages which combine individual- and group-level slope estimates, and “shrink” individual slope estimates towards group mean effect29.

### fMRI Data Acquisition and Processing

#### fMRI data acquisition

Imaging data were collected utilizing a 3 Tesla Trio MRI scanner. The BART included T2*-weighted echoplanar images (EPI; slice thickness = 3 mm; 38 slices; TR = 2 sec; TE = 25 ms; matrix = 92 × 92; FOV = 230 mm; voxel size = 2.5 × 2.5 × 3 mm3). Additionally, structural scans were acquired, including a T1* magnetization-prepared rapid-acquisition gradient echo (MPRAGE; slice thickness = 0.9 mm; 192 slices; TR = 1.9 sec; TE = 2.32 ms; matrix = 256 × 256; FOV = 230 mm; voxel size = 0.9 × 0.9 × 0.9 mm3; sagittal plane) and a T2*-weighted, matched-bandwidth (MBW), high resolution, anatomical scan (slice thickness = 3 mm; 192 slices; TR = 4 sec; TE = 64 ms; matrix = 192 × 192; FOV = 230 mm; voxel size = 1.2 × 1.2 × 3 mm3). EPI and MBW scans were obtained at an oblique axial orientation in order to maximize brain coverage and minimize dropout in orbital regions.

#### fMRI data preprocessing and analysis

Preprocessing utilized FSL FMRIBs Software Library (FSL v6.0; https://fsl.fmrib.ox.ac.uk/fsl/). Steps taken during preprocessing included correction for slice-timing using MCFLIRT; spatial smoothing using a 6 mm Gaussian kernel, full-width-at-half maximum; high-pass temporal filtering with a 128 s cutoff to remove low frequency drift across the time-series; and skull stripping of all images with BET. Functional images were re-sampled to a 2 × 2 × 2 mm space and co-registered in a two-step sequence to the MBW and the MPRAGE images using FLIRT in order to warp them into the standard stereotactic space defined by the Montreal Neurological Institute (MNI) and the International Consortium for Brain Mapping. Preprocessing was completed utilizing individual-level independent component analysis (ICA) with MELODIC combined with an automated component classifier30 (Neyman-Pearson threshold = 0.3), which was applied to filter signal originating from noise sources (e.g., motion, physiological rhythms). Global signal regression was not performed due to its tendency to increase distance-related dependencies in the strength of functional connectivity measures31.

#### Motion Correction

Prior to modeling the fMRI data further, we took several steps to reduce the influence of motion. First, as mentioned previously, we subjected each participants’ data to individual-level ICA in order to remove motion-related signal from each participants’ time-series. We also controlled for 8 nuisance regressors in the GLM and time-series analyses: 6 motion parameters generated during realignment and the average signal from both the white matter and cerebrospinal fluid masks. Finally, slices with greater than 2 mm of motion were censored from the time-series (or modeled as a junk regressor in the GLM) to remove the effects of large, sudden movements on the functional data. No participant exceeded 5% of slices being censored (range: 0–2.5%). Previous work has shown that these strategies effectively reduce the influence of motion on functional connectivity analyses31.

### Regions of Interest

To estimate how autoregressive stability in the default mode network impacts task behavior, we constructed 10 a priori regions of interest (ROIs) based on previous neuroimaging work with this network (Fig. 1). We based our ROIs off of resting-state maps of the default mode network32,33. Regions included the medial prefrontal cortex (mPFC), posterior cingulate cortex (PCC), bilateral dorsal superior frontal gyrus (dSFG), bilateral temporal poles (TP), bilateral hippocampus, and bilateral angular gyrus (AG). Regions were extracted from the templates using FSL. Most ROIs showed good separation from other DMN regions, however, in order to separate the mPFC and bilateral dSFG (which overlap in standard maps), we took additional steps to create 3 separate ROIs by zeroing voxels between the two regions with a z-stat < 6. This approach achieved appreciable separation for our masks. Individual masks were then evaluated again using the Marsbar toolbox in SPM34 and FSL to ensure that ROIs did not contain any voxels that overlapped with another mask or exceeded the boundaries of the whole-brain mask. A 3D, navigable image containing all masks superimposed onto a single brain map is available on NeuroVault (https://neurovault.org/collections/VSWQSTDA/35.

### Whole-brain Univariate Analyses

For our univariate analyses, we modeled the BART as an event-related design. Whole-brain statistical analyses were performed using the general linear model (GLM) in SPM8. Fixed effects models were constructed for each participant with several conditions of interest, including pump decisions, cash-outs, and explosion events. The jittered inter-trial period between pump decisions and between outcomes and a new balloon was not modeled and served as an implicit baseline. A parametric modulator (PM) was included for conditions of interest and corresponded to the current pump number for the current trial at the time of event. This PM serves to control for differences across pumps within a balloon. For descriptive purposes, we ran whole-brain, group-level, random effects analyses for pump decisions using GLMFlex (http://mrtools.mgh.harvard.edu/index.php/GLM_Flex). This approach corrects for variance-covariance inequality, removes outliers and sudden activation changes in the brain, partitions within- and between-person error terms, and analyzes all voxels containing data. Since this analysis was purely for visualization purposes, we thresholded the resultant statistical image at p < 0.001, with a minimum cluster size of 40 voxels. For the central univariate analyses, we extracted parameter estimates from each individual’s unthresholded fixed effects statistical map using the 10 a priori ROIs we constructed for the DMN network.

### Time-series Analysis

#### Granger Causality

Originally developed in the context of economic models36, Granger causality emerges from a vector autoregression framework (VAR), where the contemporaneous and lagged relationships between a system of variables can be examined. A weak form of causal inference (compared to experimental designs for example), Granger causality relies on the intuition that x cannot cause y if x temporally follows y (i.e., cause precedes effect)37. Based on this idea, Granger causality can be inferred if x at a previous time point (e.g., t − 1) predicts y at time (t) above and beyond the self-predictive effect of y at t − 1 on y at t. In other words, the combined information of x(t−1) and y(t−1) is more predictive of y(t) than is y(t−1) alone. Under this causal definition, it is possible for variables to Granger cause one another across time38.

#### Group Iterative Multiple Model Estimation (GIMME)

GIMME is a model-based network approach, which utilizes both individual and group-level information to derive directed functional connectivity maps39. GIMME estimates connectivity graphs using both unified SEM40 and extended unified SEM41 to assess whether the presence of a path between ROIs significantly improves the overall model fit to the time-series data. GIMME estimates both contemporaneous (e.g., ROI1 at t predicts ROI2 at t) and lagged (e.g., ROI1 at t − 1 predicts ROI2 at t) effects between ROIs, as well as the autoregressive (e.g., ROI1 at t − 1 predicts ROI1 at t) effects for each ROI time-series. Formally, a GIMME model for a set of p time-series with t measurements is:

$${\eta }_{i}(t)=({A}_{i}+{A}_{{i}^{g}}){\eta }_{i}(t)+({{\rm{\Phi }}}_{1,i}+{{\rm{\Phi }}}_{1,{i}^{g}}){\eta }_{i}(t-1)+{\zeta }_{i}(t)$$

where A represents a p × p matrix of contemporaneous paths for the individual (Ai) and group ($${A}_{{i}^{g}}$$) parameters, Φ1 is matrix of first-order lagged paths (for the individual and group respectively), and ζ is the p-length vector of errors, assumed to be a white noise process, with means of zero, a finite variance, and no sequential dependencies (i.e., all temporal information is contained within A and Φ1)42. The diagonal of Φ1 contains path estimates for the autoregressive effects (e.g., ROI1 at t − 1 predicts ROI1 at t), which represent the autocorrelations of each ROI predicting itself forward in time. GIMME assesses directional paths by testing whether a given ROI can predict another, controlling for the predicted ROI’s autoregressive effect (i.e., establishing Granger causality). GIMME has been developed for both block40 and event-related41 fMRI data, and is freely available through the open-source R platform43.

In contrast with many other functional connectivity approaches (e.g., graph theoretical approaches), GIMME constructs functional maps through a model-driven, multi-step processing of model building and pruning. First, information across all participants is used to derive a common network map that is representative of the majority of the sample. Group paths are only retained if they are significant for 70% of all individuals in the sample. All autoregressive paths are automatically estimated in order to accurately assess directionality in the between-ROI paths. Once a group map has been obtained, additional paths at the individual level are evaluated based on improvements to model fit for that individual. Unnecessary paths are pruned at the group level, and additional paths at the individual level are evaluated based on improvements to model fit for that individual. Individual-level paths are then pruned if they do not significantly improve the fit of the final model. This approach offers the unique advantage of being able to derive a group-level map that should be applicable to the majority of the sample, while still recognizing that individuals often show significant heterogeneity from the group map. This approach shows significant advantages over other methods in recovering “true” paths in simulated data while minimizing false positives44.

Our task provided two main challenges when measuring neural connectivity. First, our goal was to analyze connectivity patterns during risky decisions; however, the BART also contains feedback trials (i.e., cash-out outcomes, explosions). Secondly, our task was self-paced and as such, we needed a modeling approach that would allow for individuals to possess different amounts of data. Fortunately, GIMME is capable of handling unequal amounts of data between participants, as well as the inclusion of missing data45. Missing values are replaced with placeholder NaN values to maintain the temporal ordering of scans, and neither contemporaneous nor lagged effects are estimated based on missing values. These features make GIMME especially well-suited to estimating connectivity graphs for the BART, allowing for the self-paced nature of the task, as well as specifically examining connectivity during risk decisions, without considering connectivity during outcomes.

