text:
Q: file upload in gridview I need to add a column with a file upload control to my grid view so that I can upload files against any particular row. Is it possible to do this, ideally I need to be able to do this without putting the gridview into it's edit state.

A: You can use this within the  as follows:
   <asp:TemplateField HeaderText="UploadImage">

   <ItemTemplate>

      <asp:Image ImageUrl="~/images/1.jpg" runat="server" ID="image" /> // shown only when not in edit mode

    </ItemTemplate>

   <EditItemTemplate>

       <asp:FileUpload ID="FileUpload1" runat="server" /> // shown only in edit mode

   </EditItemTemplate>

  </asp:TemplateField>

At last  include a  as follows to enter edit mode.
       <asp:commandField showEditButton="true" showCancelButton="true">

Then add two events as follows:
   protected void GridView1_RowEditing(object sender, GridViewUpdateEventArgs e)
   {
       gvwID.EditIndex=e.NewEditIndex;
       BindGrid();
   }

    protected void GridView1_RowCancelEdit(object sender, GridViewUpdateEventArgs e)
    {
          gvwID.EditIndex=-1;
           BindGrid();
    }

The FileUpload control, will not automatically save the uploaded file. To save the file you need to use the FileUpload control's SaveAs method. Before you can use the SaveAs method you need to get the instance of the FileUpload control for the row you are editing. To get the instance of the control you can hook up to the GridView's RowUpdating event. The following code will get the instance of the FileUpload control and save the uploaded file:
   protected void GridView1_RowUpdating(object sender, GridViewUpdateEventArgs e)
   {

       int RowID=Convert.ToInt32(gvwID.DataKeys[e.RowIndex].value);

       FileUpload fileUpload = GridView1.Rows[e.RowIndex].FindControl("FileUpload1") as   FileUpload;

     if(fileUpload.HasFile)
     {

       fileUpload.SaveAs(System.IO.Path.Combine(Server.MapPath("Images"), fileUpload.FileName));

        //update db using the name of the file corresponding to RowID
      }

       gvwID.EditIndex=-1;
       BindGrid();

    }

Hope this will help...

A: The following link will help you:
http://msdn.microsoft.com/en-us/library/7tas5c80.aspx
It has a sample code for adding a DateTimePicker to datagridview cell.
You may add the fileupload control the same way...
Hope this helps...

A: Sudha have a great post with full functionality of file upload in GridView:
Fileupload control in gridview?

<asp:GridView ID="GridView1" runat="server" AutoGenerateColumns="False" DataKeyNames="ID"
DataSourceID="AccessDataSource1" Width="148px" OnRowCommand="GridView1_RowCommand">
<Columns>
<asp:BoundField DataField="ID" HeaderText="ID" ReadOnly="True" SortExpression="ID" />
<asp:BoundField DataField="Name" HeaderText="Name" SortExpression="Name" />
<asp:TemplateField>
<ItemTemplate>
<asp:Label ID="lblFileUpLoad" runat="server"></asp:Label>
</ItemTemplate>
<EditItemTemplate>
<asp:FileUpload ID="FileUpload1" runat="server" />
</EditItemTemplate>
</asp:TemplateField>
<asp:CommandField ShowEditButton="true" ShowDeleteButton="true" />
</Columns>
</asp:GridView>
....

A: The EditTemplate looks like this:
    <EditItemTemplate>

       <asp:TextBox ID="txtImage" runat="server" Text='<%# Bind("Image") %>' Visible="False"></asp:TextBox>
       <asp:FileUpload ID="FileUpload1" runat="server" />

    </EditItemTemplate>

In code behind this will upload the file on Row Update:
  protected void GridView1_RowUpdating(object sender, GridViewUpdateEventArgs e)
{
    GridViewRow row = GridView1.Rows[e.RowIndex] as GridViewRow;
    FileUpload FileUpload1 = (FileUpload)GridView1.Rows[e.RowIndex].FindControl("FileUpload1");

    if (FileUpload1 != null && FileUpload1.HasFile)
    {
        FileUpload1.SaveAs(Server.MapPath("~/uploads/" + myfilename));
    }
}

This check is in place in case no file is selected so previous name is picked. Note in edit template we have placed a textbox that has visibility set to false which binds to the image name in the DB
protected void GridView1_RowCommand(object sender, GridViewCommandEventArgs e)
{
    if (GridView1.EditIndex == -1) return;
    FileUpload fileUpLoad = GridView1.Rows[GridView1.EditIndex].FindControl("FileUpload1") as FileUpload;
    string fileName = fileUpLoad.FileName;
    TextBox txtImage = GridView1.Rows[GridView1.EditIndex].FindControl("txtImage") as TextBox;

    if (fileUpLoad != null && fileUpLoad.HasFile)
    {
        txtImage.Text = fileUpLoad.FileName;
    }
    else
    {
        txtImage.Text = txtImage.Text;
    }

}

A: <asp:ScriptManager runat="server" ID="scm"></asp:ScriptManager>

<asp:UpdatePanel runat="server" ID="upMain" UpdateMode="Conditional">
<ContentTemplate>

  <asp:GridView AutoGenerateColumns="False" runat="server" ID="dt">
<Columns>
<asp:TemplateField HeaderText="Catagory">
<ItemTemplate>
<asp:DropDownList runat="server" ID="ddlSubCat">
</asp:DropDownList>
</ItemTemplate>
</asp:TemplateField>

<asp:TemplateField HeaderText="Attachments">
<ItemTemplate>

<asp:UpdatePanel runat="server" UpdateMode="Conditional" ID="updFU">                   <ContentTemplate>
<asp:FileUpload runat="server" ID="updCon" /><asp:Button runat="server" ID="btnUpload" Text="Upload" />
</ContentTemplate>
<Triggers>
<asp:PostBackTrigger ControlID="btnUpload" />
</Triggers>
</asp:UpdatePanel>

</ItemTemplate>
</asp:TemplateField>
</Columns>
</asp:GridView>

</ContentTemplate>
</asp:UpdatePanel>

redpajama_set_name: RedPajamaStackExchange


When it comes to maintaining a healthy lifestyle, we are often recommended physical exercise, but there isn't much focus on exercising out minds. Engaging your mind in certain activities can result in amazing benefits for your brain as well as your body.
The second week of the Championship Challenge may be toughest you'll face, in large part due to the sugar detox you're experiencing. Read on to learn more about sugar's effect on your body and how you can overcome the detox experience.
Please remember to submit your Starting Measurements by 1/22 at 11:59pm eastern.
Please remember to submit your Starting Weight by 1/22 at 11:59pm eastern.
Please remember to submit your Benchmark Test results by 1/22 at 11:59pm eastern.
Recipe Challenge are open for submission until the deadline on 02/19 at 11:59pm eastern.
Performance Workout 1 results are open for submission until the deadline on 1/29 at 11:59pm eastern.
Welcome to week 2! Feeling a little off your game? If so, don't worry - you're not alone. Learn why you have a lot to look forward to and about this week's lifestyle focus: detox and de-stress.
Of course we want you to be excited about each weekly workout challenge, but the title is more than a reference to your potential emotional response. Variations of jumping and lower body exercises are the focus this week.
Week 1 Lifesytle Focus are open for submission until the deadline on 1/26 at 11:59pm eastern.
Cooking takes time, but if you want to eat what's best to reset your body's hormones and reach optimal performance level during the Challenge, it's a necessary part of the process. Meal planning can help make the most of the time you spend cooking.
As a 53-year-old aging ex-athlete with a number of physical limitations, I was failing in the weight and fitness aspects of my life. Yo-yoing in size my entire life, at my largest I was one meal away from 350lbs...most recently, Jan. 1st, 2016, I was 320lbs. With my wedding less than 6 months away to my beautiful, physically fit, personal trainer fiancé, I wanted to get a grip on my health. So for two months I did Weight Watchers, a great system but it just got stale for me, so after losing 20lbs over two months, the third month I put back on 9lbs. Sitting at 309 pounds, with a wedding shirt I bought on January 1st, 2 sizes too small, for incentive to lose weight…I was struggling. I'm not a quitter, but with the wedding less than 3 months away, fitting into my wedding shirt was looking bleak.
I finally did a smart thing, I asked my fiancé for help. She suggested the Level-X Challenge she was having some of her clients sign up for. Well, to make a long story short, I jumped on board with her guidance. I've learned not only the value of 'perimeter' shopping but how to do it and have enjoyable meals. The best video I watched on your site was how to handle the Bakery section of the grocery store where the host just turned and ran out of it!! I think of it every time I go and it helps me move along. I've played college level sports, been in (and out) of many gyms and with many weight loss attempts in my life I still couldn't grasp the full concept of healthy eating and physical activity. The Level-X Challenge has taught me a healthy balance of proper nutrition and physical activity that I actually can handle in this older, broken down body. The best part of it was I did it right at home, no clubs, gyms or partners telling me to do '1 more rep'. My pain level hovered in the 6-7 range before the challenge. Now, with the elimination of most dairy, sugar, and processed foods, my pain level has halved and my energy has doubled.
It feels good to be able to tie my shoes and not have to come up for air. I like being the 'Big Guy' I just needed not to be the 'Big Fat Guy'. Learning to make shopping lists and planning weekly menus is the key to my success, based on your principles of eating of course. Learning that I can still move, modifications of what I used to be able to do is okay and not requiring hours and hours in the gym is life changing for me. The help of my fiancé on the food preparation and shopping has been immeasurable and having the support to talk with her about the workouts was key. The nutritional information on the site was enlightening and your encouragement throughout was a big help.
With 4 days left until the Challenge ends and when I weigh myself again, I can comfortably fit in my wedding shirt and I need a belt for my pants!!! Thanks, Challenge Team, for a thoughtful, insightful and reasonable challenge that is made for not only the 'fit' but those of us that needed to be 'fit' again.
Making changes to your life is not always easy, but it can help to develop a routine. Routines can make changes become habits. Developing a routine can help you fit healthy changes into your busy life - even the smallest changes to your routine can have a big impact on your health.
Goals may now be submitted until 1/29 at 11:59pm eastern.
Benchmark Test results are open for submission until the deadline on 01/22 at 11:59pm eastern.
The Challenge has begun! Before you go off to the races, check out the first weekly update so you don't miss anything, you know what to look for, and you can learn about this week's lifestyle focus: routine.
Starting Measurements are open for submission until the deadline on 9/28 at 11:59pm eastern.
Starting Weight are open for submission until the deadline on 9/28 at 11:59pm eastern.
Enjoy daily inspiration for getting active, fit, and healthy. Our exclusive 5-week fitness plan provides 6 workouts per week, each with 3 fitness levels, so anybody can start where you are at and see daily improvement during the next several weeks. Access your Daily Fitness Plan here.


text:
"""
This module provides a client for STS.
"""

from future.utils import iteritems
import copy
import http.client
import os
import json
import logging
import shutil
from builtins import str
from builtins import bytes

import baidubce
from baidubce import bce_client_configuration
from baidubce import utils
from baidubce.auth import bce_v1_signer
from baidubce.bce_base_client import BceBaseClient
from baidubce.exception import BceClientError
from baidubce.exception import BceServerError
from baidubce.exception import BceHttpClientError
from baidubce.http import bce_http_client
from baidubce.http import handler
from baidubce.http import http_content_types
from baidubce.http import http_headers
from baidubce.http import http_methods
from baidubce.services import sts
from baidubce.utils import required

_logger = logging.getLogger(__name__)

class StsClient(BceBaseClient):
    """
    sdk client
    """
    def __init__(self, config=None):
        BceBaseClient.__init__(self, config)

    def get_session_token(self, acl, duration_seconds=None):
        """
        :type duration_seconds: int
        :param duration_seconds: None
        :token effective period

        :type id: string
        :param id: None
        :acl id

        :type acl: dict
        :param acl: None
        :acl

        :return:
            **HttpResponse**
        """

        params = None

        if duration_seconds is not None:
            if isinstance(duration_seconds, int):
                params = {b'durationSeconds': duration_seconds}

        if acl is None:
            body = None
        else:
            if not isinstance(acl, dict):
                raise TypeError(b'acl should be dict')
            if 'id' in acl:
                body = json.dumps(acl)
            else:
                body = json.dumps(acl, default=self._dump_acl_object)

        return self._send_request(
            http_methods.POST,
            body=body,
            headers={http_headers.CONTENT_TYPE: http_content_types.JSON},
            params=params)

