Show detected faces in Google Photos

Google photos knows about the people in pictures, and can give me a list of pictures every person appears in. When a picture is chosen as the default profile picture for a person, it knows how to center it properly on the face.

Is there a way when viewing a picture to know who is who on that picture (in a similar way to how Facebook highlights people)?

Contact names not showing on dialer, but do show in Recents

Have a Samsung J3 2017 (SM327). About a week ago, noticed that incoming & outgoing calls going to/from someone in my Contacts only shows the number, not the Contact name, so I often have no idea who’s calling me. But when I hang up and look in the Recent call log, the Contact name shows up correctly there. I have tried every permission, every setting possible. Spent an hour with an ATT tech – nothing. Did a factory restore – nothing. Took it to a Samsumg authorized serviced center and had it completely reflashed. Fixed the prob for 2 days, then reappeared. Any info?

Show that a set $K$ cofinal in a directed set $J$ is also a directed set

$ J$ directed means that for all pairs $ a,b$ in $ J, \exists c$ s.t. $ a,b<c$ and $ K$ is cofinal in $ J$ so for each $ j \in J$ , $ \exists d \in K$ such that $ j<d$ .

I am struggling with how to prove this. I have tried by contradiction (there exists $ x,y \in K$ such that there does not exist $ z \in K$ such that $ x,y<z$ ), but don’t know where to go from here.

Drupal 7 show terms in view filtered from another view

I have two views, first one is block, second one is page. I have vocabulary, let’s call it “food”. In the food vocabulary we have that hierarchy:

  • Fruit
    • Apple
    • Banana
  • Meat
    • Chicken
    • Pork

Both views goal is to show taxonomy terms, but first one shows ONLY parent terms, then we click on parent Fruit, we get second view, which shows ONLY terms from Fruit, so Apple and Banana. If we clicked on Meat term, we get terms Chicken and Pork. My question is how to do it?

How to show a field of the model to its inline method Django?


I have three models HomePage, Callout, FeatureContent like below:

class FeatureTip(models.Model):    feature_tip_title = models.CharField(max_length=120, null=True, blank=False)    feature_tip_description = models.TextField()    def __str__(self):       return self.feature_tip_title   class Citie(models.Model):    name = models.CharField(max_length=120, null=True, blank=False)    description = models.TextField()     def __str__(self):       return self.name   class HomePage(models.Model):    header = models.CharField(max_length=120, null=True, blank=False)    cities = models.ManyToManyField(Citie)    featured_tips = models.ManyToManyField(FeatureTip)     def __str__(self):       return 'Home Page'   class Callout(models.Model):    header = models.CharField(max_length=120, null=True, blank=False)    home_page = models.ForeignKey(HomePage,on_delete=models.CASCADE, null=True, blank=False)    def __str__(self):       return self.header     def get_city(self):       return self.cities.all()  class FeatureContent(models.Model):    title = models.CharField(max_length=120, null=True, blank=False)    home_page = models.ForeignKey(HomePage, on_delete=models.CASCADE, null=True)    def __str__(self):       return self.feature_article_title_en 

and it has been added to Admin.py like below:

class CalloutInline(admin.StackedInline):     model = Callout     fields = ['header']     extra = 4     max_num = 4     def get_queryset(self, request):         HomePage.objects.filter(name="Eminem")  class FeatureContentInline(admin.StackedInline):     model = FeatureContent     fields = ['title']     extra = 1     max_num = 1  class HomePageAdmin(admin.ModelAdmin):     filter_horizontal = ['cities', 'featured_tips']      inlines = [CalloutInline, FeatureContentInline] 

The problem is I want to show cities to show below the CalloutInline and featured_tips to show below FeatureContentInline. Currently, both are showing under Homepage. How can I do that?

Find the Taylor series of this polynomial. How do I formally show radius of convergence?

The question I am asked is this:

Find the Taylor series for $ f(x)$ centered at the given value of a. [Assume that f has a power series expansion. Do not show that $ R_n(x) -> 0.$ Also find the associated radius of convergence.

I’m having trouble finding a general formula of this Taylor series and therefore, also having problems finding the radius of convergence since I can’t perform the ratio test:

$ $ f(x) = x^6 – x^4 + 2$ $ at $ a = -2$

so:

$ $ f'(x) = 6x^5 – 4x^3$ $ $ $ f”(x) = 30x^4 – 12x^2$ $ $ $ f”'(x) = 120x^3 – 24x$ $ $ $ f^{iv}(x) = 360x^2 – 24$ $ $ $ f^{v}(x) = 720x$ $ $ $ f^{vi}(x) = 720$ $

and at a = -2 $ $ f(-2) = 50$ $ $ $ f'(-2) = -160$ $ $ $ f”(-2) = 432$ $ $ $ f”'(-2) = -912$ $ $ $ f^{iv}(-2) = 1416$ $ $ $ f^{v}(-2) = -1440$ $ $ $ f^{vi}(-2) = 720$ $

I’m having trouble finding the general formula for each term. Without it, how am I supposed to find the radius of convergence?

EDIT So the general term I have for the derivative of x is:

$ $ f^n(x) = \frac{6!x^{6-n}}{(6-n)!}$ $

So far the general term I have for the Taylor Series is:

$ $ \sum_{n=0}^{\infty} \frac{6! 2^{6-n}}{(6-n)!n!}(x+2)^n$ $

I can see why the radius of convergence is $ \infty$ , it’s because $ x$ can be anything and it’ll converge. But how do I show this formally? Can I use the ratio test?