#### Autoregressive Paths

Autoregressive pathways are estimated as the predictive effect of activity in an ROI at one time point on that ROI’s activity at the next time point. As such, stronger autoregressive path indicate that an ROI’s activation is more stable over time. In our analyses, we were specifically interested in testing whether the strength of an individual’s autoregressive paths was related to task behavior on the BART. As such, the parameter estimates from GIMME for each ROI’s autoregressive path were extracted for use in subsequent regression analysis.

### Analytic Plan

To test the hypothesis that stability versus deactivation of the default mode network would be important for feedback learning, we took two analytic approaches. First, we utilized standard univariate analyses to extract each individual’s parameter estimates of deactivation in the 10 a priori DMN ROIs during risk decisions. Secondly, we took a model-based network approach using the same 10 ROIs to estimate stability over time in DMN activation. After completing both univariate and network analyses, each participant had 10 parameter estimates of mean activation and 10 parameter estimates of stability in activation over time. To avoid concerns of multiple comparisons by regressing each of these 20 estimates on behavioral metrics of interest, we took a dimension reduction approach through principal components analysis (PCA). PCA also offers a key advantage by partitioning variance into bins: variance that is common across regions (and representative of the DMN as a network), and variance that is unique to a particular region. Because activation in a given region is likely a combination of network- and ROI-level information, partitioning this region-specific variance out helps remove noise from our estimates that originate from individual ROIs. For both sets of parameters, we extracted the first principal component from a PCA where estimates from all 10 ROIs were used as inputs. We utilized the R function, “principal” (https://cran.r-project.org/web/packages/psych/psych.pdf), to extract the first principal component utilizing the covariance matrix and varimax rotation. We ran follow-up analyses using the cross-validation function in the R function “pca” (https://cran.r-project.org/web/packages/mdatools/index.html), and results remained unchanged. These principal component scores were then used in subsequent regression analyses to predict task engagement (i.e., sensitivity to the valence and value of feedback in the task).

## Results

### Group-Level Results

#### DMN Deactivation During Risk Decisions

We first ran main effects analysis at the whole brain level, for descriptive purposes, to check for the expected deactivation of the default mode regions during decisions to pump across individuals in the sample. Consistent with prior work2, adolescents showed strong deactivation of default mode regions during risk decisions at the group level, including mPFC, PCC, STS, AG, dSFG, and hippocampus (Fig. 2; Table 1). In contrast, typical task-active regions such as the anterior cingulate, anterior insula, and motor cortex showed positive activation during risk decisions. However, substantial individual differences emerged such that not all adolescents showed strong deactivation of DMN regions, and some adolescents even showed positive activation of DMN regions during risk decisions (Fig. 3).

#### Network Map of the DMN During Risk Decisions

Next, we constructed model-based functional networks between our a priori DMN ROIs. While our focus was on the autoregessive paths, we estimated and displayed the full group model for descriptive purposes. Results show a strongly interconnected default mode network (Fig. 4). In addition to connections between bilateral complements (e.g., left and right dSFG), the PCC, left TP, and left AG show many between-region paths. Importantly for testing our hypothesis, the autoregressive pathways for each of the 10 DMN ROIs were estimated for all subjects.

#### Individual Differences in DMN Deactivation and Stability Differentially Predict Feedback Learning

Finally, our key analysis explored individual differences in deactivation (from the univariate analyses) and stability (from the functional network analyses) in DMN regions to task engagement. We took parameter estimates of univariate activation and autoregressive strength for each of the 10 a priori DMN regions and ran separate PCA analyses on each type of parameter and obtained one score per person, per analysis. For univariate analyses, this analysis resulted in a representative level of deactivation across DMN regions. For network analyses on the autoregressive paths, the PCA score reflected representative activational stability in the DMN as a whole (see Table 2 for factor loadings to each principal component).

Next, we entered both of the scores into a multiple regression analysis with adolescents’ two indices of feedback learning (i.e., sensitivity to the valence and sensitivity to the value of previous feedback) as our outcomes in two separate analyses (thresholded at p = 0.025 to correct for multiple comparisons). Results showed that DMN stability (B = 0.063, SE = 0.026, p = 0.018) but not mean deactivation (B = 0.025, SE = 0.026, p = 0.349), is associated with adolescents’ sensitivity to the valence of the previous outcome. Similarly, stability (B = 0.150, SE = 0.048, p = 0.003) but not deactivation (B = 0.033, SE = 0.048, p = 0.487) is associated with adolescents’ sensitivity to the value (i.e., magnitude) of previous feedback on the task. Furthermore, the two factor scores were uncorrelated (r = 0.112, p = 0.376), meaning that deactivation did not indicate more stability in DMN activity, nor were interactions between deactivation and stability predictive of feedback learning (p = 0.186 and p = 0.963 respectively). These results suggest that stability in DMN activity, even if that activity is positive on average (as is characteristic of some adolescents; Fig. 3), is more predictive of adolescents’ feedback learning than absolute level of (de)activation.


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# 2.2 The limit of a function  (Page 7/12)

 Page 7 / 12

## Key concepts

• A table of values or graph may be used to estimate a limit.
• If the limit of a function at a point does not exist, it is still possible that the limits from the left and right at that point may exist.
• If the limits of a function from the left and right exist and are equal, then the limit of the function is that common value.
• We may use limits to describe infinite behavior of a function at a point.

## Key equations

• Intuitive Definition of the Limit
$\underset{x\to a}{\text{lim}}f\left(x\right)=L$
• Two Important Limits
$\underset{x\to a}{\text{lim}}x=a\phantom{\rule{1em}{0ex}}\underset{x\to a}{\text{lim}}c=c$
• One-Sided Limits
$\underset{x\to {a}^{-}}{\text{lim}}f\left(x\right)=L\phantom{\rule{1em}{0ex}}\underset{x\to {a}^{+}}{\text{lim}}f\left(x\right)=L$
• Infinite Limits from the Left
$\underset{x\to {a}^{-}}{\text{lim}}f\left(x\right)=\text{+}\infty \phantom{\rule{1em}{0ex}}\underset{x\to {a}^{-}}{\text{lim}}f\left(x\right)=\text{−}\infty$
• Infinite Limits from the Right
$\underset{x\to {a}^{+}}{\text{lim}}f\left(x\right)=\text{+}\infty \phantom{\rule{1em}{0ex}}\underset{x\to {a}^{+}}{\text{lim}}f\left(x\right)=\text{−}\infty$
• Two-Sided Infinite Limits
$\underset{x\to a}{\text{lim}}f\left(x\right)=\text{+}\infty :\underset{x\to {a}^{-}}{\text{lim}}f\left(x\right)=\text{+}\infty$ and $\underset{x\to {a}^{+}}{\text{lim}}f\left(x\right)=\text{+}\infty$
$\underset{x\to a}{\text{lim}}f\left(x\right)=\text{−}\infty :\underset{x\to {a}^{-}}{\text{lim}}f\left(x\right)=\text{−}\infty$ and $\underset{x\to {a}^{+}}{\text{lim}}f\left(x\right)=\text{−}\infty$

For the following exercises, consider the function $f\left(x\right)=\frac{{x}^{2}-1}{|x-1|}.$

[T] Complete the following table for the function. Round your solutions to four decimal places.

x $f\left(x\right)$ x $f\left(x\right)$
0.9 a. 1.1 e.
0.99 b. 1.01 f.
0.999 c. 1.001 g.
0.9999 d. 1.0001 h.

What do your results in the preceding exercise indicate about the two-sided limit $\underset{x\to 1}{\text{lim}}f\left(x\right)?$ Explain your response.

$\underset{x\to 1}{\text{lim}}f\left(x\right)$ does not exist because $\underset{x\to {1}^{-}}{\text{lim}}f\left(x\right)=-2\ne \underset{x\to {1}^{+}}{\text{lim}}f\left(x\right)=2.$

For the following exercises, consider the function $f\left(x\right)={\left(1+x\right)}^{1\text{/}x}.$

[T] Make a table showing the values of f for $x=-0.01,-0.001,-0.0001,-0.00001$ and for $x=0.01,0.001,0.0001,0.00001.$ Round your solutions to five decimal places.

x $f\left(x\right)$ x $f\left(x\right)$
−0.01 a. 0.01 e.
−0.001 b. 0.001 f.
−0.0001 c. 0.0001 g.
−0.00001 d. 0.00001 h.

What does the table of values in the preceding exercise indicate about the function $f\left(x\right)={\left(1+x\right)}^{1\text{/}x}?$

$\underset{x\to 0}{\text{lim}}{\left(1+x\right)}^{1\text{/}x}=2.7183$

To which mathematical constant does the limit in the preceding exercise appear to be getting closer?

In the following exercises, use the given values to set up a table to evaluate the limits. Round your solutions to eight decimal places.

[T] $\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}2x}{x};\phantom{\rule{0.2em}{0ex}}±0.1,±0.01,±0.001,±.0001$

x $\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}2x}{x}$ x $\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}2x}{x}$
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.

a. 1.98669331; b. 1.99986667; c. 1.99999867; d. 1.99999999; e. 1.98669331; f. 1.99986667; g. 1.99999867; h. 1.99999999; $\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}2x}{x}=2$

[T] $\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}3x}{x}$ ±0.1, ±0.01, ±0.001, ±0.0001

X $\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}3x}{x}$ x $\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}3x}{x}$
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.

Use the preceding two exercises to conjecture (guess) the value of the following limit: $\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}ax}{x}$ for a , a positive real value.

$\underset{x\to 0}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}ax}{x}=a$

[T] In the following exercises, set up a table of values to find the indicated limit. Round to eight digits.

$\underset{x\to 2}{\text{lim}}\frac{{x}^{2}-4}{{x}^{2}+x-6}$

x $\frac{{x}^{2}-4}{{x}^{2}+x-6}$ x $\frac{{x}^{2}-4}{{x}^{2}+x-6}$
1.9 a. 2.1 e.
1.99 b. 2.01 f.
1.999 c. 2.001 g.
1.9999 d. 2.0001 h.