    @staticmethod
    def _dump_acl_object(acl):
        result = {}
        for k, v in iteritems(acl.__dict__):
            if not k.startswith('_'):
                result[k] = v
        return result

    def _merge_config(self, config):
        if config is None:
            return self.config
        else:
            new_config = copy.copy(self.config)
            new_config.merge_non_none_values(config)
            return new_config

    def _send_request(self,
                      http_method,
                      body=None,
                      headers=None,
                      params=None,
                      config=None,
                      body_parser=None):
        config = self._merge_config(config)
        path = sts.URL_PREFIX + b"sessionToken"
        if body_parser is None:
            body_parser = handler.parse_json

        return bce_http_client.send_request(
            config, bce_v1_signer.sign, [handler.parse_error, body_parser],
            http_method, path, body, headers, params)

redpajama_set_name: RedPajamaGithub


End of quarter is a highly stressful time during campaigns. You have to reach important fundraising goals, which often means a lot of pressure on finance and web teams. To relieve some of the stress, listed below are great tips to ease things for the people in charge of communications.
Running out of good ideas is easy when you're sending emails every weekday. It's always a smart move to keep some ideas at hand when you don't have time to do much thinking. These are some ideas to start you off on the right path.
Keep people up to date on supporter activity (Facebook likes, volunteers, email subscribers, etc.) How much money was already raised for the current quarter and how many donations already received. People always like feeling as if they are a part of something huge and that momentum is increasing in the campaign.
A thermometer graphic that shows the "temperature" of your objective can provide people with a visual sense of the progress your campaign is making. If the numbers indicate that the contribution average is lower than possible, let your supporters know and kindly request that people donate the average amount in helping to further the goals for the EOQ.
Lastly, reward yourself (and your team). Running campaigns is tough work. Congratulate the whole team by throwing a party. The quarter's end is a very intense time for ever person in your department, so organizing activities for your team like pizza parties or maybe laser tag or even a potluck can relieve a ton of stress. There are countless ways your team can take time to celebrate the successful quarter and unwind. Having a fun event to look forward to can be the perfect motivation. Team events are also the perfect time to discuss important things like what didn't work and what gets the best results.
NGP VAN is a United States-based company that is dedicated to making sure that organizations running campaigns can effectively use technology in reaching campaign goals. In 2009, NGP Van was the leading provider of campaign software provider. NGP Vans applications are used by many Congress members. Their software has been utilized in the Obama campaign, Hillary Clinton, Bernie Sanders, Liberal Party Of Canada and the British Liberal Democrats campaigns.


text:
Mercury Mouth: Are Amalgam Fillings Safe?
The FDA has deemed them harmless, but concern over the safety of amalgam dental fillings — those silver-colored fillings that contain a mixture of liquid mercury and a powdered silver, tin, and copper alloy — persists. Here's what you should know.
Wyatt Myers
Few dental health issues have gotten more attention recently than mercury dental fillings, commonly called "silver" or amalgam fillings.
In a statement released by the U.S. Food and Drug Administration (FDA) in 2009, these dental fillings, which contain mercury along with other metals, were pronounced safe for most adults and children over 6 years of age.
However, in December 2010, an FDA advisory panel of scientists and other dental health experts met again to review the safety information of amalgam fillings. The panel upheld the 2009 ruling, but it also urged the FDA to continue to review and evaluate their safety, suggesting the issue still isn't definitively settled.
Since then, a whole lot of confusion has ensued. Many people are wondering: Are my fillings really safe?
The History of Mercury Dental Fillings
Mercury dental fillings have been the mainstay of cavity fillings for more than 150 years. These dental fillings are a mixture of roughly half liquid mercury and half a combination of silver, tin, and copper.
The reason that amalgam fillings are ideal is that they are soft and malleable. "Mercury is the binding agent that holds the other metals in the amalgam together," says Vincent C. Mayher, DMD, past president of the Academy of General Dentistry. "When manipulating the material to place in the tooth, dentists can sculpt the filling so that it fits properly and ensures that the patient's bite is correct."
FDA Green-Lights Marketing of New Hybrid Closed-Loop System for Type 1 Diabetes
FDA Approves New Migraine Drug for Fast Symptom Relief
But Are Mercury Fillings Risky?
Why are there concerns over amalgam fillings? Because of their mercury content.
Critics believe that mercury dental fillings may cause neurological problems in fetuses and young children. "The concern centers on the health effects of toxicity or allergy that may be associated with mercury exposure, particularly as a potential cause of chronic illnesses, autoimmune disorders, neurodegenerative diseases, birth defects, oral lesions, and mental disorders," says Jeffrey Gross, DDS, an associate clinical professor at the Case Western Reserve University School of Dental Medicine in Cleveland.
The FDA acknowledges that the fillings release low levels of mercury vapor that can be inhaled. But the 2009 FDA findings determined that these levels were far too low to have any adverse health effects on adults and children ages 6 and above. They also concluded that the amount of mercury in breast milk attributable to amalgam fillings was safe for infants. It is noteworthy, however, that the FDA reclassified dental mercury from a class I device to a class II device that requires special considerations and the need to warn patients of the potential risks involved.
While there are still public concerns about the safety of amalgam fillings, the position of both the FDA and the American Dental Association (ADA) is this: They are safe. "There are some who feel the mercury contained in these fillings is toxic and could potentially jeopardize a person's physical health," says Dr. Mayher. "Currently neither the FDA nor the ADA supports these claims, which are usually anecdotal and not based on hard or proven science. It is difficult to determine why some individuals feel that these fillings are harming them. It's possible that these claims first started as a result of situations of true metal allergies that, while rare, are a valid cause for concern."
Alternatives to Amalgam Fillings
If you still don't feel safe about having mercury in your mouth, there are dental fillings that can serve as suitable alternatives to amalgam fillings. In fact, the newer type of filling now represent about 70 percent of the fillings used today.
10 Plant-Based Recipes You Can Make in 30 Minutes or Less
Following a plant-based diet may help you lose weight and lower your risk for chronic disease. Peruse 10 quick and flavorful recipes that are good to make if you're trying to eat more fruits, veggies, whole grains, nuts, and seeds. Learn More
"The newer composite resin materials are a viable option," explains Mayher. "Many dentists only place fillings of this type. In addition, because composites are tooth-colored, they present a much more aesthetic alternative to the silver-colored composites."
If you already have amalgam fillings and are concerned about their potential adverse impact on your health, ask your dentist about composite resin fillings.
How Do You Get Tonsil Stones? And What to Do About Them
What You Need to Know About Contaminants Found in Zantac
The Latest in Dental Health
A Detailed Guide to Canker Sores: Causes, Prevention, Remedies, Treatment, and More
Most canker sores are harmless and common, but they can be painful and recurrent. The more you know about these lesions, the easier it'll be to treat ...
By Valencia Higuera October 25, 2018
Home Remedies and Treatments for Canker Sores
Many canker sores will go away without treatment, but a number of home remedies can speed healing and offer pain relief.
Drinking Alcohol Increases Disease-Causing Mouth Bacteria
Changes in oral microbes may lead to gum disease, heart problems, and cancer, study says.
By Don Rauf April 24, 2018
Want Home Remedies For Cleaning False Teeth?
Are you looking for home remedies for cleaning false teeth? Learn a number of home remedies for cleaning dentures.

redpajama_set_name: RedPajamaCommonCrawl


Not the version you're looking for? See A Donation of Runecloth.
<name>, you've been a tremendous contributor to our cloth drive. As we work hard to replenish our reserves, one form of cloth remains critically low - runecloth. We are in absolutely dire need of runecloth, and we hope that you will be able to help us as you have in the past.
If you are willing, please bring me what runecloth you can spare. We'll initially accept a single donation of sixty, and then we'll go from there.
Fantastic - these will go right to use, I assure you. Thank you again for your assistance! We'd certainly be lost without you.
With all the donations you've made, I would certainly process any additional runecloth donations you care to make in the future. Just be sure to check in with me, and I'll personally make sure that your good deeds won't go unrecognized!
Patch 1.7.0 (2005-09-22): All 8 versions of the "A Donation of Runecloth" quest have had their quest text clarified.


Lil Pump And Zaytoven Keep The Trap Boomin' With New Song "Designer"
Lil Pump linked up with the trap gawd.
Lil Pump has been loved and hated by many this year, but regardless of what anyone has to say about him, the Florida rapper keeps a smile on his face as he parties his way to the top. After receiving some backlash for his wildly popular "Gucci Gang" music video, Pump reaches back into his bag for a new cut with trap-gawd Zaytoven.
As expected, he uses the rowdy instrumental to floss and brag about his riches. Listen to "Designer" below.


The Directors Guild of America Award for Outstanding Directorial Achievement in Motion Pictures is one of the annual awards given by Directors Guild of America. With 3 wins out of 11 nominations, Steven Spielberg is both the most awarded and most nominated director for this category in the history of DGA. Additionally, Alejandro G. Iñárritu is the only director to win consecutive awards. He was awarded in 2015 and 2016 for his directorial work in Birdman or (The Unexpected Virtue of Ignorance) and The Revenant, respectively.
The Directors Guild of America Award for Feature Film has traditionally been a near perfect barometer for the Academy Award for Best Director. Only seven times since the DGA Award's inception has the DGA Award winner not won the Academy Award; in 1968 (Carol Reed won the Oscar for directing Oliver!); 1972 (Bob Fosse won the Oscar for directing Cabaret); 1985 (Sydney Pollack won the Oscar for directing Out of Africa, as Steven Spielberg was not nominated for directing The Color Purple); 1995 (Mel Gibson won for directing Braveheart, as Ron Howard was not nominated for directing Apollo 13); 2000 (Steven Soderbergh won the Oscar for directing Traffic); 2002 (Roman Polanski won the Oscar for directing The Pianist); and 2012 (Ang Lee won the Oscar for directing Life of Pi as Ben Affleck was not nominated for directing Argo). This does not count 1948–49, when the DGA used a non-calendar award year (unlike the Academy Awards); both 1948 and 1949 DGA winners competed for the 1949 Academy Award for Best Director, which went to the 1948 DGA winner (Joseph L. Mankiewicz for A Letter to Three Wives).


Our challenge was to promote and instill the classic adventurous spirit of Land Rover in the next generation of prospective owners. And encourage them to discover life beyond their office cubicles.
Every prospective Land Rover adventurer has a day job and spends hours working on their PC every day. We decided to surprise them with the Land Rover LR4 Escape Key. A small object that lets them personalise their keyboard in a special way.
Now when they're at work and routine drives them up the wall, the key will serve as a gentle reminder to escape the everyday. And consider test driving the Land Rover LR4.
Little boxes of LR4 Escape Keys made their way to potential Land Rover customers in and around the region. Since most PC keyboards come with either square, round or rectangle pegs, we provided all three connectors along with the box.
We have sent out three batches of 800 pieces. The number of queries almost tripled. Test drive's up by 208%. But most importantly, it made Land Rover LR4 the top-of-mind 4x4 vehicle of our prospective customers.
The project was featured in Creativity-Online where it was made 'Pick of the Day', became the top weekly winner at Best Ads on TV, and talked about in popular blogs and news sites such as Brainstorm9, I Believe in Advertising, The Inspiration Room, Creative Guerrilla Marketing and Creative Criminals among others.
The custom made LR4 Escape Keys are made of plastic resin coated in metallic paint. Making it light and durable for constant office abuse.


Oolong teas compose one of the most varied categories of tea because they cover the entire range of oxidation levels between green and black tea. Newcomers to this category might be surprised to find that two teas called "oolong" could be completely different in aroma, flavor, and even leaf shape. In fact, there are four main types of oolong, hailing from four geographically isolated regions near the eastern coast of China. Each region has its own unique terroir and specialized varieties of Camellia sinensis, but the easiest influence to observe in the final flavors of the tea is each region's specific crafting style. And one aspect of the finished tea that gives us a clue to crafting style is leaf shape.
Some oolong teas consist of compressed, or "balled" tea leaves. Though high quality versions will expand into large, unbroken leaves, these oolongs are always tightly compressed into small, round pebbles. Commonly used in both Anxi County, Fujian and across the strait in Taiwan, this technique reduces the amount of surface area exposed to air, and is preferred for most lightly oxidized oolongs, which remain green in color. More oxidized leaves can be rolled as well, and some rolled teas from both regions are heavily roasted, resulting in richer flavor profiles and a darker leaf.
Green-style oolongs from both Taiwan and China are well known for intense floral aromas, but Taiwanese teas are also notable for the creamy textures that develop at high elevations. Oolong teas from Taiwan are typically named for the mountain they are grown on, like Tung Ting (Cold Summit), Alishan, or Lishan (Pear Mountain). Lightly oxidized styles are a popular recent development in Anxi County, where traditional nong xiang Tieguanyin are still valued for their rich flavor and nutty finish.
Just as with rolled oolongs, twisted oolongs are produced in two main regions, but these two styles have much less in common than the two rolled types. In the southern province of Guangdong, oolongs from the Phoenix Mountains are twisted and oxidized up to 30% to bring forth natural flavor notes of fruit or flowers. These teas are finished with a light roast to emphasize their bright aromas and subtle flavor notes. Meanwhile, in the Wuyi Mountains of northern Fujian, leaves are bruised and allowed to oxidize until they are nearly 70% brown before undergoing heavy roasting, traditionally over charcoal wood. These teas emerge bold and toasty, fortified with the flavor of the roast.
Again, color is a clue to the specific style, but this difference is more subtle than in rolled oolongs. Instead, the aroma and flavor notes are the greatest distinguishing factor when observing these two styles of tea. Wuyi oolongs will appear darker, almost black, and smell smoky or toasted, while Phoenix oolongs will lighten to a mottled red and green when rehydrated, and smell more honey-like or fruity.
Unfortunately, leaf shape alone tells us little about the actual flavor of the tea, which depends more on factors like harvest date and oxidation level. But paying attention to leaf shape along with other details can help identify specific styles that you like (or don't like), and choose new teas accordingly.