$\underset{x\to 1}{\text{lim}}\left(1-2x\right)$

x $1-2x$ x $1-2x$
0.9 a. 1.1 e.
0.99 b. 1.01 f.
0.999 c. 1.001 g.
0.9999 d. 1.0001 h.

a. −0.80000000; b. −0.98000000; c. −0.99800000; d. −0.99980000; e. −1.2000000; f. −1.0200000; g. −1.0020000; h. −1.0002000; $\underset{x\to 1}{\text{lim}}\left(1-2x\right)=-1$

$\underset{x\to 0}{\text{lim}}\frac{5}{1-{e}^{1\text{/}x}}$

x $\frac{5}{1-{e}^{1\text{/}x}}$ x $\frac{5}{1-{e}^{1\text{/}x}}$
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.

$\underset{z\to 0}{\text{lim}}\frac{z-1}{{z}^{2}\left(z+3\right)}$

z $\frac{z-1}{{z}^{2}\left(z+3\right)}$ z $\frac{z-1}{{z}^{2}\left(z+3\right)}$
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.

a. −37.931934; b. −3377.9264; c. −333,777.93; d. −33,337,778; e. −29.032258; f. −3289.0365; g. −332,889.04; h. −33,328,889 $\underset{x\to 0}{\text{lim}}\frac{z-1}{{z}^{2}\left(z+3\right)}=\text{−}\infty$

What is a independent variable
a variable that does not depend on another.
Andrew
solve number one step by step
x-xcosx/sinsq.3x
Hasnain
x-xcosx/sin^23x
Hasnain
how to prove 1-sinx/cos x= cos x/-1+sin x?
1-sin x/cos x= cos x/-1+sin x
Rochel
how to prove 1-sun x/cos x= cos x / -1+sin x?
Rochel
how to prove tan^2 x=csc^2 x tan^2 x-1?
divide by tan^2 x giving 1=csc^2 x -1/tan^2 x, rewrite as: 1=1/sin^2 x -cos^2 x/sin^2 x, multiply by sin^2 x giving: sin^2 x=1-cos^2x. rewrite as the familiar sin^2 x + cos^2x=1 QED
Barnabas
how to prove sin x - sin x cos^2 x=sin^3x?
sin x - sin x cos^2 x sin x (1-cos^2 x) note the identity:sin^2 x + cos^2 x = 1 thus, sin^2 x = 1 - cos^2 x now substitute this into the above: sin x (sin^2 x), now multiply, yielding: sin^3 x Q.E.D.
Andrew
take sin x common. you are left with 1-cos^2x which is sin^2x. multiply back sinx and you get sin^3x.
navin
Left side=sinx-sinx cos^2x =sinx-sinx(1+sin^2x) =sinx-sinx+sin^3x =sin^3x thats proved.
Alif
how to prove tan^2 x/tan^2 x+1= sin^2 x
Rochel
Salim
what is function.
what is polynomial
Nawaz
an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
Alif
a term/algebraic expression raised to a non-negative integer power and a multiple of co-efficient,,,,,, T^n where n is a non-negative,,,,, 4x^2
joe
An expression in which power of all the variables are whole number . such as 2x+3 5 is also a polynomial of degree 0 and can be written as 5x^0
Nawaz
what is hyperbolic function
find volume of solid about y axis and y=x^3, x=0,y=1
3 pi/5
vector
what is the power rule
Is a rule used to find a derivative. For example the derivative of y(x)= a(x)^n is y'(x)= a*n*x^n-1.
Timothy
how do i deal with infinity in limits?
f(x)=7x-x g(x)=5-x
Awon
5x-5
Verna
what is domain
difference btwn domain co- domain and range
Cabdalla
x
Verna
The set of inputs of a function. x goes in the function, y comes out.
Verna
where u from verna
Arfan
If you differentiate then answer is not x
Raymond
domain is the set of values of independent variable and the range is the corresponding set of values of dependent variable
Champro
what is functions
give different types of functions.
Paul
how would u find slope of tangent line to its inverse function, if the equation is x^5+3x^3-4x-8 at the point(-8,1)
pls solve it i Want to see the answer
Sodiq
ok
Friendz
differentiate each term
Friendz
why do we need to study functions?
to understand how to model one variable as a direct relationship to another variable
Andrew

format: plain


text:
# How to compute a 2D distribution function from its density?

Suppose that we have two random variables $$X, Y$$ with a joint probability density function

$$f(x,y)=1,\ -y\lt x\lt y,\, 0\lt y\lt 1.$$

How can I calculate the cumulative joint probability function $$F(x,y)$$? I think it depends on how we choose the intervals of the domain.

The graph of the domain is in the figure:

• From joint probability density function you want to calculate joint probability density function..? What is f(x, y), what is F(x, y) ? – Tim Jan 23 at 9:58
• F(x,y) is the cumulative joint probabilty function – George Jan 23 at 10:03
• @Stubborn Because the triangle has unit area, the total probability is $1,$ so $f$ is a valid pdf. – whuber Jan 23 at 14:28
• @whuber Yes it is valid. Somehow I thought $X$ and $Y$ were independent as OP mentions when they are clearly not. – StubbornAtom Jan 23 at 14:34
• @StubbornAtom Good point! I overlooked the "independent" part altogether, because it's so obviously incorrect (and superfluous, anyway). – whuber Jan 23 at 14:48

By definition, $$F(x,y)$$ is the chance that both $$X\le x$$ and $$Y \le y.$$ It is the measure of the lower left quadrant bounded by the point $$(x,y).$$ In the present case, that measure will equal the area of the triangle within that quadrant, shown here by their overlap:

Evidently all the action occurs where $$0\lt y \lt 1:$$ increasing $$y$$ above $$1$$ will not change the overlap area, nor will decreasing $$y$$ below $$0.$$ As a convenience, let's write this process of "clamping" $$y$$ to the interval $$[0,1]$$ as

$$[y] = \min(1,\max(0,y)).$$

Given such a $$y$$, consider scanning from left to right by increasing $$x.$$ As you do, you sweep out the area of the triangle below the line of height $$y.$$ This is exactly the same process performed by univariate integration, except now the computation of "area below the curve" is replaced by "area above the curve." Thus, our problem is reduced to integrating a triangular function.

At https://stats.stackexchange.com/a/43075/919, I explain several ways to compute and write down this integral. One of the more convenient to express breaks the triangular function into a sum of functions that are shifted versions of

$$\theta_1(x) = \begin{array}{ll} \left\{ \begin{array}{ll} 0 & x \lt 0 \\ x & x\ge 0. \end{array}\right. \end{array}$$

(This is the integral of the $$\theta$$ defined in the referenced thread.)

At a level of $$y,$$ the triangle drops with a slope of $$-1$$ beginning at $$x=-y,$$ then rises with a slope of $$1=-1+2$$ beginning at $$x=0,$$ and finally stops (with a slope of $$-1+2-1=0$$) at $$x=y.$$ From this description it is evident we will be integrating the linear combination

$$f_y(x) = \theta_1(x+y) - 2\theta_1(x) + \theta_1(x-y).$$

The integral of $$\theta_1$$ is elementary, equal to

$$\psi(x) = \int_{-\infty}^x \theta_1(x)\mathrm{d}x = \begin{array}{ll} \left\{ \begin{array}{ll} 0 & x \lt 0 \\ \frac{x^2}{2} & x\ge 0. \end{array}\right. \end{array}$$

The integral of $$f_y$$ is the sum of the integrals of its terms, whence

$$F(x,y) = \psi(x+[y]) - 2\psi(x) + \psi(x-[y]).$$

Some plots of $$F$$ are illustrative. In the contour plot I have sketched out the "bounding box" of the triangle, because beyond this box it is evident the contours must all be linear. As expected, $$F$$ rises from zero below and to the left of the triangle up to $$1$$ above and to its right.

Here is a Mathematica implementation of the foregoing:

clamp[y_] := Min[1, Max[0, y]];
\[Psi][x_] := HeavisideTheta[x] x^2/2;
ff[x_, y_] := With[{v = clamp[y]}, \[Psi][x + v] - 2 \[Psi][x] + \[Psi][x - v]];

It produced the figures in this post. As a check, here is a plot of a numerical derivative of ff (implementing the density $$f(x,y) = \frac{\partial^2}{\partial x\partial y} F(x,y)$$):

Exactly as intended, it has constant value within the triangle (showing the density is uniform) and otherwise is zero.

• +1 I edited what I think is a small typo in your definition of $\theta(x)$. – Jarle Tufto Jan 24 at 15:59
• @Jarle Thank you; you are correct. That tells me I should not have used "$\theta$" to name this function, because it's really the integral of the $\theta$ defined in the post I referenced! I had better change that. – whuber Jan 24 at 16:00


text:
# Inverse system of a system with integral

I am trying to find the inverse system of the following (I tried finding the mathematical inverse function but since it is not the same I am not so sure) . Can someone help me find it?

$$y(t)=\int_{t-1}^{t+1}\cos\left(\frac{\pi\tau}{8}\right)x(\tau)d\tau$$

It seems that we have the same homework. You probably are Greek.

After having the same question, I came to the conclusion that in order for this system to be invertible, you have to prove that for any given $$x_1(t), \, x_2(t)$$ the following sentence is correct: $$x_1(t) \ne x_2(t) \,\, {\Rightarrow} \,\, y_1(t) \ne y_2(t)$$

As like you proved when a math function is $"1-1"$.

For this very example, if you have the following the following two signals:

$$\ x_1(t) = δ(t-4) \$$

and

$$x_2(t)=0$$

it turns out that while initially $x_1(t) \ne x_2(t)$, we have $cos(πt/8)x_1(t) = cos(πt/8)x_2(t)$, for any $t$. This means that

$$\int_{t-1}^{t+1}\cos\left(\frac{\pi\tau}{8}\right)x_1(\tau)d\tau = \int_{t-1}^{t+1}\cos\left(\frac{\pi\tau}{8}\right)x_2(\tau)d\tau$$

and the system is not invertible.

• Thank you , you are very right . Ευχαριστω φιλε , καλο κουραγιο με τις υπολοιπες – Maverick98 Jan 14 '18 at 9:50

HINT:

The given input-output relation can be interpreted as modulation followed by filtering with an LTI system:

$$\tilde{x}(t)=x(t)\cos(\pi t/8)\\y(t)=\int_{t-1}^{t+1}\tilde{x}(\tau)d\tau$$

First you need to figure out the impulse response of the LTI system. Then find its frequency response. Next, see if that frequency response can be inverted or not. The modulation can be undone by appropriate demodulation. So the main question to solve is if the LTI system can be inverted. You can see this by checking if you can divide by its frequency response.


text:
Lemma 77.23.2. Assumptions and notation as in Lemmas 77.20.2 and 77.20.3. The $2$-commutative diagram of Lemma 77.20.3 is a $2$-coequalizer in the following sense: Given

1. a stack in groupoids $\mathcal{X}$ over $(\mathit{Sch}/S)_{fppf}$,

2. a $1$-morphism $f : \mathcal{S}_ U \to \mathcal{X}$, and

3. a $2$-arrow $\beta : f \circ s \to f \circ t$

such that

$\beta \star \text{id}_ c = (\beta \star \text{id}_{\text{pr}_0}) \circ (\beta \star \text{id}_{\text{pr}_1})$

then there exists a $1$-morphism $[U/R] \to \mathcal{X}$ which makes the diagram

$\xymatrix{ \mathcal{S}_ R \ar[r]_ s \ar[d]^ t & \mathcal{S}_ U \ar[d] \ar[ddr]^ f \\ \mathcal{S}_ U \ar[r] \ar[rrd]_ f & [U/R] \ar[rd] \\ & & \mathcal{X} }$

$2$-commute.

Proof. Suppose given $\mathcal{X}$, $f$ and $\beta$ as in the lemma. By Stacks, Lemma 8.9.2 it suffices to construct a $1$-morphism $g : [U/_{\! p}R] \to \mathcal{X}$. First we note that the $1$-morphism $\mathcal{S}_ U \to [U/_{\! p}R]$ is bijective on objects. Hence on objects we can set $g(x) = f(x)$ for $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U) = \mathop{\mathrm{Ob}}\nolimits ([U/_{\! p}R])$. A morphism $\varphi : x \to y$ of $[U/_{\! p}R]$ arises from a commutative diagram

$\xymatrix{ S_2 \ar[dd]_ h \ar[r]_ x \ar[dr]_\varphi & U \\ & R \ar[u]_ s \ar[d]^ t \\ S_1 \ar[r]^ y & U. }$

Thus we can set $g(\varphi )$ equal to the composition

$\xymatrix{ f(x) \ar@{=}[r] \ar[rrrrrd] & f(s \circ \varphi ) \ar@{=}[r] & (f \circ s)(\varphi ) \ar[r]^\beta & (f \circ t)(\varphi ) \ar@{=}[r] & f(t \circ \varphi ) \ar@{=}[r] & f(y \circ h) \ar[d] \\ & & & & & f(y). }$

The vertical arrow is the result of applying the functor $f$ to the canonical morphism $y \circ h \to y$ in $\mathcal{S}_ U$ (namely, the strongly cartesian morphism lifting $h$ with target $y$). Let us verify that $f$ so defined is compatible with composition, at least on fibre categories. So let $S'$ be a scheme over $S$, and let $a : S' \to R \times _{s, U, t} R$ be a morphism. In this situation we set $x = s \circ \text{pr}_1 \circ a = s \circ c \circ a$, $y = t \circ \text{pr}_1 \circ a = s \circ \text{pr}_0 \circ a$, and $z = t \circ \text{pr}_0 \circ a = t \circ \text{pr}_0 \circ c$ to get a commutative diagram

$\xymatrix{ x \ar[rr]_{c \circ a} \ar[rd]_{\text{pr}_1 \circ a} & & z \\ & y \ar[ru]_{\text{pr}_0 \circ a} }$

in the fibre category $[U/_{\! p}R]_{S'}$. Moreover, any commutative triangle in this fibre category has this form. Then we see by our definitions above that $f$ maps this to a commutative diagram if and only if the diagram

$\xymatrix{ & (f \circ s)(c \circ a) \ar[r]_-{\beta } & (f \circ t)(c \circ a) \ar@{=}[rd] & \\ (f \circ s)(\text{pr}_1 \circ a) \ar[rd]^\beta \ar@{=}[ru] & & & (f \circ t)(\text{pr}_0 \circ a) \\ & (f \circ t)(\text{pr}_1 \circ a) \ar@{=}[r] & (f \circ s)(\text{pr}_0 \circ a) \ar[ru]^\beta }$

is commutative which is exactly the condition expressed by the formula in the lemma. We omit the verification that $f$ maps identities to identities and is compatible with composition for arbitrary morphisms. $\square$

Comment #2535 by wpb on

$f(\varphi \circ t)$ in the second diagram in the proof should be $f(t \circ \varphi)$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


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# TLM : The Falling out with D3D follow up

67 views

Hmmm, maybe I should goto town on D3D more, I seem to get more comments that way [grin]

Anyways, the comments about using PIX has been taken onboard and I'll be looking into that in a bit, but I figured before I do I'd make an entry to reply to a few points from the comments [smile]

@FReY
Last place I'll blame things on is a driver bug, the fact that ATI's R2VB samples work fine also kinda rule this out and point to me doing something wrong.

@jollyjeffers
The problem is that I'm NOT doing any sampling in the PS at all. The only thing being passed down is interpolated values which are then written out to the framebuffer.

The values themselves are coming from textures but these textures have been bound to vertex streams (1&2) to act as VBs.

Hmm, although you do have to bind 'fake' VBs to the streams as well to make sure D3D doesn't cry and I'm reusing the same one for all 3 sources (because I know it'll be big enuff to avoid the debug runtime yelling at me), I wonder if that'll effect things....

As for the RT stuff, it's not that it isn't simple it's just that it's annoyingly verbose to use and you have to execute the same series of function each time.
Let me compare for you (as I have it setup)
D3D:
Setup :
- Create textures at some point
Rendering:
- save old view port
- save old colour target
- get surface from texture
- bind to colour buffer
- set new view port based on surface size
- render
- bind back old colour target
- restore viewport
- safe release surface
- safe release old colour buffer surface

OGL:
Setup :
- Create texture
- Create FBO
- (optional) create render target for depth
- (optional) bind to FBO
- bind texture to FBO
- unbind FBO
Rendering:
- save old viewport
- set new viewport
- bind FBO
- render
- unbind FBO
- restore viewport

Two steps less over all and alot less in the rendering loop. FBOs also remember render target state so if you setup up say a 4 target RT in setup it'll automagically be configured when you bind the FBO. Give me something like this in D3D and I'd be happy [smile]

I'm sure that decls are fine with a decent FX system, however they just seem to cause pain when you are just knocking up a proof of concept and fiddling with things and don't have/want/need an FX system.

@Mars_999
JCD's GUI is nice and I've looked at it before, however like all GUIs it suffers from the 'zomg, use my classes, kk!' problem. Being forced to use someone elses texture and tuple classes when they could have just inforced an interface instead and let me use my own makes me twitch. I'm all for code reuse but not when it bloats out of control because you have 4 texture classes for 4 different libs all in different namespaces.

Worse comes to worse I'll just end up hacking it apart and re-engineering the bugger for interface instead of instances of classes. (Templates a-go-go).

The other thing which I couldn't work out was runtime binding of events, it all seemed somewhat XML hardcoded, which again made me twitch.

Anyways, I'm done, hopefully later tonight I'll have success to report... it's that or I'll be wacking D3D about the head until it's laying bleeding on the floor... [wink]

## 1 Comment

Quote:
 Two steps less over all and alot less in the rendering loop. FBOs also remember render target state so if you setup up say a 4 target RT in setup it'll automagically be configured when you bind the FBO. Give me something like this in D3D and I'd be happy
Yeah, the verbosity of RtT in D3D is a pain. I wrote myself a wrapper class that handles all of it though such that code looked something like:

{
RENDER_TO_TEXTURE( pNewRenderTarget, pNewDepthStencil );

// Do render to texture stuff here
}

The RENDER_TO_TEXTURE() was just a macro to a class on the stack that did all of the defensive copying and state blocks in the constructor and then undid the changes in the destructor (the explicit scope block forces the stack object to be destroyed at the right time).