text:
\section{Introduction}

Let $\mathbb{F}_q$ denote the finite field of characteristic $p$
with $q$ elements ($q=p^e, e\in \mathbb{N}$, the set of positive integers), and $\mathbb{F}_q^*$
the non-zero elements of $\mathbb{F}_q$. Let $\mathbb{F}_q[x]$ be the polynomial ring with indeterminate $x$. For every positive integer $s$ and a positive divisor $t$ of $s$, the {\it relative trace map} from $\mathbb{F}_{q^s}$ to $\mathbb{F}_{q^t}$ is defined as
\begin{equation}\label{f-1}
   {\rm Tr}^s_t(c)=c+c^{2^t}+c^{2^{2t}}+\cdots+c^{2^{s-t}}, \forall c\in \mathbb{F}_{q^s}.
\end{equation}
The {\it absolute trace map} is defined by
\begin{equation*}
    {\rm Tr}(c)=c+c^p+c^{p^2}+\cdots+c^{p^{e-1}}, \forall c\in \mathbb{F}_q,
\end{equation*}
and the function maps from $\mathbb{F}_q$ to $\mathcal{C}^*$, the set of nonzero complex numbers, defined by
\begin{equation*}
    \chi_a(c)=e^{2\pi \sqrt{-1}{\rm Tr}(ac)/p}, \forall c\in \mathbb{F}_q,
\end{equation*}
is called an {\it additive character} of  $\mathbb{F}_q$.
Let $\chi$ be an additive character of $\mathbb{F}_q$, and $f(x)\in \mathbb{F}_q[x]$. The sum
\begin{equation*}
    \mathcal{S}(f)=\sum_{c\in \mathbb{F}_q}\chi(f(c))
\end{equation*}
is called a {\it Weil Sum}.
Let $g$ be a generator of the cyclic group $\mathbb{F}_q^*$, the function maps from $\mathbb{F}_q$ to $\mathcal{C}^*$ defined by
\begin{equation*}
   \psi(g^k)=e^{2\pi \sqrt{-1}k/(q-1)}, \mbox{ for } k=0,1,2,\cdots,q-2,
\end{equation*}
is called a {\it multiplicative character} of $\mathbb{F}_q$. It is easy to see that $\psi$ is a generator of the characteristic group of $\mathbb{F}_q^*$. For every multiplicative character $\psi$ of $\mathbb{F}_q$ and a polynomial $f(x)\in \mathbb{F}_q[x]$, one can also define the following exponential sum
\begin{equation*}
   \mathcal{T}(f)=\sum_{c\in \mathbb{F}_q}\psi(f(c)),
\end{equation*}
here we extend the definition of $\psi$ to the set $\mathbb{F}_q$ by setting $\psi(0)=0$.

The problem of
explicitly evaluating these sums, $\mathcal{S}(f), \mathcal{T}(f)$, is quite often difficult. Results giving estimates
for the absolute value of the sums are more common and such results have been
regularly appearing for many years. Lidl and Niederreiter gave an
overview of this area of research in the concluding remarks of Chapter 5 in \cite{lidl}. See also, \cite{coulter,lison, moiso,vlugt,wan,weil1,weil2} for instance.

Using the technique of L-functions, one can prove the following results:

{\thm \label{thm-1}\cite[p.~220, Theorem 5.36]{lidl} Let $f(x)\in \mathbb{F}_q[x]$ be of degree $n\geq 2$ with $\gcd(n,q)=1$ and let $\chi$ be a nontrivial additive character of $\mathbb{F}_q$. Then there exist complex numbers $\omega_1,\omega_2,\cdots, \omega_{n-1}$, only depending on $f$ and $\chi$, such that for any positive integer $s$ we have
\begin{equation}\label{f-2}
    \sum_{\gamma\in \mathbb{F}_{q^s}}\chi^{(s)}(f(\gamma))=-\omega_1^s-\omega_2^s-\cdots\omega_{n-1}^s,
\end{equation}
where $\chi^{(s)}(x)=\chi({\rm Tr}^s_1(x))$ for all $x\in \mathbb{F}_{q^s}$ is the lifting of $\chi$ from $\mathbb{F}_q$ to $\mathbb{F}_{q^s}$.}

{\thm \label{thm-2} \cite[p.223, Theorem 5.39]{lidl} Let $\psi$ be a multiplicative character of $\mathbb{F}_q$ of order $m>1$ and $f\in \mathbb{F}_q[x]$ be a monic polynomial of positive degree that is not an $m$-th power of a polynomial. Let $d$ be the number of distinct roots of $f$ in its splitting field over $\mathbb{F}_q$ and suppose that $d\geq 2$. Then there exist complex numbers $\theta_1,\theta_2,\cdots,\theta_{d-1}$, only depending on $f$ and $\psi$, such that for every positive integer $s$ we have
\begin{equation*}
   \sum_{\gamma\in \mathbb{F}_{q^s}}\psi^{(s)}(f(\gamma))=-\theta_1^s-\theta_2^s-\cdots-\theta_{d-1}^s,
\end{equation*}
where $\psi^s(x)=\psi({\rm Norm}_1^s(x))$ for all $x\in \mathbb{F}_{q^s}$ is the lifting of $\psi$ from $\mathbb{F}_q$ to $\mathbb{F}_{q^s}$, and
\begin{equation*}
    {\rm Norm}_1^s(x)=x^{1+q+q^2+\cdots+q^{s-1}}.
\end{equation*}}

For convenience, we make the following notations:
\begin{equation}\label{f-0}
    \mathcal{S}_s(f)= \sum_{\gamma\in \mathbb{F}_{q^s}}\chi^{(s)}(f(\gamma)), \mathcal{T}_s(f)= \sum_{\gamma\in \mathbb{F}_{q^s}}\psi^{(s)}(f(\gamma)).
\end{equation}

In this note, we establish some recursive formulae about $\mathcal{S}_s(f)$ and $\mathcal{T}_s(f)$. Some results abut the exponential sum $G_u^{(s)}(a,b)=\sum_{c\in \mathbb{F}^*_{q^s}}\chi^{(s)}(ac^u+bc^{-1})$ are obtained, where $u$ is a positive integer, $a,b\in \mathbb{F}_q$.

The organization of the rest of the paper is as follows: In Sect. 2, we introduce some preliminaries results which will be used in the sequel. In Sect. 3, we give a recursive formula and an estimate for the exponential sum $G_u(a,b)=\sum_{c\in \mathbb{F}_q^*}\chi(ac+bc^{-1})$. In Sect. 4, we provide some properties of the sequences constructed from these exponential sums.
\section{Preliminaries}

In this section, we introduce the concept of Dickson polynomials, Newton's identities and an useful tool for evaluating exponential sums.

\subsection{Dickson polynomials}

Before going to state our main results, we need the concept of Dickson polynomials.

Consider a polynomial

\begin{equation}\label{f-3}
    r(c_1,c_2,\cdots,c_k,x)=x^{k+1}-c_1x^k+c_2x^{k-1}+\cdots+(-1)^kc_kx+(-1)^{k+1}a\in \mathbb{F}_q[x].
\end{equation}
This polynomial has $k+1$ not necessarily distinct roots $\beta_1,\cdots,\beta_{k+1}$ in a suitable extension of $\mathbb{F}_q$. Now, let $n\in \mathbb{N}$, the set of positive integers, and set
\begin{equation*}
  r_n(c_1,\cdots,c_k,x)=(x-\beta_1^n)\cdots (x-\beta_{k+1}^n).
\end{equation*}
Define
\begin{eqnarray*}
e_0(x_1,x_2,\cdots,x_{k+1})&=&1\\
  e_1(x_1,x_2,\cdots,x_{k+1}) &=& \sum_{i=1}^{k+1}x_i \\
  e_2(x_1,x_2,\cdots,x_{k+1})  &=& \sum_{1\leq i<j\leq k+1}^{k+1}x_ix_j \\
 \dots &\cdots& \cdots \\
 e_{k+1}(x_1,x_2,\cdots,x_{k+1})  &=& x_1x_2\cdots x_{k+1}\\
 e_m(x_1,x_2,\cdots,x_{k+1})  &=&0, \mbox{ for }m>k+1.
\end{eqnarray*}
We know that the coefficients of $r_n$ are elementary symmetric functions $e_i(\beta_1^n,\cdots,\beta_{k+1}^n),$ $i=1,2,\cdots,k+1$. Since $e_i$ is symmetric in the indeterminates $x_1,\cdots,x_{k+1}$, there exist integral polynomials $D^{(1)}_n, \cdots, D_n^{(k+1)}$ in $k+1$ indeterminates such that
\begin{equation*}
   e_i(x_1^n,\cdots, x_{k+1}^n)=D_n^{(i)}(e_1(x_1,\cdots,x_{k+1}),\cdots,e_{k+1}(x_1,\cdots,x_{k+1}))
\end{equation*}
for $1\leq i\leq k+1$.
Since $\beta_1,\cdots,\beta_{k+1}$ are the roots of the polynomial $r$ we have $e_i(\beta_1,\cdots,\beta_{k+1})=c_i$ for $1\leq i\leq k$ and $e_{k+1}(\beta_1,\cdots, \beta_{k+1})=a$. Thus we have
\begin{equation*}
   D_n^{(i)}(c_1,\cdots,c_k,a)=e_i(\beta_1^n,\cdots,\beta_{k+1}^n).
\end{equation*}
Therefore,
\begin{equation*}
    r_n(c_1,\cdots,c_k,x)=x^{k+1}+\sum_{1\leq i\leq k}(-1)^iD_n^{(i)}(c_1,\cdots,c_k,a)x^{k+1-i}+(-1)^{k+1}a^n.
\end{equation*}

\begin{defn} The Dickson polynomials of the first kind $D_n^{(i)}(x_1,\cdots,x_k,a)$, $1\leq i\leq k$, are given by the functional equations
\begin{equation*}
   D_n^{(i)}(x_1,\cdots,x_k,a)=e_i(u_1^n,\cdots,u_{k+1}^n), 1\leq i\leq k,
\end{equation*}
where $x_i=e_i(u_1,\cdots,u_{k+1})$ and $u_1\cdots u_{k+1}=a$.\end{defn}

Thus, particularly, for $i=1$, we have
\begin{equation}\label{f-4}
    D_n^{(1)}(e_1,e_2,\cdots, e_{k+1})=u_1^n+u_2^n+\cdots+u_{k+1}^n,
\end{equation}
where $e_j=e_j(u_1,\cdots,u_{k+1})$.

Waring's formula also gives the explicit expression of $D_n^{(1)}(x_1,\cdots,x_k,a)$ as
\begin{eqnarray*}
    &&D_n^{(1)}(x_1,\cdots,x_k,a) \\
    &=& \sum_{i_1=0}^{\lfloor \frac{n}{2}\rfloor}\cdots \sum_{i_k=0}^{\lfloor \frac{n}{k+1}\rfloor}\frac{n}{n-i_1-2i_2-\cdots-ki_k}{n-i_1-2i_2-\cdots-ki_k\choose i_1+\cdots+i_{k-1}}\\
    &&{i_1+\cdots+i_k\choose i_1+\cdots+i_{k-1}}\cdots{i_1+i_2\choose i_1}a^{i_k}(-1)^{i_1+2i_2+\cdots+ki_k}\\
    &&x_1^{n-2i_1-\cdots-(k+1)i_k}x_2^{i_1}\cdots x_k^{i_{k-1}},
\end{eqnarray*}
where $\lfloor x \rfloor$ denotes the largest integer $\leq x$. See Lidl \cite[~p.19]{LMT} for details.

Moreover, we have the following results.

{\lem \cite[~p.19]{LMT}The Dickson polynomials of the first kind $D_n^{(1)}(x_1,\cdots,x_k,a)$ satisfy the generating function
\begin{equation*}
   \sum_{n=0}^{\infty}D_n^{(1)}(x_1,\cdots,x_k,a)z^n=\frac{\sum_{i=0}^k(k+1-i)(-1)^ix_iz^i}{\sum_{i=0}^{k+1}(-1)^ix_iz^i} \mbox{ for $n\geq 0$}
\end{equation*}
and the recurrence relation
\begin{equation}\label{f-5}
  D_{n+k+1}^{(1)}-x_1D_{n+k}^{(1)}+\cdots+(-1)^kx_kD_{n+1}^{(1)}+(-1)^{k+1}aD_n^{(1)}=0
\end{equation}
with the $k+1$ initial values
\begin{equation*}
   D_0^{(1)}=k+1,D_j^{(1)}=\sum_{t=1}^j(-1)^{t-1}x_tD_{j-t}^{(1)}+(-1)^j(k+1-j)x_j \quad\mbox{for $0<j\leq k$}.
\end{equation*}
}

\subsection{Newton's identities}
Let $x_1,\cdots, x_{k+1}$ be $k+1$ unnecessarily distinct numbers, denote for $n \geq 1$ by $p_n(x_1,\cdots,x_{k+1})$ the $n$-th power sum:
\begin{equation*}
    p_n(x_1,\cdots,x_{k+1})=x_1^n+x_2^n+\cdots+x_{k+1}^n.
\end{equation*}
Then Newton's identities can be stated as
\begin{equation}\label{f-6}
    me_m(x_1,\cdots,x_{k+1})=\sum_{j=1}^m(-1)^{j-1}e_{m-j}(x_1,\cdots,x_{k+1})p_j(x_1,\cdots,x_{k+1})
\end{equation}
valid for all $m\geq 1$. See ${\rm http://en.wikipedia.\ org/ wiki/ Newton's\ identities}$ for details.