I did mean to post the above code in my journal, but it appears I never got around to it. I've got a whole bunch of these things kicking around on my HDD [rolleyes]

Quote:
 I'm sure that decls are fine with a decent FX system, however they just seem to cause pain when you are just knocking up a proof of concept and fiddling with things and don't have/want/need an FX system.
Yup, can't disagree with you there. Now that D3D10 creates all its objects from struct's the code is getting more verbose unless you abstract it all away with some sort of caching/manager system - which is a bit of a bugger for PoC and simple apps [headshake]

Quote:
 it's that or I'll be wacking D3D about the head until it's laying bleeding on the floor...
I know where you live... [evil]

## Create an account

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text: Say we’re trying to simulate an unfair coin that we know only lands heads 20% of the time. Now let's flip a coin twice in succession. For example: A game that has coins tossed by the computer needs to have 2 values lets say either a 0 or 1. It is fun to see the history of the Super Bowl coin results. This post discusses the problem of the gambler's ruin. Thus, if a trader sees a stock that has moved 3 standard deviations, the odds of that event are only 1% (or 1 in 100), meaning that stock is showing a major move from a statistical standpoint that is outside the realm of normal statistical expectations. 45 pˆ= 48 100 =0. The one represents the desired outcome, and the four represents the amount of tries used. "After the first flip is known, you have the same thing. If you understand the definition of probability this idea *should* make intuitive sense example: If you flip a coin you won't get heads or tails exactly half the time. 4096 number of possible sequences of heads & tails. to flip a coin to get three heads in a row? AP Computer Science Principles In this assignment you will write a computer simulation to answer the question "On average, how many times do you need to flip a coin to get three heads in a row?" Simulations are one way to solve problems. Probability of flipping a coin 3 times and getting 5 heads in a row; Probability of getting 5 heads when flipping 3 coins together; A coin is tossed 3 times, find the probability that at least 5 are heads? If you flip a fair coin 3 times what is the probability that you will get exactly 5 heads?. Why Throwing 92 Heads in a Row is not Surprising Martin Smith, University of Edinburgh When we first meet the title characters of Tom Stoppards Rosencrantz and Guildenstern Are Dead, they are betting on coin throws. Is one more likely to appear first? If so, which one and with what probability?. Suppose that a person tosses a coin 6 times and gets a head each time. It's not that hard, for some people. The difference is in that in the second case we can easily differentiate between the coins: one is the first, the other second. With R we can play games of chance - say, rolling a die or flipping a coin. This is one of those last-possession, coin-flip games. Probability of flipping a coin 3 times and getting 5 heads in a row; Probability of getting 5 heads when flipping 3 coins together; A coin is tossed 3 times, find the probability that at least 5 are heads? If you flip a fair coin 3 times what is the probability that you will get exactly 5 heads?. For example if you flip a coin the odds are 1/2 for heads lets say. The toss of a coin, throwing dice and lottery draws are all examples of random events. The probability of flipping 91 coins and getting a head 91 times in a row is 1/291 or (1/2)91 or 4. When a coin is tossed, there lie two possible outcomes i. But now when we look at the sequence of tosses starting at position two, we have to throw out the outcomes where we had two heads at toss one - we've already seen two heads, so we can't continue flipping coins in those outcomes. So in 50 seasons, with a 32 team league, the chances are 45. People tend to confuse that with the odds of multiple coin tosses in a row coming up all heads. To do this, we need a probability distribution…for a coin toss, we have the following. =( What is the probability of getting heads 5 times in a row when flipping a fair coin?. com Go into any casino, or start any conversation about gambling and you are bound to hear people talking about winning or losing streaks. Stats on NFL coin flips aren’t kept, but media coverage identified some even better streaks in recent years. Now getting it to land heads up 100 times in a row, well that's an entirely different problem. But the coin has not changed - if it's a "fair" coin, the probability of getting tails is still 0. The exact probability of the coin landing heads exactly 2 times is? Where p is the probability, n is the total events and. The odds of getting tails twice in a row are 1/2 * 1/2 = 1/4. 1 True or false. So there is a probability of one that either of these will happen. , SAS, SPSS, Stata) who would like to transition to R. Odds of Ten Reds in a Row. Players form two teams. The chance of that would be 6. I can easily find the number of heads out of 100 and the chances of coin flipping heads out of 100 flips. So the probability of a success of 4 or more heads in a row for every 10 coin flips is 251/1,024 = 0. If that event is repeated ten thousand different times, it is. Even unlikely events happen given enough opportunities. That is because each time you flip the coin, the odds remain 1/2; the two flips are independent of each other. Basically, I calculate if the current flip in a 10 flip session is equal to the prior flip, and if it is, I increment a counter. Jacque Wilson, in a piece for CNN titled: Calculating the odds: 12 sons in a row reports the odds at about 1 in 4000. Because of all the random factors beyond our control that enter the flipping process (force with which the coin is flipped, motion of the air in the room, position of our hand when we catch the coin) we therefore expect a probability of 1/2 for heads, and 1/2 for tails. game theory written in collaboration with Oskar Morgenstern entitled Theory of Games and Economic Behavior, 1944. Flip a single coin 20 times in a row. 0625 but the probability at each point when you flip the coin of the next result being a head will remain at 0. On the European wheel, there are 37 numbers. It's going to be their free throw percentage-- in this case, LeBron's was 75%-- to the number of free throws that we want to get in a row. Despite having the latest technology at their disposal, the Democrat Party hierarchy decided the best course of action would be to go back to basics and settle the dispute with a coin toss! A supporter in the Clinton camp correctly called "heads" meaning that Clinton picked up five of the eight delegates in precinct 2-4 Ames. This is an application of Bayes' theorem. And only one of these tosses yielded two heads in a row, so the probability of not seeing two heads after two tosses is 3/4. teacher’s preference, you can write this probability as a fraction or a decimal. So what are the odds someone will win six coin tosses in a row? Let’s do math. "I was doing so well and then suddenly lost 6 times in a row and was bankrupt. Nothing remarkable. First compare the highest red to the highest blue. 11 1 26 12 •= A 60% free throw shooter making 3 free throws in a row 0. You may have flipped 10 heads in a row but didnt know it. Note that this answer works for any odd number of coin flips. seven 2-digit random numbers. The first player gets a point for a toss of heads. Bernie Sanders' voters will not be amused to learn that the Democratic National Committee awarded six deadlocked precincts, out of 99 precincts total, to Hillary Clinton with a literal coin toss. It is fun to see the history of the Super Bowl coin results. The defining characteristic of a Markov chain is that no matter how the process arrived at its present state, the possible future states are fixed. The odds of getting tails twice in a row are 1/2 * 1/2 = 1/4. The chance on the first toss is 50%, and on the 42nd toss it's 50%, and on the 100th toss it's 50%. 5%, and the chance for the fourth and final toss to be heads would be 6. 11 1 26 12 •= A 60% free throw shooter making 3 free throws in a row 0. What is the probability of flipping a coin four times in a row and. In the case of coins, heads and tails each have the same probability of 1/2. Other discussions of the theory of games relevant for our present purposes may be found in the text book,Game Theory by Guillermo Owen, 2nd edition, Academic Press, 1982, and the expository book, Game Theory and Strategy by. 6, then 60% of the time, the coin lands heads). The # of tosses determines how many coins are tossed in each run of the experiment. Life is full of random events! You need to get a "feel" for them to be a smart and successful person. Tossing a Biased Coin Michael Mitzenmacher When we talk about a coin toss, we think of it as unbiased: with probability one-half it comes up heads, and with probability one-half it comes up tails. The number of ways you can have 7 heads in 10 flips is (10 choose 7). coin toss probability calculator,monte carlo coin toss trials. Source (s): Maths. When the coin is thrown in the air, it should rotate several times before landing on the ground, or caught and inverted by a chosen person. What were the Odds of Having Such a Terrible Streak at the Casino? February 2009 [email protected] A probability of one means that the event is certain. That is because each time you flip the coin, the odds remain 1/2; the two flips are independent of each other. In an empirical study,we per-form an experiment many times, keep records of the results, and then analyze these results. The probability of flipping 91 coins and getting a head 91 times in a row is 1/291 or (1/2)91 or 4. Probability, physics, and the coin toss L. The Texans struggled against a team with a strong secondary last week, and the Chargers have that with Casey Hayward and Desmond King. He calculates the odds at 8,912 to 1. For example, it strikes most people as unexpected if heads comes up four times in a row during a series of coin flips. An ideal unbiased coin might not correctly model a real coin, which could be biased slightly one way or another. The probability of getting a given value for the total on the dice may be calculated by taking the total number of ways that value can be produced and dividing it by the total number of distinguishable outcomes. Assume that the coin is fair. "After the first flip is known, you have the same thing. The coin can only land on one side or the other (event) but there are two possible outcomes: heads or tails. But yeah, to echo an earlier poster, 5 head tosses in a row indicates the coin is not fair and I would not assign 50/50 odds to the next coin toss. The probability of getting the three or more heads in a row is 0. Recently heads hit 5 years in a row, followed by a 4 year on tails winning the toss. So, for example, if a mutual fund manager has had three above-average years in a row, many people will conclude that the fund manager is better than average, even though this conclusion does not follow from such a small amount of data. 5 assuming fairness - they are different probabilities. Probability of getting exactly 8 heads in tossing a coin 12 times is 495/4096. For example: A game that has coins tossed by the computer needs to have 2 values lets say either a 0 or 1. Flip a single coin 20 times in a row. Theoretically, a team will start the season with seven winning coin tosses once every four years. Flip a biased coin and get an arbitrary amount of heads in a row for an arbitrary probability by Netwinder Last Updated October 05, 2019 00:20 AM 0 Votes 8 Views. Are you looking for the expected number of times out of the 1000 flips that five flips in a row will be uh, heads? tails? I don't know what "lose" means! Also, if I "lose" 6 in a row, does that count as one or two times I lost 5 in a row? We have to be specific in probability questions. At the beginning of the game, player A has 1 coin and player B has 3 coins. Stats on NFL coin flips aren’t kept, but media coverage identified some even better streaks in recent years. The odds of the first are dependent. So, for example, if a mutual fund manager has had three above-average years in a row, many people will conclude that the fund manager is better than average, even though this conclusion does not follow from such a small amount of data. Theoretically, the odds of a coin landing on heads are 1 in 2 and that, since either the NFC or AFC will be represented by heads, the odds of either conference winning the coin flip are equal. I wrote a C++ program to solve the problem. This serves as a nice. The posted odds of a horse actually overestimate the horse’s chance of winning to ensure that the punter is underpaid for a win. It can be calculated by dividing the number of possible occurrence by the total number of options. 25=(1/4) thus you would expect to have to flip four times before you would get two consecutive heads. What is the probability of at least 5 consecutive heads? I thought it was like this: Those 5 heads can start at spots 1-6 in 10 flips, so there are 6 possibilities. Probability is the measure of the possibility that a given event will occur. We wager you never want in the direction of miss this fairly Excellent probability. Now getting it to land heads up 100 times in a row, well that's an entirely different problem. This is a classic case of independence—the fact that certain events have occurred (in this case, nine heads being tossed in a row) has no impact on the probability of a subsequent event. to flip a coin to get three heads in a row? AP Computer Science Principles In this assignment you will write a computer simulation to answer the question "On average, how many times do you need to flip a coin to get three heads in a row?" Simulations are one way to solve problems. So in 50 seasons, with a 32 team league, the chances are 45. Java Coin Flip. 2451171875 Basically, I want to know the procedure for solving this type of problem (formulas - that type of thing), as opposed to working out every success out of all the possible outcomes. Playing Blackjack can be a frustrating experience. A probability of zero is a result which cannot ever occur: the probability of getting five heads in four flips is zero. For example, it strikes most people as unexpected if heads comes up four times in a row during a series of coin flips. This feature is not available right now. Since the probability to flip a head is the same as the probability to flip a tail, the probability of outcome (i) must be equal to the probability of outcome (ii). That is because there is a 1% chance of picking the two-headed coin, which has a 100% of getting 10 heads, and a 99% of picking a fair coin, which has a (1/2) 10 chance of flipping 10 heads in a row. What is the probability of obtaining heads and a two? asked by Sabrine on December 12, 2008; probability. If your second toss is a T (making your first toss H), then you get 1. What is the minimum number weighings that can always guarantee to determine the different coin. A probability of one represents certainty: if you flip a coin, the probability you'll get heads or tails is one (assuming it can't land on the rim, fall into a black hole, or some such). Coin Toss Probability Calculator. In other words, if you flip a coin 20 times over and over again and mark whether or not you get 4 heads in a row, you will find that half of the time it occurs, and it does not occur the other half of the time (assuming you run this test many, many times). These examples can help to illustrate the more general rules of probability that will be discussed subsequently. Boy did she get a scalping with the DNA test results. Usually, this consists of events in a sequence, such as flipping "heads" twice in a row on a coin toss, but the events may also be concurrent. Math: Odds of coin landing on Heads 10 times? What are the odds of flipping a coin and it coming up heads 10 times in a row? How many times would you need to flip a coin on heads in a row to get the odds 1:10,000,000,000?. Mathematical probability, on the other hand, has to do with the number of possible outcomes of an event. The best example of probability would be tossing a coin, where the probability of resulting in head is. You cover the flipped coins and pull them out, the only thing thats changed is you are randomly discovering the results of your flipping. 