For convenience, we denote $p_j(x_1,\cdots,x_{k+1})$ simply by $p_j$, and denote $e_j(x_1,\cdots,x_{k+1})$ by $e_j$.
By (\ref{f-6}), we know that
\begin{equation}\label{f-7}
   p_m=\sum_{j=1}^{m-1}(-1)^{j-1}e_jp_{m-j}+(-1)^{m-1}me_m, m\geq 2.
\end{equation}

It is easily seen that both (\ref{f-7}) and (\ref{f-5}) tell the something.

By (\ref{f-7}) and Cramer's rule, we obtain  that
\begin{eqnarray}\label{f-8}
 p_m=\left|\begin{array}{ccccc}
             e_1 & 1 & 0 & \cdots &  \\
            2e_2 & e_1 & 1 & 0& \cdots \\
             3e_3 & e_2 & e_1 &\ddots &  \\
             \vdots & \vdots & \vdots & \ddots &  \\
             me_m & e_{m-1} & \cdots & & e_1
           \end{array}
 \right|, \end{eqnarray}and\begin{eqnarray}\label{f-9}e_m=\frac{1}{m!}\left|\begin{array}{ccccc}
             p_1 & 1 & 0 & \cdots &  \\
            p_2 & p_1 & 2 & 0& \cdots \\
             \vdots & \vdots & \ddots& \ddots &  \\
             p_{m-1}&p_{m-2}&\cdots&p_1&n-1\\
             p_m & p_{m-1} & \cdots & p_2& p_1
           \end{array}\right|.
\end{eqnarray}

As an application of Newton's identities, we show that (\ref{f-5}) is sufficient to define the Dickson polynomial of the first kind $D_n^{(1)}$.

{\prop If $D_n$ satisfies (\ref{f-5}) with the initial values, then as a sequence, $D_n$ is just $D_n^{(1)}$.}

 \begin{proof}
 As a sequence, $D_n$ has the characteristic polynomial as
\begin{equation*}
     r(x_1,x_2,\cdots,x_k,x)=x^{k+1}-x_1x^k+x_2x^{k-1}+\cdots+(-1)^kx_kx+(-1)^{k+1}a.
\end{equation*}
Let the roots of $ r(x_1,x_2,\cdots,x_k,x)=0$ be $\beta_1,\cdots,\beta_{k+1}$. Then we have $x_i=e_i(\beta_1,\cdots,\beta_{k+1})$, and for every positive integer $n$ $$D_n^{(1)}=l_1\beta_1^n+\cdots+l_{k+1}\beta_{k+1}^n$$ holds for some constants $l_1,\cdots,l_{k+1}$. Now by the initial values of $D_j^{(1)}$, we get the following system of equations.
\begin{equation*}
\left\{ \begin{array}{ll}
 l_1+\cdots+l_{k+1} = k+1& \\
  \sum_{v=1}^{k+1}l_v\beta_v^j=\sum_{t=1}^j(-1)^{t-1}x_t\sum_{v=1}^{k+1}l_v\beta_v^{j-t}+(-1)^j(k+1-j)x_j \quad\mbox{for $0<j\leq k$}.&
 \end{array}\right.
\end{equation*}
Simplifying this system, we have
\begin{equation*}
\left\{ \begin{array}{ll}
 l_1+\cdots+l_{k+1} = k+1& \\
  jx_j=\sum_{t=1}^j(-1)^{t-1}x_{j-t}\sum_{v=1}^{k+1}l_v\beta_v^t\quad\mbox{for $0<j\leq k$}.&
 \end{array}\right.
\end{equation*}
Comparing with (\ref{f-6}), we obtain $l_1=\cdots=l_{k+1}=1$ is a solution to this system. Furthermore, $D_n$ is uniquely determined by its initial values. Thus we obtain $l_1=\cdots=l_{k+1}=1$ is the unique solution to this system and $D^{(1)}=\beta_1^n+\cdots+\beta_{k+1}^n$.
This completes the proof.\end{proof}

Thus we obtain an equivalent definition of Dickson polynomials of the first kind.

Now we consider the exponential sums defined in Theorem \ref{thm-1} and Theorem \ref{thm-2}. Using the relations between the power sums and the elementary functions, we obtain the following recursive formulae:

{\prop (1) Let $f$ and $\mathcal{S}_s(f)$ be defined as in Theorem \ref{thm-1} and (\ref{f-0}).Then for every integer $s\geq 2$, we have
\begin{equation*}
   \mathcal{S}_s(f)=\sum_{j=1}^{s-1}(-1)^{j-1}e_{j}(\omega_1,\cdots,\omega_{n-1})\mathcal{S}_{s-j}(f)+(-1)^sse_s(\omega_1,\cdots,\omega_{n-1}).
\end{equation*}
(2) Let $f$ and $\mathcal{T}_s(f)$ be defined as in Theorem \ref{thm-2} and (\ref{f-0}). Then for every integer $s\geq 2$, we have
\begin{equation*}
    \mathcal{T}_s(f)=\sum_{j=1}^{s-1}(-1)^{j-1}e_{j}(\theta_1,\cdots,\theta_{d-1})\mathcal{T}_{s-j}(f)+(-1)^sse_s(\theta_1,\cdots,\theta_{d-1}).
\end{equation*}

}

\subsection{An useful tool for computing exponential sums}
In order to determine some exponential sums over finite fields, we use the idea of $L-$functions: The canonical L-function of the exponential sum $\mathcal{S}_s(f)=\sum_{c\in \mathbb{F}_{q^s}}\chi^{(s)}(f(c))$ is defined by $L^*(f,t)=\exp\left(\sum_{s=1}^{\infty}\mathcal{S}_s(f)\frac{t^s}{s}\right)$. By a famous theorem of Dwork and Grothendieck, $L^*(f, t)$ is a rational function.

Let $\Phi$ be the set of monic polynomials over $\mathbb{F}_q$, and let $\lambda$ be a complex-valued function on $\Phi$ which is
multiplicative in the sense that
\begin{equation}\label{f-10}
    \lambda(gh)=\lambda(g)\lambda(h) \mbox{ for all }g,h\in\Phi,
\end{equation}
and which satisfies $|\lambda(g)|\leq 1$ for all $g\in \Phi$, and $\lambda(1)=1$. Denote by $\Phi_k$ the subset of $\Phi$ containing the polynomials of degree $k$.

Consider the power series
\begin{equation}\label{f-11}
    L(z)=\sum_{k=0}^{\infty}\left(\sum_{g\in \Phi_k}\lambda(g)\right)z^k.
\end{equation}
It is easily seen that
\begin{equation*}
    L(z)=\prod_{f\ irr}(1-\lambda(f)z^{\deg(f)})^{-1},
\end{equation*}
where the product is taken over all monic irreducible polynomials $f$ in $\mathbb{F}_q[x]$. Now apply logarithmic differentiation and multiply by $z$ to get
\begin{equation}\label{f-12}
    z\frac{d\log L(z)}{dz}=\sum_{s=1}^{\infty}L_sz^s
\end{equation}
with
\begin{equation}\label{f-13}
   L_s=\sum_{f}\deg(f)\lambda(f)z^{s/\deg(f)}
\end{equation}
where the sum is extended over all monic irreducible polynomials $f$ in $\mathbb{F}_q[x]$ with degree dividing $s$.

Now suppose that there exists a positive integer $t$ such that
\begin{equation}\label{f-14}
    \sum_{g\in \Phi_k}\lambda(g)=0 \mbox{ for all } k>t.
\end{equation}
Then $L(z)$ is a complex polynomial of degree $\leq t$ with constant $1$, so that we have
\begin{equation}\label{f-15}
  L(z)=(1-\omega_1z)(1-\omega_2z)\cdots(1-\omega_tz)
\end{equation}
with complex numbers $\omega_1,\omega_2,\cdots,\omega_t$. It follows that
\begin{equation}\label{f-16}
    L_s=-\omega_1^s-\omega_2^s-\cdots-\omega_t^s \mbox{ for all }s\geq 1.
\end{equation}

\section{A recursive formula and an estimate for a specific exponential sum}

Now we consider the exponential sums
\begin{equation}\label{f-17}
   G_u(a,b)=\sum_{c\in \mathbb{F}^*_q}\chi(ac^u+bc^{-1}), a\in \mathbb{F}^*_q,b\in \mathbb{F}_q,
\end{equation}
here $u\geq 1$ is a positive integer.

If $u=1$, then $G_u(a,b)$ is the well-known Kloosterman sum. If $u=3$, then $G_u(a,b)$ is called the {\it inverse cubic sum}. It is easily seen that $G_u(a,b)=G_u(ab^u,1)$ if $b\neq 0$.

In what follows, we proceed to give a recursive formula and an estimate of the exponential sum $G_u(a,b)$ by using the ideal introduced in subsection 2.3. To this end, we should define a multiplicative function $\lambda$ from the set $\Phi$ of all the monic polynomials in $\mathbb{F}_q[x]$ to the set of complex numbers.

We put $\lambda(1)=1$ and if $g\in \Phi_k$, $k\geq 1$, say
\begin{equation*}
    g(x)=\sum_{j=0}^k(-1)^jc_jx^{k-j} \mbox{ with }c_0=1,
\end{equation*}
suppose that $g$ factors as
\begin{equation*}
    g(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_k)
\end{equation*}
in its splitting field, we set
\begin{equation*}
    \tilde{g}(x)=(x-\alpha_1^u)(x-\alpha_2^u)\cdots(x-\alpha_k^u)=\sum_{j=0}^k(-1)^j\tilde{c}_jx^{k-j} \mbox{ with }\tilde{c}_0=1.
\end{equation*}
It is easily seen that the numbers
\begin{equation*}
     \tilde{c}_1=\sum_{j=1}^k\alpha_j^u, c_{k-1}c_k^{-1}=\sum_{j=1}^k\alpha_j^{-1}
\end{equation*}
are invariant under the Frobenius automorphism $x\mapsto x^q$, thus $\tilde{c}_1, c_{k-1}c_k^{-1}\in \mathbb{F}_q$, and we can define
\begin{equation*}
    \lambda(g)=\chi(a\tilde{c}_1+bc_{k-1}c_k^{-1}) \mbox{ if } c_k\neq 0,
\end{equation*}
and $\lambda(g)=0$ if $c_k=0$.

Now we check that $\lambda(gh)=\lambda(g)\lambda(h)$ holds for all $g,h\in \Phi$. Suppose that
\begin{equation*}
     g(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_k), h(x)=(x-\gamma_1)(x-\gamma_2)\cdots(x-\gamma_l),
\end{equation*}
and then
\begin{eqnarray*}
  && \tilde{g}(x)=(x-\alpha_1^u)(x-\alpha_2^u)\cdots(x-\alpha_k^u) = \sum_{j=0}^k(-1)^j\tilde{c}_jx^{k-j} \\
  &&\tilde{h}(x)=(x-\gamma_1^u)(x-\gamma_2^u)\cdots(x-\gamma_k^u) = \sum_{i=0}^l(-1)^i\tilde{d}_ix^{k-i}
\end{eqnarray*}
with
\begin{equation*}
   \tilde{ c}_1=\sum_{j=1}^k\alpha_j^u, \tilde{d}_1=\sum_{i=1}^l\gamma_i^u; c_{k-1}c_k^{-1}=\sum_{j=1}^k\alpha_j^{-1}, d_{l-1}d_l^{-1}=\sum_{i=1}^l\gamma_i^{-1}.
\end{equation*}
Therefore, if $$g(x)h(x)=\sum_{m=0}^{k+l}(-1)^mE_mx^{k+l-m}, \tilde{g}(x)\tilde{h}(x)=\sum_{m=0}^{k+l}(-1)^m\tilde{E}_mx^{k+l-m},$$
we have,
\begin{equation*}
    \tilde{E}_1=\tilde{c}_1+\tilde{d}_1, E_{k+l-1}E_{k+l}^{-1}=\sum_{j=1}^k\alpha_j^{-1}+\sum_{i=1}^l\gamma_i^{-1}=c_{k-1}c_k^{-1}+ d_{l-1}d_l^{-1}.
\end{equation*}
Thus we get that $\lambda(gh)=\lambda(g)\lambda(h)$ holds for all $g,h\in \Phi$.

By Newton's identities (\ref{f-7}), we have
\begin{equation*}
   \tilde{c}_1=\sum_{j=1}^k\alpha_j^u=p_u=\sum_{j=1}^{u-1}(-1)^je_jp_{u-j}+(-1)^{u-1}ue_u, u\geq 2,
\end{equation*}
where $p_j=\alpha_1^j+\cdots+\alpha_k^n$ is the power sum, since $p_j$ is a symmetric function on $\alpha_1,\cdots,\alpha_k$, we know that $p_j\in \mathbb{F}_q$, and $e_j=e_j(\alpha_1,\cdots,\alpha_k)=c_j$ for $j\leq k$ and $e_j=0$ for $j>k$. Therefore,
\begin{equation}\label{f-26}
   \tilde{c}_1=\left\{\begin{array}{ll}
                        \sum_{j=1}^k(-1)^jc_jp_{u-j} & \mbox{ if }k<u, \\
                        \sum_{j=1}^{u-1}(-1)^{j-1}c_jp_{u-j}+(-1)^{u-1}uc_u & \mbox{ if }k\geq u.
                      \end{array}
   \right.
\end{equation}
It is easily seen that $\sum_{j=1}^{u-1}(-1)^{j-1}c_jp_{u-j}=\pi(c_1,\cdots,c_{u-1})$ is a function on $c_1,\cdots,c_{u-1}$.

Hence, for $k>u+1$, if $\gcd(u,q)=1$, we can write
\begin{eqnarray*}
  \sum_{g\in \Phi_k}\lambda(g)&=& \sum_{c_1,c_2,\cdots,c_{k-1}\in \mathbb{F}_q}\sum_{c_k\in \mathbb{F}_q^*}\chi(a\tilde{c}_1+bc_{k-1}c_k^{-1})\\
  &=&q^{k-u-1}\left(\sum_{c_u\in \mathbb{F}_q}\chi(a(-1)^{u-1}uc_u)\right)\cdot\\
  & &\left(\sum_{c_k\in \mathbb{F}_q^*}\sum_{c_1, \cdots,c_{u-1}, c_{k-1}\in \mathbb{F}_q}\chi(a\sum_{j=1}^{u-1}(-1)^{j-1}c_jp_{u-j})\chi(bc_k^{-1}c_{k-1})\right)\\
  &=&0;
\end{eqnarray*}
If $b\neq 0$, we can write
\begin{eqnarray*}
  \sum_{g\in \Phi_k}\lambda(g)&=& \sum_{c_1,c_2,\cdots,c_{k-1}\in \mathbb{F}_q}\sum_{c_k\in \mathbb{F}_q^*}\chi(a\tilde{c}_1+bc_{k-1}c_k^{-1})\\
  &=&q^{k-u-1}\sum_{c_k\in \mathbb{F}_q^*}\sum_{c_1, \cdots,c_{u}\in \mathbb{F}_q}\chi\left(a\sum_{j=1}^{u-1}(-1)^{j-1}c_jp_{u-j}+(-1)^{u-1}uc_u\right)\\
  & &\left(\sum_{c_{k-1}\in \mathbb{F}_q}\chi(bc_k^{-1}c_{k-1})\right)\\
  &=&0;
\end{eqnarray*}
We note that the aim of presupposition $k>u+1$ is to separate $c_u$ and $c_{k-1}$ such that $c_u$ or $c_{k-1}$ is free in the above summation.

Therefore (\ref{f-14}) holds with $t=u+1$. From (\ref{f-15}) we obtain
\begin{equation*}
    L(z)=(1-\omega_1z)(1-\omega_2z)\cdots(1-\omega_{u+1}z)
\end{equation*}
for some complex numbers $\omega_1,\cdots,\omega_{u+1}$.

Now we calculate $L_s$ from (\ref{f-13}). Let $g$ be a monic irreducible polynomial in $\mathbb{F}_q[x]$ whose degree
divides $s$, and let $\gamma\in E=\mathbb{F}_{q^s}$ be a root of $g$. Then $g^{s/{\deg(g)}}$ is the characteristic polynomial of $\gamma$ over
$\mathbb{F}_q$; that is,
\begin{equation*}
    g(x)^{s/\deg(g)}=(x-\gamma)(x-\gamma^q)\cdots(x-\gamma^{q^{s-1}})=x^s-c_1x^{s-1}+\cdots+(-1)^{s-1}c_{s-1}x+(-1)^sc_s,
\end{equation*}
say. Then $c_1={\rm Tr}_{E/\mathbb{F}_q}(\gamma)$, $c_s=\gamma \gamma^q\cdots\gamma^{q^{s-1}}$, and $$c_{s-1}c_s^{-1}=\gamma^{-1}+\gamma^{-q}+\cdots+\gamma^{-q^{s-1}}={\rm Tr}_{E/\mathbb{F}_q}(\gamma^{-1}).$$
Thus we have
\begin{eqnarray*}
  \lambda\left(g(x)^{s/\deg(g)}\right)&=& \chi\left(a(\gamma^u+\gamma^{uq}+\cdots+\gamma^{uq^{s-1}})+b(c_{s-1}c_s^{-1}) \right)\\
  &=&\chi\left(a{\rm Tr}_{E/\mathbb{F}_q}(\gamma^u)+b{\rm Tr}_{E/\mathbb{F}_q}(\gamma^{-1})\right)\\
  &=&\chi^{(s)}(a\gamma^u+b\gamma^{-1}),
\end{eqnarray*}
and so
\begin{equation*}
    L_s=\sum_{g}\ ^*\deg(g)\lambda(g^{s/\deg(g)})=\sum_{g}\ ^*\sum_{\gamma\in E,g(\gamma)=0}\chi^{(s)}(a\gamma^u+b\gamma^{-1}),
\end{equation*}
where $^*$ means the summation is extend to all monic irreducible polynomial $g(x)\in \mathbb{F}_q[x]$
with $g(x)=x$ is excluded. If $g$ runs through the range of summation above, then $\gamma$ runs exactly through all elements of $E^*$. Thus
\begin{equation*}
    L_s=\sum_{\gamma\in E^*}\chi^{(s)}(a\gamma^u+b\gamma^{-1}).
\end{equation*}
Therefore, we have proved the following:

{\thm \label{thm-3} Let $\mathbb{F}_q$ be a finite field and $u$ be a positive integer. Let $s$ be any positive integer, and
$\chi^{(s)}$ be the lifting of the additive character $\chi$ of $\mathbb{F}_q$. For any two elements $a,b\in \mathbb{F}_q$ with $a\neq 0$, define
\begin{equation}\label{f-27}
   G_u^{(s)}(a,b)=\sum_{c\in \mathbb{F}^*_{q^s}}\chi^{(s)}(ac^u+bc^{-1}).
\end{equation}
Then, if either $\gcd(u,q)=1$ or $b\neq 0$, there exist complex numbers $\omega_1,\omega_2,\cdots, \omega_{u+1}$, only depending on $u,a,b$ and $\chi$, such that for any positive integer $s$ we have
\begin{equation}\label{f-2}
   G_u^{(s)}(a,b) =-\omega_1^s-\omega_2^s-\cdots\omega_{u+1}^s.
\end{equation}}

Moreover, for the complex numbers $\omega_1,\cdots,\omega_{u+1}$, we have the following:

{\thm \label{thm-5}Let $u$ be a positive integer with $\gcd(u,q)=\gcd(u+1,q)=1$ and $ab\neq 0$. Then the complex numbers $\omega_1,\cdots,\omega_{u+1}$ in Theorem \ref{thm-3} are all of absolute $\leq \sqrt{q}$.}

Thus we have the following:

{\cor Let $u$ be a positive integer with $\gcd(u,q)=\gcd(u+1,q)=1$ and $ab\neq 0$, and let the exponential sum $G_u^{(s)}(a,b)$ be defined as in Theorem \ref{thm-3}. Then we have
\begin{equation*}
    |G_u^{(s)}(a,b)|\leq (u+1)\sqrt{q} \mbox{ for all }s=1,2,\cdots.
\end{equation*}}

For the proof of Theorem \ref{thm-5}, we need the following lemma:

{\lem \label{lem-1} \cite[Lemma 6.55, p.310]{lidl} Let $\omega_1,\cdots,\omega_n$ be complex numbers, and let $B>0$, $C>0$ be constants such that
\begin{equation}\label{f-19}
   |\omega_1^s+\cdots+\omega_n^s|\leq CB^s \mbox{ for } s=1,2,\cdots.
\end{equation}
Then $|\omega_j|\leq B$ for $j=1,2,\cdots,n.$}

Now we give the proof of Theorem \ref{thm-5}.

\begin{proof}For any $a, b\in \mathbb{F}_q$ with $a\neq 0$, we write
\begin{equation*}
   \sum_{c\in E^*}\chi^{(s)}(ac^u+bc^{-1})=\sum_{c\in E^*}\chi\left({\rm Tr}_{E/\mathbb{F}_q}(ac^u+bc^{-1})\right)=\sum_{v\in \mathbb{F}_q}N(v)\chi(v),
\end{equation*}
where
\begin{equation*}
    N(v)=|\{c\in E^*|{\rm Tr}_{E/\mathbb{F}_q}(ac^u+bc^{-1})=v\}|.
\end{equation*}
If $\beta_0$ is a fixed element in $E$ with ${\rm Tr}_{E/\mathbb{F}_q}(\beta_0)=v$, then by Hilbert's Theorem 90, we know that
${\rm Tr}_{E/\mathbb{F}_q}(ac^u+bc^{-1})=v$ if and only if $ac^u+bc^{-1}=\beta^q-\beta+\beta_0$ for some $\beta\in E$, (see \cite[Theorem 2.25]{lidl}), or equivalently, $ac^{u+1}-(\beta^q-\beta+\beta_0)c+b=0$. Let $N$ be the number of solutions of
\begin{equation}\label{f-21}
  f(x,y)= ay^{u+1}-(x^q-x+\beta_0)y+b=0
\end{equation}
in $E^2$.

We proceed to show that $f(x,y)$ is absolutely irreducible over $\mathbb{F}_{q^s}$. We note that a polynomial is called {\it absolute irreducible} over a field $L$ means that this polynomial is irreducible over every extension field of $L$.

If there is an extension field $K$ of $\mathbb{F}_{q^s}$ such that $f(x,y)$ is reducible in $K[x,y]$. Let
  \begin{equation*}
    \Omega=\{g(x,y)| g(x,y) \mbox{ is an irreducible factor of } f(x,y) \mbox{ in } K[x,y]\}.
\end{equation*}
We define the action of the additive group of $\mathbb{F}_q$ on $\Omega$ as follows.
If $f_1(x,y)=a_0(y)+a_1(y)x+\cdots+a_t(y)x^t \in \Omega$, for every element $\xi\in \mathbb{F}_q$, define the action of $\xi$ on $f_1(x,y)$ by $\sigma_\xi: x\mapsto x+\xi,y\mapsto y$. It is clear that this is a group action, so either

Case (1) $f_1(x,y)$ is invariant under the action of every $\sigma_\xi, \xi\in \mathbb{F}_q$, or

Case (2) there are some orbits, such that the product of those polynomials in each orbit is invariant under the action of every $\sigma_\xi, \xi\in \mathbb{F}_q$.

Thus, both in case (1) and case (2), we always have a factor of $f(x,y)$, say $f_1(x,y)$, which is invariant under the action of $\mathbb{F}_q$.

We denote by $\partial_x(f_1(x,y))$ (resp. $\partial_y(f_1(x,y))$) the highest degree of $x$ (resp. $y$) among all monomials in $f_1(x,y)$, then all the polynomials in an orbit have the same $\partial_x$, and the same $\partial_y$.

We proceed to prove that $\partial_x(f_1(x,y))=q$.

Since $f_1(x,y)$ is invariant under the action of $\sigma_\xi,\xi\in \mathbb{F}_q$, considered as a polynomial in indeterminate $z$, the following polynomial
\begin{equation*}
    T(z)=f_1(x_0+z,y_0)-f_1(x_0,y_0)
\end{equation*}
has at least $q$ roots, namely, all the elements in $\mathbb{F}_q$, where $x_0,y_0$ are prescribed two elements arbitrarily. Thus $T(z)$ is divisible by $\prod_{\xi\in \mathbb{F}_q}(z-\xi)=z^q-z$. In other words,
\begin{equation*}
    T(z)=f_1(x_0+z,y_0)-f_1(x_0,y_0)=(z^q-z)q(z) \mbox{ for a } q(z)\in K[z].
\end{equation*}
Comparing the degrees of both sides yields that $\partial_z(q(z))=0$.
Hence $f_1(x,y)$ should have the form $g_1(y)+(x^q-x)g_2(y)$ for some polynomials $g_1(y),g_2(y)$. Suppose that $f(x,y)$ factors as $f(x,y)=f_1(x,y)f_2(y)$ with $\deg(f_2(y))\geq 1$.
Then we have
\begin{equation*}
    f(x,y)=ay^{u+1}-(x^q-x+\beta_0)y+b=(g_1(y)+(x^q-x)g_2(y))f_2(y),
\end{equation*}
which yields that $g_2(y)$ is a constant and $\deg(f_2(y))=1$. Suppose that $g_2(y)=\alpha_0, f_2(y)=\epsilon_0+\epsilon_1y$, then
\begin{equation*}
    ay^{u+1}-(x^q-x+\beta_0)y+b=g_1(y)(\epsilon_0+\epsilon_1y)+(x^q-x)\alpha_0(\epsilon_0+\epsilon_1y)
\end{equation*}
which implies that $\epsilon_0=0$ and thus $b=0$. Contrary to the fact that $b\neq 0$.

It follows that $f(x,y)$ can not have a proper factor which is invariant under the action of $\mathbb{F}_q$. Therefore, $f(x,y)$ can not factor as $f_1(x,y)g(y)$ for a $g(y)\in K[y]$ with $\deg(g(y))\geq 1$. By the same reason, $f(x,y)$ can not factor as $f_1(x,y)g(x)$ for a $g(x)\in K[x]$ with $\deg(g(x))\geq 1$, for if it does, it is obvious that $\partial_y(\sigma_\xi(f_1(x,y)))=\partial_y(f_1(x,y))$, then $f_1(x,y)$ is a proper factor of $f(x,y)$ which is invariant under the action of $\mathbb{F}_q$, a contradiction.