00781 Interpret this probability. When a coin is tossed, there lie two possible outcomes i. You've got 27 coins, each of them is 10g, except for 1. Bernie Sanders' voters will not be amused to learn that the Democratic National Committee awarded six deadlocked precincts, out of 99 precincts total, to Hillary Clinton with a literal coin toss. (c) What is the probability of tossing 3 heads in a row? Explain. Probability Theory on Coin Toss Space 1 Finite Probability Spaces 2 Random Variables, Distributions, and Expectations 3 Conditional Expectations. Wiki User Although very slight you have a better chance of getting heads when the coin is flipped from a position where. If we think physically that all those coin flips have nothing to do with each other, information about the fifth and sixth coin flip are not going to change what we expect from the first three. So the probability of flipping it is 1/1024 = 0. , the next flip—will revert to 50:50. The odds of winning seven coin tosses in a row are 1 in 128. In simple words, the probability of either head or tails is one. Thus the probability was ¼ or 0. You can get the answer by just comparing groups of coins. “Fair” means, technically, that the probability of heads on a given flip is 50%, and the probability of tails on a given flip is 50%. Since each coin toss has a probability of heads equal to 1/2, I simply need to multiply together 1/2 eleven times. Step 1 was a little time consuming, so for the rest of the (24) trials, flip all 20 coins at once and count the number of heads you get. Math: Odds of coin landing on Heads 10 times? What are the odds of flipping a coin and it coming up heads 10 times in a row? How many times would you need to flip a coin on heads in a row to get the odds 1:10,000,000,000?. Find the probability that both heads and tails occurs. ) If a coin is flipped two times, one hundred different times, it is expected that two tails in a row would occur about 25 times. 77% no team would lose the first 11 coin tosses in a row, or 54. Sometimes, people use words, such as unlikely, possible, certain likely, and impossible. Tossing a Biased Coin Michael Mitzenmacher When we talk about a coin toss, we think of it as unbiased: with probability one-half it comes up heads, and with probability one-half it comes up tails. (Multiply 2 by itself 10 times. It does not matter how many times it has come up heads (or tails) before. 963% of the time (7,987,316 there will be five heads or more in a row only, 7,987,316 times five tails or more in a row only, and 2,467,930 times both five heads or. What is the probability, P(k), of obtaining k heads? There are 16 different ways the coins might land; each is equally probable. Then I tossed H H H H H H H H H H. This serves as a nice. s air force of non-u. Mathematician: Want an exact answer without all the hard work and really nasty formulas?Computers were invented for a reason, people. PROBABILITY. the probability of. If the probability of getting at least one contract is 4/5, what is the probability that he will get both the contracts ?. We could demonstrate that by explaining the probability numbers behind it. Which of these sequences is less likely to happen? H H H H H H T H H T H T Almost everyone would answer that the first sequence is less likely. Since the rows are assumed to be independent, you can then compute the probability of seeing the event in any of the 12 rows. Problem: A coin is biased so that it has 60% chance of landing on heads. The first column is the ID of a particular Coin Flipper. If the question is "What is the probability of getting either five heads in a row or five tails in a row with 25 tosses, then I totally agree that one or the other will occur 54. Equally Likely Outcomes. The denominator is always all the possible events. The sacred coin flip exhibits (at minimum) a whopping 1% bias, and possibly much more. But we need a few more rules to get very far. To find the probability of heads and then tails, walk along the appropriate branches multiplying 0. A little better than 1/512. What's the difference between Bayesian and non-Bayesian statistics? Monday November 11, 2013. Have a look! History of Coin Flipping. Yet recent research into coin flips has discovered that the laws of mechanics determine the outcome of coin tosses: The startling finding is that they aren't random. The table below, which associates each outcome with its probability, is an example of a probability distribution. The Super Bowl coin toss has landed on tails four years in a row. And it’s the same for any “even chance” bet including highs and lows, and odds and evens. the probability that you get heads on any given toss is 0. ) Of these 1024, only 2 of them are "on the same side": * All heads * All tails So, if every coin in the world were perfect, a. [email protected] There are 10 coin flips total. A jar has 1000 coins, of which 999 are fair and 1 is double headed. PROBABILITY. On the European wheel, there are 37 numbers. The # of tosses determines how many coins are tossed in each run of the experiment. Experimental Probability When asked about the probability of a coin landing on heads, you would probably answer that the chance is ½ or 50%. I know they are independent from there I don't know. (b) Create a probability histogram for the variable X. Is the event Dependent or Independent? Rolling a die twice. teacher's preference, you can write this probability as a fraction or a decimal. Toss three red and two blue dice, look at the top faces and count the dots on each. - coinflip. The exact probability of the coin landing heads exactly 2 times is. And it’s the same for any “even chance” bet including highs and lows, and odds and evens. A fair coin is tossed 5 times. Before you move on to it, you must know all about coin flipping. Each outcome has a fixed probability, the same from trial to trial. 72%, while for n = 10 it goes down to 0. 45 pˆ= 48 100 =0. Now think about the probability of getting two heads in a row - half the time you'll get that first heads, and then if that is successful half the time you'll get a second head. Record the results of each flip (head/tail) in the data table below. To rephrase the question: How many heads in a row need to be flipped in order to get the odds one in 175 million?. PROBABILITY. com Go into any casino, or start any conversation about gambling and you are bound to hear people talking about winning or losing streaks. Imagine that you toss that same coin 20 times. At the head of the lines, the "referee" sits between the two players with a coin. answer: the probability of finding the grain of sand is about 1 in 1024, whereas the probability of flipping a coin and getting 100 heads in a row is about 1 in 1025. 24 to find that there's a 24% chance of this combination. What if you flipped two coins repeatedly, so that one option would win as soon as two heads showed up in a row on that coin, and one option would win as soon as heads was immediately followed by tails on the. If you get four heads in a row, you win. In that case, your answer may be something like 95% for heads, where the remaining 5% account for the chance that the coin is only somewhat biased and tails are still possible. An ideal unbiased coin might not correctly model a real coin, which could be biased slightly one way or another. seven 2-digit random numbers. But yeah, to echo an earlier poster, 5 head tosses in a row indicates the coin is not fair and I would not assign 50/50 odds to the next coin toss. What's not so obvious is that the probability of a coin that has come up heads for the past 19 flips also landing heads up on the 20th throw is also 50 per cent. Game Theory (Part 9) John Baez. Life is full of random events! You need to get a "feel" for them to be a smart and successful person. One may toss two coins simultaneously, or one after the other. Below is a table of the odds for winning a pass line bet up to 11 times in a row. A verbal outline of the derivtion - first consider the log likelihood function as a curve (surface) where the base is $$\theta$$. And it's the same for any "even chance" bet including highs and lows, and odds and evens. Reid has always seemed lucky, but this is crazy. 18 are black, and 18 are red. coin toss probability calculator,monte carlo coin toss trials. Published on June 14, 2016. 24 to find that there's a 24% chance of this combination. In a probability model, sample points represent outcomes and combine to make up events; probabilities of events can be computed by applying the Addition and Multiplication Rules. 24 to find that there's a 24% chance of this combination. , the next flip—will revert to 50:50. Game Theory (Part 9) John Baez. The higher the probability of an event, the more certain that the event will occur. Show Step-by-step Solutions. The answer is related to Pascal's triangle. Nickerson 5 Gleason Road Bedford, MA 01730 r. Here's just how unlikely Hillary Clinton's 6-for-6 coin-toss victories would have been. He then simply showed the last 10 flips of the film on TV, claiming that he influenced the outcome of each flip to get 10 heads first time. This applies for any game of chance, including roulette. The Rams have not won a coin toss in 1992. Similar situations were reported elsewhere, including at a precinct in Des Moines, at another precinct in Des Moines , in Newton, in West Branch and in Davenport. Stats on NFL coin flips aren’t kept, but media coverage identified some even better streaks in recent years. These examples can help to illustrate the more general rules of probability that will be discussed subsequently. At the beginning of the game, player A has 1 coin and player B has 3 coins. " to describe events that are random. 5 we get this probability by assuming that the coin is fair, or heads and tails are equally likely. What's more, you can take advantage of this little-known fact to give yourself an edge in all future coin-flip battles. "I was doing so well and then suddenly lost 6 times in a row and was bankrupt. Even if you obtained five heads in a row, the odds of heads resulting from a sixth flip remain at ½. So, each time the coin is tossed there is (ideally) a 50% chance of it landing heads and a 50% chance of it landing tails. The odds of winning nine coin tosses in a row sits at 1/512. The article goes on to state:. Part of what makes Stoppard's scene so compelling is that it plays to the audience's skepticism that someone could win 92 tosses in a row by betting heads. The number of ways you can have 7 heads in 10 flips is (10 choose 7). Before you move on to it, you must know all about coin flipping. So the probability of a success of 4 or more heads in a row for every 10 coin flips is 251/1,024 = 0. She was asked, “If I have flipped a coin ten times and each time was a head, what is the probability of getting a head on the eleventh flip?” Her response, “Let me look at the coin. Pupils use simulated data to find the experimental probability of flipping a coin. Another experiment, popular in various board and gambling games, consists of throwing a single die. If the two indistinguishable coins are tossed simultaneously, there are just three possible outcomes, {H, H}, {H, T}, and {T, T}. " to describe events that are random. Do you own an iOS or Android device? Check out our app! Dice Roller. Source (s): Maths. This section looks at the mathematics of flipping coins, and the next section looks at the mathematics of rolling dice. Luck Of The Flip: New England Patriots Defy Probability With Coin Toss Wins The New England Patriots have recently been very lucky. The probability of a coin toss being a tail is 1/2. Pick a coin at random, and toss it 10 times. 2 What is the. 0625 Algebra -> Probability-and-statistics -> SOLUTION: what is the probability that you will get heads four times is a row when flipping a fair coin?. " If I toss 48 heads on 100 flips, then pˆ pˆ= 45 100 =0. The Super Bowl coin toss has landed on tails four years in a row. Toss a coin and roll a die: 2009-09-21: From Celeste: Can you please tell me how to set up this problem to find the answer. The Derren Brown Coin Flipping Scam. Show Step-by-step Solutions. The probability of flipping a head after having already flipped 20 heads in a row is 1 / 2. The probability of flipping 10 heads in a row, assuming a randomly picked coin, is (1/100)*1 + (99/100)*(1/2) 10. Probability. answer: the probability of finding the grain of sand is about 1 in 1024, whereas the probability of flipping a coin and getting 100 heads in a row is about 1 in 1025. But that's not the only surprising thing about them. , What is the probability of getting 2 heads when you toss 2 coins? The probability of tossing a head in a toss of a coin is ½ or 0. easier, as the probability of getting the same side of the coin as the one getting in the previous flip is always ½, so the average number of tosses needed would be 3 (1/2 of probability on the second flip, ½ on the third) Since the probability of tossing two consecutive heads is. We can collect these three numbers into a vector of probabilities. It's a rare feat, happening only twice in the last 20 years. With an honest coin, the chances of winning or losing are 50% and consequently, coin flipping is used to decide such momentous events like who kicks off in a football game. If that event is repeated ten thousand different times, it is. What is the probability that the coin shows a head or the die has a 5 on the upward face? Answered by Penny Nom. It's a rare feat, happening only twice in the last 20 years. So the probability of the event is written as : the numerator (3) is the number of favorable outcomes, and the denominator (8) is the number of possible outcomes. Each flip is independent in the sense that the probability of heads on one flip does not affect the probability of heads on any other flip. 2 What is the. This is a classic case of independence—the fact that certain events have occurred (in this case, nine heads being tossed in a row) has no impact on the probability of a subsequent event. Game Theory (Part 9) John Baez. Thus, we get 1/2. So these events are independent. The Predictive Power Of The Super Bowl Coin Toss it is so unlikely for a coin to land heads 11 times in a row it must be more likely that the next flip will land tails. But we need a few more rules to get very far. At odds of 25 for 1, that bet has an expected return of 0. The odds when flipping a coin & getting heads 10 times in a row is exactly 1/1024. Does that mean heads is due? Super Bowl 2018 prop bets: How wagers on the coin toss explain the concept of spreads — Quartz. When a coin is tossed, there lie two possible outcomes i. One over two is a half, or 50 per cent. The probability of getting 10 in a row is so low, that there must be something up with the coin. Answer: Still 50%! While the initial nine heads in a row is quite unlikely—given that is has already occurred—and that each coin toss is an independent event, the outcome of the previous coin flips have no impact on the subsequent tenth flip. 45 pˆ= 48 100 =0. Haven't read other responses yet, but if it's a fair coin, 10th flip ( and all flips) are 50:50, regardless of past results. “Fair” means, technically, that the probability of heads on a given flip is 50%, and the probability of tails on a given flip is 50%. For the ignorant observer, even if the coin's tendency is known, the probability of the first flip observed—i. Math: Odds of coin landing on Heads 10 times? What are the odds of flipping a coin and it coming up heads 10 times in a row? How many times would you need to flip a coin on heads in a row to get the odds 1:10,000,000,000?. If the coin is weighted so that the probability of tails is 25% and the probability of heads is 75%, then Shannon assigns an entropy of 0. The probability of a plane crashing depends on whether the plane is flying or on the ground. And we showed this happened in our excel generated random coin tosses. The more times you flip the coins together, the closer you will come to these ideal ratios. We could demonstrate that by explaining the probability numbers behind it. Winning chances. To find the probability of heads and then tails, walk along the appropriate branches multiplying 0. In this case the probability of getting. Flip a coin until you get a head OR have 3 tails in a row. A fair coin is tossed two times. At the beginning of the game, player A has 1 coin and player B has 3 coins. "After the first flip is known, you have the same thing. Experimental probability is defined as the ratio of the number of times an event occurs to the total number of times the activity is performed. 25% because there's a 50% chance for the first flip, then 25% percent chance for the second throw, the third is 12. NBC Sports notes how 'remarkable' winning seven coin tosses in a row is. The odds of red or black spinning in a row. Python function for computing the likelihood that, out of n consecutive coin flips, you will have a string of at least k of the same coin flip in a row. What are the odds of getting 6,7,8,9,10,11 heads in a row on a coin toss? What is the ratio? like 100:1 etc?? It doesn't have to be exact, near enough is good enough. Probability: Independent Events. The probability of LEGITIMATELY flipping heads 100 times in a row on a fair coin is. And it’s the same for any “even chance” bet including highs and lows, and odds and evens. I have three columns of data. Thus, we get 1/2. 50) for the operation of remote gambling. I absolutely suck at math. We focus on some counterintuitive aspects of sequences that coin-tossing produces.


text:
# Defining the Schrodinger Equation for Pauli Spin Matrix along the y-axis

uxioq99
Homework Statement:
Determine a wave function as a column vector of two states that solves the Schrodinger equation given by the Hamiltonian ##H=-\omega S_y##.
Relevant Equations:
##H=-\omega S_y##
By the statement of the question, a solution must take the form
##\begin{pmatrix} \Psi_1 \\ \Psi_2 \end{pmatrix}##
and the energy operator will be as per usual ##\hat{E} = i\hbar \frac{\partial}{\partial t}##. I am confused by the fact that
##S_y = \frac{\hbar}{2} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}##
because the Schrodinger equation implies
##\begin{pmatrix}\frac{\partial \Psi_1}{\partial t}\\\frac{\partial \Psi_2}{\partial t}\end{pmatrix}=\frac{\omega}{2}\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}
\begin{pmatrix}
\Psi_1\\
\Psi_2
\end{pmatrix}
##

So that exponential solutions as opposed to oscillatory solutions are observed.

##
\begin{pmatrix}
\Psi_1(x) \\
\Psi_2(x)
\end{pmatrix}
=
\frac{1}{\sqrt{e^{\omega/2 t} + e^{-\omega/2 t}}}
\begin{pmatrix}
e^{-\omega/2 t} \\
e^{\omega/2 t}
\end{pmatrix}
##

Is this really a physically possible solution? Is it akin to when the energy of a bound state in a finite square well is below the depth of the well leading to an exponential decay in the solution? Now the decay appears to be in time, suggesting that the final state of the system will simply be ##\Psi_1 = 0## and ##\Psi_2 = 1##.