Therefore, we know that the group action $(\mathbb{F}_q,\Omega)$ has just one orbit, since the product of polynomials in an orbit is invariant under the action of $\mathbb{F}_q$. Thus we know that the group action $(\mathbb{F}_q,\Omega)$ is transitive and $|\Omega|$ divides $q$. Thus for every $f_1(x,y),f_2(x,y)\in \Omega$, we have $\partial_x(f_1(x,y))=\partial_x(f_2(x,y))=d_x$, $\partial_y(f_1(x,y))=\partial_y(f_1(x,y))=d_y$. So that $u+1=d_y |\Omega|$ and $q=d_x|\Omega|$. By the assumption that $\gcd(u+1,q)=1$, we get at last that $|\Omega|=1$ and $f(x,y)$ is irreducible in $K[x,y]$. This contradiction means that $f(x,y)$ is absolutely irreducible.

Now, by a famous result of Weil\footnote{It says that if a polynomial $f(x,y)\in \mathbb{F}_q[x]$ is absolutely irreducible, then there exists a constant $C$ such that the number $N$ of the solutions of the equation $f(x,y)=0$ satisfies $|N-q|\leq C\sqrt{q}$.}, see Weil \cite{weil1,weil2},
we have,
\begin{equation}\label{f-24}
    |N-q^s|\leq Cq^{s/2} \mbox{ for some constant }C,
\end{equation}
where the constant $C$ is independent on $s$.

On the other hand, for each fixed $y\in E$ satisfies (\ref{f-21}), there are $q$ choices of $x$, namely by adding any element of $\mathbb{F}_q$, thus we have
\begin{equation}\label{f-23}
  N=qN(v)
\end{equation}

From (\ref{f-24}),(\ref{f-23}), it follows that
\begin{equation*}
  | N(v)-q^{s-1}|\leq Cq^{s/2-1}.
\end{equation*}
Denote by $R(v)$ the number $N(v)-q^{s-1}$, then $|R(v)|\leq Cq^{s/2-1}$, hence
\begin{eqnarray*}
  |G_u^{(s)}(a,b)|&=&|-\omega_1^s-\cdots-\omega_{u+1}^s| \\
 &=& \left|\sum_{c\in E^*}\chi^{(s)}(ac^u+bc^{-1})\right|=\left|\sum_{v\in \mathbb{F}_q}N(v)\chi(v)\right|\\
 &=&\left|\sum_{v\in \mathbb{F}_q}(q^{s-1}+R(v))\chi(v)\right|=\left|\sum_{v\in \mathbb{F}_q}R(v)\chi(v)\right|\\
 &\leq& Cq^{s/2}.
\end{eqnarray*}
The desired result then follows with Lemma \ref{lem-1}. This completes the proof.
\end{proof}

Denote $G_u^{(s)}(a,b)$ by $G^{(s)}$, then by Newton's identities, or by Dickson polynomials, we have the following:

{\cor \label{cor-3.5}Let $G^{(s)}$ and the numbers $\omega_1,\cdots,\omega_{u+1}$ be defined as above. Then
\begin{equation*}
    G^{(s)}=\sum_{j=1}^{s-1}(-1)^{j-1}e_j(\omega_1,\cdots,\omega_{u+1})G^{(s-j)}+(-1)^se_s(\omega_1,\cdots,\omega_{u+1}) \mbox{ for all }s\geq 2.
\end{equation*}

}
Because $e_s(\omega_1,\cdots,\omega_{u+1})=0$ whenever $s>u+1$. In order to setting the recursive formulae to work, we only need to know the $u+1$ initial values $G^{(1)},\cdots,G^{(u+1)}$, since we can find the elementary functions values $e_1,\cdots,e_{u+1}$ by (\ref{f-9}) under the presupposition that the characteristic of the finite field is bigger than $u+1$. On the other hand, if we can find the values of $\sum_{g\in \Phi_1}\lambda(g), \cdots,\sum_{g\in \Phi_{u+1}}\lambda(g)$, we can also determine the recursive formula without any restriction on the characteristic of the finite field. For example, when $u=1$, by Theorem \ref{thm-1}, we obtain the following well-known results about the Kloosterman sums. See \cite[p.226]{lidl} for instance.

{\cor Let $a,b$ be two elements in a finite field $\mathbb{F}_q$. Then there are two complex numbers $\omega_1,\omega_2$, depending on $ab$ and $\chi$,
 such that for every positive integer $s$, we have
 \begin{equation}\label{f-25}
    k(\chi^{(s)},a,b)=\sum_{c\in \mathbb{F}^*_{q^s}}\chi^{(s)}(ac+bc^{-1})=-\omega_1^s-\omega_2^s.
 \end{equation}
 Moreover, we have $|\omega_1|=|\omega_2|=\sqrt{q}$.}

 Since $\omega_1+\omega_2=-k(\chi,a,b)$, $ \omega_1\omega_2=q$, we have $e_1(\omega_1,\omega_2)=-k(\chi,a,b),e_2(\omega_1,\omega_2)=q$, by
 Newton's identities or by property of Dickson polynomials, we get the following:
 {\cor \cite[p.229]{lidl} Denote $k(\chi^{(s)},a,b)$ by $k^{(s)}$. Then
 \begin{equation}\label{f-26}
    k^{(s)}=-k^{(s-1)}k-qk^{(s-2)} \mbox{ for }s\geq 2,
 \end{equation}
 and
 \begin{equation}\label{f-27}
   k^{(s)}=-\left(\omega_1^s+\left(\frac{q}{\omega_1}\right)^s\right)=-D_s^{(1)}(\omega_1+\frac{q}{\omega_1},q)=-D_s^{(1)}(-k,q).
 \end{equation} where we put $k^{(0)}=-2, k^{(1)}=k=k(\chi,a,b)$, $D_s^{(1)}(.,.)$ is the Dickson polynomial.}
\vskip 0.3 cm
 If $u=2$ and $q$ is even, we proceed to compute the corresponding function $L(z)$ as follows.

 Obviously, if $ab\neq 0$, then
 \begin{equation*}
    \sum_{g\in \Phi_1}\lambda(g)=\sum_{g=x-c,c\neq 0}\lambda(g)=\sum_{c\neq 0}\chi(ac^2+bc^{-1})=G_2(a,b).
 \end{equation*}
 Moreover,
 \begin{eqnarray*}

 \sum_{g\in \Phi_2}\lambda(g)&=&\sum_{c_1\in \mathbb{F}_q,c_2\in \mathbb{F}^*_q}\chi(a(\alpha_1^2+\alpha_2^2)+bc_1c_2^{-1})\\
  &=&\sum_{c_1\in \mathbb{F}_q,c_2\in \mathbb{F}^*_q}\chi(a(\alpha_1+\alpha_2)^2+bc_1c_2^{-1})\\
  &=&\sum_{c_1\in \mathbb{F}_q,c_2\in \mathbb{F}^*_q}\chi(ac_1^2+bc_1c_2^{-1}),
 \end{eqnarray*}
where $\alpha_1,\alpha_2$ are the roots of $x^2-c_1x+c_2=0$ in an extension of the finite field $\mathbb{F}_q$. Thus
\begin{equation*}
    \sum_{g\in \Phi_2}\lambda(g)=q.
\end{equation*}
Furthermore,
\begin{eqnarray*}
   \sum_{g\in \Phi_3}\lambda(g)&=& \sum_{g(x)=x^3-c_1x^2+c_2x-c_3, c_3\neq 0}\lambda(g) \\
  &=& \sum_{c_1,c_2\in \mathbb{F}_q}\sum_{c_3\in \mathbb{F}^*_q}\chi(a(\alpha_1^2+\alpha_2^2+\alpha_3^2)+bc_2c_3^{-1})\\
  &=&\sum_{c_1,c_2\in \mathbb{F}_q}\sum_{c_3\in \mathbb{F}^*_q}\chi(ac_1^2+bc_2c_3^{-1})=0
\end{eqnarray*}
Thus, we have
\begin{equation*}
    L(z)=1+G_2(a,b)z+qz^2.
\end{equation*}
If $L(z)=(1-\omega_1z)(1-\omega_2z)$, then $\omega_1\omega_2=q$. By Theorem \ref{thm-5}, we know that $|\omega_1|=|\omega_2|=\sqrt{q}$.

By Corollary \ref{cor-3.5}, we have the following:
{\prop Let $q$ be a power of $2$, $\mathbb{F}_q$ be a finite field.  For any integer $s$, define
\begin{equation*}
    G^{(s)}(a,b)=\sum_{c\in \mathbb{F}^*_{q^s}}\chi^{(s)}(ac^2+bc^{-1}).
\end{equation*}
Then,

(1) for every elements $a,b\in \mathbb{F}^*_q$, we have
\begin{equation*}
  G^{(s)}(a,b)=-G_2(a,b)G^{(s-1)}(a,b)-qG^{(s-2)} \mbox{ for all }s>3,
\end{equation*}
where $G_u(a,b)$ is defined by (\ref{f-17}).

(2) If $1+G_2(a,b)z+qz^2=(1-\omega_1z)(1-\omega_2z)$, then for every positive integer $s$,
\begin{equation*}
    G^{(s)}(a,b)=-\omega_1^s-\omega_2^s.
\end{equation*}

(3) $|\omega_1|=|\omega_2|=\sqrt{q}$.
}
\vskip 0.1 cm
In the case of $u=2$ and $q$ is odd, the situation is slightly different, we have
 \begin{equation*}
    \sum_{g\in \Phi_1}\lambda(g)=\sum_{g=x-c,c\neq 0}\lambda(g)=\sum_{c\neq 0}\chi(ac^2+bc^{-1})=G_2(a,b),
 \end{equation*}
and
\begin{eqnarray*}
  \sum_{g\in \Phi_2}\lambda(g)&=&\sum_{c_1,c_2\in \mathbb{F}_q,c_2\neq 0}\chi(a(\alpha_1^2+\alpha_2^2)+bc_1c_2^{-1}) \\
 &=&\sum_{c_2\neq 0}\chi(-2ac_2)\sum_{c_1\in \mathbb{F}_q}\chi(ac_1^2+bc_1c_2^{-1}).
\end{eqnarray*}
Since $ac_1^2+bc_1c_2^{-1}$ is a quadratic function in $c_1$, the inner sum is
\begin{equation*}
    \sum_{c_1\in \mathbb{F}_q}\chi(ac_1^2+bc_1c_2^{-1})=\chi(-4^{-1}b^2c_2^{-2}a^{-1})\eta(a)g(\eta,\chi),
\end{equation*}
where $\eta$ is the quadratic charter, and $g(\eta,\chi)$ is the Gaussian sum
of the quadratic character $\eta$ and the additive character $\chi$. This Gaussian sum has been determined explicitly, see \cite[Theorem 5.12, Theorem 5.15]{lidl} for instance.
Hence we have
\begin{eqnarray*}
   \sum_{g\in \Phi_2}\lambda(g)&=& \eta(a)g(\eta,\chi)\sum_{c_2\neq 0}\chi(-2ac_2)\chi(-4^{-1}b^2c_2^{-2}a^{-1}) \\
 &=& \eta(a)g(\eta,\chi)\sum_{c_2\neq 0}\chi(-4^{-1}a^{-1}b^2c_2^2-2ac_2^{-1})\\
 &=&\eta(a)g(\eta,\chi)G_2(-\frac{b^2}{4a},-2a).
\end{eqnarray*}
For $\sum_{g\in \Phi_3}\lambda(g)$, we have
\begin{eqnarray*}
 \sum_{g\in \Phi_3}\lambda(g)&=&\sum_{c_1,c_2,c_3\in \mathbb{F}_q,c_3\neq 0}\chi(a(\alpha_1^2+\alpha_2^2+\alpha_3^2)+bc_2c_3^{-1}) \\
 &=&\sum_{c_1,c_2,c_3\in \mathbb{F}_q,c_3\neq 0}\chi(a(c_1^2-2c_2)+bc_2c_3^{-1})\\
 &=&\sum_{c_1,c_2\in \mathbb{F}_q}\chi(ac_1^3+3c_3)\sum_{c_3\in \mathbb{F}^*_q}\chi((bc_2c_3^{-1})\\
 &=&q\sum_{c_1\in \mathbb{F}_q}\chi(ac_1^2)\\
 &=&q\eta(a)g(\eta,\chi).
\end{eqnarray*}
Thus, if $L(z)=(1-\omega_1z)(1-\omega_2z)(1-\omega_3z)$, then $|\omega_1\omega_2\omega_3|=q|\eta(a)||g(\eta,x)|=q^{3/2}$. By Theorem \ref{thm-5}, we know that $|\omega_1|=|\omega_2|=|\omega_3|=\sqrt{q}$.
Therefore, we obtain the following result.

{\prop Let $\mathbb{F}_q$ be a finite field with ${\rm char}(\mathbb{F}_q)>2$.  For any integer, define
\begin{equation*}
    G^{(s)}(a,b)=\sum_{c\in \mathbb{F}^*_{q^s}}\chi^{(s)}(ac^2+bc^{-1}).
\end{equation*}
Then,

(1) for every elements $a,b\in \mathbb{F}^*_q$, we have
\begin{eqnarray*}
  G^{(s)}(a,b)&=&-G_2(a,b)G^{(s-1)}(a,b)-\eta(a)g(\eta,\chi)G_2(-\frac{b^2}{4a},-2a)G^{(s-2)}(a,b)\\
  & &-q\eta(a)g(\eta,\chi)
\end{eqnarray*}
 for all $s>3$.