In general, if ##A \in \text{GL}_2(\mathbb R)## and

##
i\hbar\begin{pmatrix}
\frac{\partial \Psi_1}{\partial t}\\
\frac{\partial \Psi_2}{\partial t}
\end{pmatrix}=
i\hbar
\begin{pmatrix}a & b \\ c & d\end{pmatrix}
\begin{pmatrix}
\Psi_1\\
\Psi_2
\end{pmatrix}
##

Could one perform an eigendecomposition to exponentiate the spectrum of the matrix? The corresponding characteristic polynomial will be

##\lambda^2 - \lambda(a + d) + |A| = 0##

yielding

##\lambda_\pm = \frac{a+d}{2} \pm \sqrt{\frac{a+d}{2}^2 - |A|}##

The corresponding eigenvector matrix ##S## is too tedious to include here, but can be readily solved from the expression for the eigenvalues. Does that imply then, that the general solution is

##\begin{pmatrix}\Psi_1 \\ \Psi_2 \end{pmatrix} = S e^{\Lambda t} S^{-1}##

where ##\Lambda## is the diagonal matrix formed from ##\lambda_+## and ##\lambda_-##?

For either of these evolution equations, how would somebody specify the initial conditions so that the spin is taken along some vector in ##\mathbb{R}^3##?

Last edited:

Homework Helper
2022 Award
Homework Statement:: Determine a wave function as a column vector of two states that solves the Schrodinger equation given by the Hamiltonian ##H=-\omega S_y##.
Relevant Equations:: ##H=-\omega S_y##

By the statement of the question, a solution must take the form
##\begin{pmatrix} \Psi_1 \\ \Psi_2 \end{pmatrix}##
and the energy operator will be as per usual ##\hat{E} = i\hbar \frac{\partial}{\partial t}##. I am confused by the fact that
##S_y = \frac{\hbar}{2} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}##
because the Schrodinger equation implies
##\begin{pmatrix}\frac{\partial \Psi_1}{\partial t}\\\frac{\partial \Psi_2}{\partial t}\end{pmatrix}=\frac{\omega}{2}\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}##

So that exponential solutions as opposed to oscillatory solutions are observed.

##
\begin{pmatrix}
\Psi_1(x) \\
\Psi_2(x)
\end{pmatrix}
=
\frac{1}{\sqrt{e^{\omega/2 t} + e^{-\omega/2 t}}}
\begin{pmatrix}
e^{-\omega/2 t} \\
e^{\omega/2 t}
\end{pmatrix}
##

Is this really a physically possible solution? Is it akin to when the energy of a bound state in a finite square well is below the depth of the well leading to an exponential decay in the solution? Now the decay appears to be in time, suggesting that the final state of the system will simply be ##\Psi_1 = 0## and ##\Psi_2 = 1##.

In general, if ##A \in \text{GL}_2(\mathbb R)## and

##
i\hbar\begin{pmatrix}
\frac{\partial \Psi_1}{\partial t}\\
\frac{\partial \Psi_2}{\partial t}
\end{pmatrix}=
i\hbar
\begin{pmatrix}a & b \\ c & d\end{pmatrix}
##

Could one perform an eigendecomposition to exponentiate the spectrum of the matrix? The corresponding characteristic polynomial will be

##\lambda^2 - \lambda(a + d) + |A| = 0##

yielding

##\lambda_\pm = \frac{a+d}{2} \pm \sqrt{\frac{a+d}{2}^2 - |A|}##

The corresponding eigenvector matrix ##S## is too tedious to include here, but can be readily solved from the expression for the eigenvalues. Does that imply then, that the general solution is

##\begin{pmatrix}\Psi_1 \\ \Psi_2 \end{pmatrix} = S e^{\Lambda t} S^{-1}##

where ##\Lambda## is the diagonal matrix formed from ##\lambda_+## and ##\lambda_-##?

For either of these evolution equations, how would somebody specify the initial conditions so that the spin is taken along some vector in ##\mathbb{R}^3##?
##\begin{pmatrix}\frac{\partial \Psi_1}{\partial t}\\\frac{\partial \Psi_2}{\partial t}\end{pmatrix}=\frac{\omega}{2}\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}##
What happened to the ##\Psi## column vector on the Right Hand Side?

topsquark and uxioq99
uxioq99
Thanks, I added it back into the equation, right at the end of the editing window.

Homework Helper
2022 Award
The rest is just incorrect. The solution you write down is not a solution. The exponent is not dimensionless and then it won't work anyway . At that point I stopped.
Get a copy of any good quantum text and work it through. Give it another attempt.
It is not nonsense......just full of mistakes Might be good to number your equations.

topsquark
uxioq99
@hutchphd Sorry, I realized when I typed "\exp(-\omega/2t)" it produced ##\exp(-\omega/2t)## instead of ##\exp(-\frac{\omega}{2} t)## Wouldn't the units of ##\omega## cancel those of time given that it is an angular frequency? Would

##
\begin{pmatrix}
\Psi_1(x,t) \\
\Psi_2(x,t)
\end{pmatrix}
=
\begin{pmatrix}
\exp(\frac{\omega}{2} t) \\
\exp(-\frac{\omega}{2} t)
\end{pmatrix}
##

Be a valid solution of ##E\vec\Psi = H \vec \Psi## where ##H = -\omega S_y##?

Wouldn't

##
-i\hbar
\begin{pmatrix}
\frac{\partial \Psi_1}{\partial t} \\
\frac{\partial \Psi_2}{\partial t}
\end{pmatrix} =
-\frac{\hbar \omega}{2} \begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix}
\begin{pmatrix}
\Psi_1 (x,t) \\
\Psi_2 (x,t)
\end{pmatrix}
##

because

##
-i\hbar \frac{\partial}{\partial t} \left(e^{\pm\frac{\omega}{2} t}\right) =
\mp i \frac{\hbar \omega}{2} e^{\pm\frac{\omega}{2} t}
##

and

##
-\frac{\hbar \omega}{2}
\begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix}
\begin{pmatrix}
e^{-\frac{\omega}{2} t} \\
e^{\frac{\omega}{2} t}
\end{pmatrix} =
i\frac{\hbar \omega}{2}
\begin{pmatrix}
-e^{\frac{\omega}{2} t} \\
e^{-\frac{\omega}{2} t}
\end{pmatrix}
##

I believe that a solution to the Schrodinger equation cannot always be real, so my selected form for ##\Psi(x,t)## does not appear to be correct. Additionally, the normalization factor I mentioned earlier

##\frac{1}{\sqrt{e^{\omega t} + e^{-\omega t}}}##

is time dependent. I know that I neglected the appropriate derivatives in the energy operator, but I was unsure of what else to do.

When I consider the similar problem for ##H = -\omega S_z##, I had that the Hamiltonian became

##
H = -\frac{\hbar \omega}{2} \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
##

##
\begin{pmatrix}
\Psi_1(x,t) \\
\Psi_2(x,t)
\end{pmatrix} =
\frac{1}{\sqrt{2}}\begin{pmatrix}
e^{i\frac{\omega}{2} t} \\
e^{-i\frac{\omega}{2} t}
\end{pmatrix}
##

The normalization ##\frac{1}{\sqrt{2}}## remained constant in time because of the presence of these oscillators. I am confused by the presence of the imaginary unit ##i## in the ##S_y##. It must be present by definition, but it cancels the ##i## in the energy operator ##E = i \hbar \frac{\partial}{\partial t}##. Then, the solutions appear to be exponential instead of oscillatory. This fact conflicted my intuition because -- as said previously -- the wavefunction is real and does not appear to be easily normalizable.

Also regarding my generalization to any 2 by 2 matrix. I am aware that the Pauli matrices span ##\mathbb{C}^{2\times 2}## and so the general solution could be written by decomposing the Hamiltonian into these constituent matrices. I had hoped that an alternate characterization in purely the language of introductory linear algebra could be fashioned by examining the equation as a system of ODEs and considering the spectrum. If
##
\frac{d\vec x}{dt} = A \vec x(t)
##
then
##
\vec x (t) = S\exp(\Lambda t) S^{-1} \vec x_0
##
by considering ##\exp## as a matrix power series.

Last edited:
Homework Helper
Gold Member
the Schrodinger equation implies
##\begin{pmatrix}\frac{\partial \Psi_1}{\partial t}\\\frac{\partial \Psi_2}{\partial t}\end{pmatrix}=\frac{\omega}{2}\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}
\begin{pmatrix}
\Psi_1\\
\Psi_2
\end{pmatrix}
##
On the right side, check to see if you missed an overall minus sign.

If you multiply out the right hand side, what column vector do you get?

uxioq99
uxioq99
@TSny Thank you, dividing by ##i## introduces the negative sign. I make more mistakes when I type things in Latex. Is the solution still supposed to be exponential, or should I have instead looked for a oscillatory behavior? I don't see how that is possible, given that the matrix defining the system becomes real. (Unless, I forgot an ##i## in addition to a negative sign.) My solution appears wrong because it no longer carries a phase.

Homework Helper
Gold Member
You shouldn't get exponentially growing or damping solutions.

What do you get after performing the matrix multiplication on the right-hand side of the equation below?

##\begin{pmatrix}\frac{\partial \Psi_1}{\partial t}\\\frac{\partial \Psi_2}{\partial t}\end{pmatrix}=\frac{\omega}{2}\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\begin{pmatrix}\Psi_1\\\Psi_2\end{pmatrix}##

hutchphd
uxioq99
I'm quite embarrassed. The values are on the off-diagonal. I apologize for asking this in the first place. I've been studying too long and now am just making stupid mistakes. Even by my standards, this was an incredible mistake -- sorry.

hutchphd
Homework Helper
Gold Member
No problem. I’m glad it’s resolved.

hutchphd