(2) If $1+G_2(a,b)z+\eta(a)g(\eta,\chi)G_2(-\frac{b^2}{4a},-2a)z^2+q\eta(a)g(\eta,\chi)z^3=(1-\omega_1z)(1-\omega_2z)(1-\omega_3z)$, then for every positive integer $s$,
\begin{equation*}
    G^{(s)}(a,b)=-\omega_1^s-\omega_2^s-\omega_3^s.
\end{equation*}

(3) $|\omega_1|=|\omega_2|=|\omega_3|=\sqrt{q}$.

Where $G_u(a,b)$ is defined by (\ref{f-17}).

}

\section{Sequences constructed from the exponential sums}
In this section, we restrict the characteristic of the finite field being $2$, and we denote $G_u(a,1)$ simply by $G_u(a)$. For every element $a\in \mathbb{F}^*_q$, we define a sequence $\mathcal{G}_a$ by
\begin{equation*}
    \mathcal{G}_a=\left(G_u(ax)\right)_{x\in \mathbb{F}^*_q},
\end{equation*}
The correlation of two such sequences is defined
as
\begin{equation}\label{f-41}
   \mathcal{C}_{a,b}=
   C_{\mathcal{G}_a,\mathcal{G}_b}=\sum_{x\in
   \mathbb{F}^*_{q}}G_u(ax)G_u(bx).
\end{equation}
The autocorrelation of a sequence $\mathcal{G}_a$ is defined as
\begin{equation}\label{f-42}
   \mathcal{A}_{\mathcal{G}_a}(h)= \sum_{x\in
   \mathbb{F}^*_{q}}G_u(ax)G_u(ahx), h\in \mathbb{F}^*_{q}.
\end{equation}

About the distribution of values of the autocorrelation of the sequence $\mathcal{G}_a$, we have the following:

{\prop For every $a\in \mathbb{F}^*_q$ and a positive integer $u$ coprime with $(q-1)$, we have
\begin{equation*}
    \mathcal{A}_{\mathcal{G}_a}(h)=\left\{\begin{array}{ll}
                                            -q-1, & \mbox{ if $h\neq 1$} \\
                                            q^2-q-1, & \mbox{ if $h=1$.}
                                          \end{array}
    \right.
\end{equation*}}
Therefore, $\mathcal{G}_a$ is a two-valued correlation sequence which has some applications in communications, for example, Code Division Multiple Access (CDMA) systems, etc.
\begin{proof}Direct evaluation shows that
\begin{eqnarray*}
   \mathcal{A}_{\mathcal{G}_a}(h)&=&\sum_{x\in \mathbb{F}^*_q}G_u(ax)G_u(ahx)\\
 &=& \sum_{x\in \mathbb{F}^*_q}\sum_{c\in \mathbb{F}^*_q}\chi(axc^u+c^{-1})\sum_{d\in \mathbb{F}^*_q}\chi(ahxd^u+d^{-1})\\
 &=&\sum_{c,d\in \mathbb{F}^*_q}\chi(c^{-1}+d^{-1}) \sum_{x\in \mathbb{F}^*_q}\chi(ax(c^u+hd^u))\\
 &=&\sum_{c,d\in \mathbb{F}^*_q}\chi(c^{-1}+d^{-1})\left( \sum_{x\in \mathbb{F}_q}\chi(ax(c^u+hd^u))-1\right)\\
 &=&q\sum_{d\in \mathbb{F}^*_q}\chi((h^{-u'}+1)d^{-1})-\sum_{c,d\in \mathbb{F}_q^*}\chi(c^{-1}+d^{-1}) \quad (uu'\equiv 1\ {\rm mod}\ q-1)\\
 &=&\left\{\begin{array}{ll}
                                            -q-1, & \mbox{ if $h\neq 1$} \\
                                            q^2-q-1, & \mbox{ if $h=1$,}
                                          \end{array}
    \right.
\end{eqnarray*}
which is the desired result.\end{proof}
Hence we obtain the following corollary:
{\cor\label{cor-43} For every pair of $(a,b)\in (\mathbb{F}^*_{q})^2$, and a positive integer $u$ coprime with $(q-1)$, we have
\begin{equation}\label{f-44}
   \sum_{x\in \mathbb{F}^*_{q}}G_u(ax)G_u(bx)=\left\{\begin{array}{ll}
              q^2-q-1,& \mbox{ if } a=b \\
              -q-1, & \mbox{ otherwise. }
            \end{array}
  \right.
\end{equation}}
We also
have the following proposition.
{\prop \label{p-3}
For every $a,b,c\in \mathbb{F}^*_{q}$, and a positive integer $u$ coprime with $(q-1)$, we
have
\begin{equation}\label{f-46}
   \sum_{x\in
   \mathbb{F}^*_{q}}G_u(ax)G_u(b(c-x))=q G_u(c(a^{u'}+b^{u'})^u)+G_u(bc),
\end{equation}
where $G_u(a)=G_u(a,1)$ is defined by (\ref{f-17}) and $uu'\equiv 1{\rm mod }\ q-1$.}
\begin{proof} By definition, we have
\begin{eqnarray*}
  &&\sum_{x\in
   \mathbb{F}^*_{q}}G_u(ax)G_u(b(c-x))
    = \sum_{x\in
   \mathbb{F}^*_q}\sum_{y,z\in
   \mathbb{F}^*_{q}}\chi\left(axy^u+y^{-1}+b(c-x)z^u+z^{-1}\right)\\
   &=&  \sum_{y,z\in
   \mathbb{F}^*_{q}}\chi(y^{-1}+z^{-1}+bcz^u)\left(\sum_{x\in \mathbb{F}_q}\chi\left(x(ay^u-bz^u)\right)-1\right)\\
   &=&q \sum_{z\in
   \mathbb{F}^*_{q}}\chi\left(bcz^u+(a^{u'}{b^{-u'}}+1)z^{-1}\right)-\sum_{z\in \mathbb{F}_q^*}\chi(bcz^u+z^{-1})\sum_{y\in \mathbb{F}_q^*}\chi(y^{-1})\\
   &=&q G_u(c(a^{u'}+b^{u'})^u)+G_u(bc).
\end{eqnarray*}This completes the proof.\end{proof}

\section*{Concluding remarks}

(1) For the exponential sum
   \begin{equation*}
    G_u(a,b)=\sum_{c\in \mathbb{F}^*_{q}}\chi(ac^u+bc^{-1}),
\end{equation*}
it is easily seen that
\begin{equation*}
    G_u(a,b)=\sum_{c\in \mathbb{F}^*_{q}}\chi(ac^{q-1-u}+bc).
\end{equation*}
However, in the extension field of $\mathbb{F}_q$, we should have
\begin{equation*}
  \sum_{c\in \mathbb{F}^*_{q^s}}\chi(ac^{q-1-u}+bc)\neq \sum_{c\in \mathbb{F}^*_{q^s}}\chi(ac^u+bc^{-1}).
\end{equation*}
Thus, by Theorem \ref{thm-1}, one can only obtain that there are $q-2-u$ complex numbers $\omega_1,\cdots,\omega_{q-2-u}$ such that $|\omega_j|=\sqrt{q}, j=1,\cdots,q-2-u$, and
\begin{equation*}
    \sum_{c\in \mathbb{F}_{q^s}}\chi^{(s)}(ac^{q-1-u}+bc)=-\omega_1^s-\cdots-\omega_{q-2-u}^s
\end{equation*}
for all $s=1,2,\cdots.$
This is obviously different with Theorem \ref{thm-3}. In other words, Theorem \ref{thm-3} is not covered by Theorem \ref{thm-1}.

(2) After we finished this paper, we found that one can generalize Theorem \ref{thm-3} to a more generic case as follows:

{\thm \label{thm-6}Let $\mathbb{F}_q$ be a finite field and $f(x),g(x)\in \mathbb{F}_q[x]$ be polynomials with degree $m\geq 1$ and $n\geq 1$, respectively. Let $s$ be any positive integer, and
$\chi^{(s)}$ be the lifting of the additive character $\chi$ of $\mathbb{F}_q$. Define
\begin{equation*}
   G^{(s)}(f,g)=\sum_{c\in \mathbb{F}_{q^s}}\chi^{(s)}(f(c)+g(c^{-1})).
\end{equation*}
Suppose that either $\gcd(m,q)=1$ or $\gcd(n,q)=1$. Then there exist complex numbers $\omega_1,\omega_2,\cdots, \omega_{m+n}$, only depending on $f,g$ and $\chi$, such that for any positive integer $s$ we have
\begin{equation*}
   G^{(s)}(f,g) =-\omega_1^s-\omega_2^s-\cdots-\omega_{m+n}^s.
\end{equation*}}
{\thm \label{thm-7} Let $f(x),g(x)\in \mathbb{F}_q[x]$ be two polynomials with degree $m\geq 1$ and $n\geq 1$, respectively. If either $\gcd(m,q)=1$ or $\gcd(n,q)=1$,
then the complex numbers $\omega_1,\cdots,\omega_{m+n}$ in Theorem \ref{thm-6} are all of absolute $\leq \sqrt{q}$ provided that $\gcd(m+n,q)=1$.}

(3) For Theorem \ref{thm-6} and Theorem \ref{thm-7},
we find that, in 1998, Shanbhag,  Kumar and Helleseth \cite{skt} proved an upper bound on the characteristic sum
\begin{equation*}
    K_{e,m}( f_1, f_2)=\sum_{x\in \mathcal{T}^*_{e,m}}\psi_{e,m}(f_1(x)+f_2(x^{-1}))
\end{equation*}
over Galois ring $R_{e,m}=GR(p^e,m)$, where $f_1(x),f_2(x)$ are polynomials over $R_{e,m}$ with no monomial term in these polynomials has degree which is a multiple of $p$, and $T_{e,m}^*$ is the Teichmuller set of $R_{e,m}$. The main result of \cite{skt} is as follows.

{\lem \label{lem-2} Let $f_1(x),f_2(x)\in R_{e,m}[x]$ be non-degenerate and have weighted degree $D_{e,f_1} D_{e,f_2}$, respectively. Assume that the reduction of $f_1$ is nonzero and $f_2\not\equiv 0$. Then
\begin{equation}\label{f-48}
    \left|\sum_{x\in \mathcal{T}^*_{e,m}}\psi_{e,m}(f_1(x)+f_2(x^{-1}))\right|\leq (D_{e,f_1}+D_{e,f_2})\sqrt{p^m}.
\end{equation}}

We should say there are some difference between \cite{skt} and our results.

Firstly, the restrictions on the polynomials $f_1(x),f_2(x)$ is different, our condition is obviously weaker than that in Lemma \ref{lem-2}.

Secondly, the methods to prove the upper bound are different. Our method is elementary, while the method in \cite{skt} is deeper and involved.

Thirdly, using Theorem \ref{thm-3}, we established a recursive formula for $\mathcal{S}_s(f)$. However, since the main idea of \cite{skt} is to provide the upper bound, Shanbhag,  Kumar and Helleseth did not give the recursive formula.

Finally, one of our contribution is providing some properties of the sequence constructed from the exponential sums, see Section 4, we proved that the correlation of the sequences are two-valued.

\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


Unparalleled luxury awaits at this light-filled, full floor residence at the iconic Carlyle hotel.
Perched high above the Upper East Side on the 26th floor with spectacular panoramic Central Park views to the west and expansive city views from every room, this 4-bedroom, 4.5-bath residence, masterfully renovated by William Sofield, is waiting for a most discerning buyer. Step out of the keyed elevator into the exquisitely frescoed private landing before entering the residence's expansive gallery, ideal for entertaining and/or showcasing a world class art collection. A den (or fourth bedroom) and large living room showcase dramatic views. In the sleeping quarters, each bedroom is accompanied by a luxurious, spa-like, en-suite bath. With a seamless flow yet with separate and distinct entertaining spaces and sleeping quarters, the apartment functions equally well as a grand venue for hosting events as it does a private, luxurious retreat. Residents of The Carlyle enjoy the white-glove services and amenities of the legendary 5-star hotel combined with the privacy of a home on the prestigious Upper East Side of Manhattan. The extraordinary building services include twice-daily maid service, laundry and valet, in-home catering, and dedicated staff including elevator operators, doormen and a concierge ready to attend to your every need. The storied hotel is rich with amenities including the world class Sisley-Paris Spa, Yves Durif Salon, Cafe Carlyle, the Carlyle Restaurant and the fabled Bemelman's Bar as well as a 24-hour fitness center, garage, business center and florist to name a few. A New York City landmark built in 1930, The Carlyle is known for its discretion and security. Most notably, the hotel has housed some of the world's most distinguished guests including U.
. presidents, Royalty and other Heads of State.


As far as antipasti are concerned, we think Italians do it better! A typical antipasti spread in Italy is often a meal in its own right and always looks amazing! We have put together a selection of artisan Italian antipasti treats which makes a great gift for gourmet food lovers or anyone who loves to entertain.
Sicilian Shredded Aubergines (290g). Tasty mediterranean aubergines, seasoned with red pepper, garlic and oregano and preserved in olive oil. This is wonderful antipasto which also works well as a side dish or as an ingredient in your pasta recipes.
Sun dried Tomatoes in Olive Oil (314g). Our tomatoes have been dried in the Mediterranean sun and have an intense, sweet flavour. They are wonderful straight from the jar or alongside some mozzarella as part of a colourful antipasti spread.
Grilled Peppers with Garlic and Oregano (212g) These colourful sweet grilled peppers have been flavoured with oregano and garlic and are wonderful served as part of an antipasti platter or on a bruschetta.
Sun dried Tomato, Almond and Pistachio Paté (180g). Hand made in Southern Italy from top quality local ingredients, this is a delicious topping for bruschette and crackers.
La Bumba (212g). This super spicy spread is a combination of chopped aubergines, peppers, carrots, celery and green olives mixed with chilli!
Cracked Green Olives (314g). These lightly spiced Sicilian olives have a bold firm flavour and make a great addition to any antipasto platter.
Our Gourmet Antipasto Platter is presented in strong box tied with a lovely ribbon. A gift message option is available at the checkout. Please note that if we are out of stock of any item we reserve the right to substitute it with the most similar available item. Dishes are not included.


text:
T: 01252 597765 | [email protected]h.co.uk
Kings Avenue, Farnham
Surrey, GU10 1AX
£335,000 Guide Price
One of two similar properties set on a residential road within a popular semi rural village on the Surrey border. This modern three bedroom semi detached family home features a spacious lounge diner, a refitted kitchen, a well tended garden and off street parking.
Location Tongham is a small Surrey village located close to the north east Hampshire and Surrey border. The village lies in a triangle between the A31 and the A331 with neighbouring villages including Ash and Badshot Lea.
The village is thought to have a long history as Stone Age relics have been found here, both the Saxons and the Romans are known to have been here and the village has evolved throughout the ages. Some of the village's heritage is evident through the old character properties in amongst the newer homes.
Within the village centre sits St.Paul's Church built of chalk stone in the mid 19th-century as well as the old clock tower, the old school and a peaceful tree lined churchyard. The village is home to a traditional pub, bakery, salon, jewellers and convenience shop whilst the village cricket and recreation grounds provide a space for the community to enjoy activities together.
Ground Floor Accommodation A double glazed UPVC door opens into an entrance hall that accesses the downstairs cloakroom. A door opens into a reception hall where stairs rise to the first floor with an under stairs cupboard and there is access to the kitchen and to the spacious lounge diner that features a box bay window over looking the front of the property. From here, sliding patio doors open into a conservatory with a tiled floor, an electric heater and double doors to the garden.
Fitted with a quality range of floor and wall mounted cabinets and a tiled floor, the rear aspect kitchen Includes an integrated electric oven, an electric hob with an extractor, a dishwasher and a fridge. A door opens into an adjacent utility room with a side access door, a door to the patio and space and plumbing for further appliances.
First Floor Accommodation To the landing are doors to all rooms, an airing cupboard and access to the loft space. The master bedroom has views out to light woodland and the rear garden and benefits form built in wardrobes. There are two further bedrooms and a family bathroom fitted with a white suite comprising of a panel enclosed bath, a shower cubicle, wash hand basin and a low level W.C.
Outside A driveway to the front of the property provides off road parking for up to three vehicles. The rear garden features a raised veranda followed by an expanse of lawn and there is a side gate.


text:
Campaign URL Copy
Mundus maris newsletter: September 2020
Dear <<First Name>>
The global number of people succumbing to the coronavirus exceeds a million and many more are affected by the pandemic socially, economically and, of course, health-wise.
Small-scale fishers and other artisans operating along value chains relying on natural resources have been particularly impacted by the ensuing restrictions while perhaps understanding better than many how important it is to keep the ocean, nature and its ecosystems in good shape and health. Their livelihoods are intertwined more closely than many others with functioning and productive natural systems. They need our care and protection to keep delivering the goods and services so important to all. In tropical regions these systems are already feeling the heat of climate change which makes fish move polewards to remain within suitable temperature ranges and sees Mediterranean fish species arrive in the North Sea.
The small-scale fishers, men and women, are trying to adapt but can not by themselves change that. Turning back the tsunami of climate change and biodiversity loss requires efforts of everybody, but particularly those responsible for the biggest drivers of biodiversity loss and the fossil guzzlers together with a massive effort in favour of renewables. It's a matter of blue justice..
Likewise, small-scale fishers suffer overproportionately from IUU fishing and harmful subsidies, 85 percent of which go to industrials and "pull the rug under their feet" or the fish from under their boats. They need a level playing field to continue providing nutritious and valuable seafood to continue meeting the demand of hundreds of millions of people..
So, limited in our field activities as a result of corona, we are using electronic media to keep sharing information and opportunities. In this vein, we have completed the Wolof sound track of a short video on the Small-scale Fisheries Guidelines. This way we hope to reach those in Senegal and neighbouring countries where the language is spoken and many depend on the trade and need to know, but often do not speak the official UN languages..
One word again on developing the Book of the Marine World – check out the invitation in five languages – now open to everyone without age restrictions
DEEN ESFR IT
Hurry up with your contribution so that it's ready for sharing at World Fisheries Day, 21 November. Join all the others to make this a truly remarkable celebration of living sustainably with the ocean.
And we have more to celebrate. For our 10th anniversary we start sharing now until the end of the year video messages and statements of what inspired members and others to support Mundus maris. Get inspired too! Become a supporting member or regular donor of Mundus maris to make even bigger waves to protect and connect to the ocean.
Mundus maris asbl, Belfius Bank, Rue de Linthout 224, 1040 Brussels, Belgium
IBAN: BE54 0688 9178 6297 BIC: GKCCBEBB
Contact us any time at info@mundusmaris.org. We'll shortly send you a questionnaire so that you can have your say in what we tackle together in the future and how best to go about it. Thanks in advance for your feedback.
Cornelia E. Nauen and the entire Mundus maris team
Support our work with a donation
Activities around the world
Animation of SSF Guidelines in Wolof
We are very happy to announce that we have been able to create a Wolof sound track for an existing French animation about the essential elements of the Small-Scale Fisheries Guidelines produced earlier through a collaboration of IFAD, ICSF, World Forum of Fish Harvesters and Fish Workers, World Forum of Fisher People and Peoples' Food Sovreignty. We noted in our field work that few of the most concerned women and men knew about them. So Mundus maris asbl made good on a promise at the launch of the Small-Scale Fisheries Academy in Dakar to provide for a version in Wolof to facilitate access.
Tenth anniversary celebrations of Mundus maris
We are receiving a number of lovely statements from members and people following and supporting our work about what inspires them about Mundus maris. Here is one by former intern Maxime Lambert to get a flavour:
"Working with Mundus maris allowed me to witness how this remarkable organisation blends the spheres of science, society, and art into a rich knowledge network that values and protects the oceans, its ecosystems, and the societies that depend on the sea. What an honor it is to have been a part of their mission!"
And here is a video message by Carolina Castellanos Arévalo in Buenos Aires, Argentina.
Watch the message
Cross talk over the big pond from Nigeria to Argentina
Mundus maris Vice-President Prof. Stella Williams felt honoured to have been requested by the Vice President for Latin America - Professor Marcelo Morales Yokobori to interact with his students in Argentina. They asked pointed questions about fisheries and aquaculture in Nigeria specifically, but also wanted an update on West Africa. This is the first in a series of online and outreach activities planned by Marcelo for his students in different places. This 30-minutes video demonstes the interest of students in Argentina in learning about regional trends and of programmes put in place by Mundus maris asbl.
A sculpture showing a burning kangoroo with the world map on it, saying "Australia is everywhere".
Marching together to protect the climate, Germany
25 September 2020. No more than 1.5°C increase max! - in some 480 towns and cities in Germany alone Fridays for Future had called to strikes and demonstrations to ask for immediate climate protection measures keeping in mind the necessary distancing between protesters and mandatory face masks - corona obliging. Mundus maris was at the manifestation in Düsseldorf in front of the Parliament building to underscore that climate protection and ocean protection go hand in hand. An estimated 200,000 citizens of all ages participated.
Copyright © 2020 Mundus maris asbl, All rights reserved.
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Jono Carroll (Boxer) Wiki, Biography, Age, Wife, Family, Net Worth
Jono Carroll is an Irish professional boxer who challenged for the IBF junior-lightweight title in 2019.
Explore Jono Carroll Wiki Age, Height, Biography as Wikipedia, Wife, Family relation. There is no question Jono Carroll is the most famous & most loved celebrity of all the time. You can find out how much net worth Jono has this year and how he spent his expenses. Also find out how he got rich at the age of 29. He has a pure loving kind heart personality. Scroll Down and find everything about him.
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Date of Birth 12 April 1992
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Famously known by the Family name Jonathan Beresford, is a great boxer. He was born on 12 April 1992, in Dublin, Ireland.Dublin is a beautiful and populous city located in Dublin, Ireland Ireland.
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Jonathan Beresford has a net worth of $5.00 million (Estimated) which he earned from his occupation as boxer. Popularly known as the boxer of Ireland. He is seen as one of the most successful boxer of all times. Jonathan Beresford Net Worth & Basic source of earning is being a successful Irish boxer.
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The boxer with a large number of Twitter followers, with whom he shares his life experiences. Jono is gaining More popularity of his Profession on Twitter these days. You can read today's latest tweets and post from Jono Carroll's official Twitter account below, where you can know what he is saying in his previous tweet. Read top and most recent tweets from his Twitter account here…
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Born on 12 April 1992, the boxer Jono Carroll is arguably the world's most influential social media star. Jono is an ideal celebrity influencer. With his large number of social media fans, he often posts many personal photos and videos to interact with his huge fan base on social media platforms. Personal touch and engage with his followers. You can scroll down for information about his Social media profiles.
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Carroll made his Irish professional debut in November 2014, when he faced Declan Geraghty at the 3Arena. On the undercard of Matthew Macklin vs Jorge Sebastian Heiland, Eddie Hearn confirmed that the winner would earn a slot at the following month's Prizefighter tournament in London. Carroll and Geraghty went on to produce 'Fight of the Night', in a bout which ended in the fourth round when Geraghty was disqualified for persistent use of the head, that same fight would go on to win Irish fight of the year. .
Carroll began his pro career while in Australia in 2012. He eventually returned home to Dublin where he began fighting under Paschal "Packie" Collins, out of Celtic Warrior Gym, Blanchardstown.
Jonathan Beresford (born 12 April 1992), better known as Jono Carroll, is an Irish professional boxer who challenged for the IBF junior-lightweight title in 2019.


Cait Etherington, writer and education/training consultant with over two decades of experience writes, "In late October, Inside Higher Ed released its 2017 report on faculty attitudes and uses of technology".
The report, a collaboration between Inside Higher Ed and Gallup, found that while faculty are warming up to technology, change is slow in the ivory tower and that many faculty continue to express concerns about public institutions partnering too closely with educational technology companies.
...What's most promising are current trends. With more and more faculty reporting that they have taught a course online or in a blended format each year, it seems likely that by 2020, a majority of faculty will have attempted teaching online on at least one occasion. Given the high demand among students for more online and blended learning opportunities, this is no doubt good news for higher education.


text:
Ela Meh
Eviction of Africa House, Calais, 2011. Photo by calaismigrantsolidarity.wordpress.com.
There is a sense of fear in the air in the recent months. Fear of different kinds: some is well-founded, some completely invented, some is spoken about openly and some is pushed deep down, denied and transformed into hate and violence or bravery and compassion.
The most obvious of all fears seems to be the one spread by the securitarian discourse: migrants are a threat and you should fear them. Personally I feel fear connected with the normalisation of such a discourse, as well as by the current rise of the extreme right and its political and grass-root influence. The newly arrived, labeled as "migrants", have good reasons for different types of fears. There is increased structural and police violence: as of the 8th of March, the humanitarian corridor is officially closed down. The recent illegalisation of travel has already caused the first deaths: three people drowned trying to cross a river in Macedonia, just days after the closure of the Greek-Macedonian border. The sense of uncertainty as to what will happen to those who have arrived to the EU in the past several months is frightening too. To be able to stay and qualify for refugee status, asylum procedures oblige people to prove that they face "well-founded fear" persecution in the country of their flight – and if they cannot prove it, well then theirs is the well-founded fear they might face detention or deportation. And of course, there is the increase of racist attacks: in Germany, the arson attack on a lagers, where migrants are accommodated, are a regular occurence. All these fears also affect those who solidarise and support the newly arrived – the supporters often living with the fear that they will be criminalised for helping those, who have been illegalised.
The recent violence, but also solidarity and resistance in the jungle of Calais, which is currently being evicted, reminds me of the eviction I witnessed in the summer of 2011. Eviction after eviction, from the individual squats, to whole sections of the jungle, the violence is constant, increasingly brutal and well-documented, but remains unsanctioned and endorsed by the local and national authorities.
The current eviction of the jungle is much more large scale and even more brutal than the one I witnessed in the summer of 2011. That time it was the French riot police (CRS) that descended on and brutally evicted the "Africa House", a self-organised squat in which we were staying with well over a hundred people from Sudan, Erytrea, Ethiopia, Afghanistan, Pakistan, and Iran. The evening before the eviction was announced, I was paralysed by fear: I was afraid of the police violence, of repression, of what will happen to all my friends who will find themselves homeless. Even though the Africa House, an old, huge derelict building, was slowly falling apart and in many places did not have all four walls intact, even though it was raided by the police a couple of times a week, it was still a space of mutual aid and solidarity, a place where you could gather some strength and know that a comrade is watching for the police raid and that they would wake you up with whistle-blows before the CRS riot cops reach the front gate.
Sitting next to my friend, who was preparing some food and singing quietly to herself, I asked her how come she is not afraid at all about the events of the next day. Her answer stayed with me till today.
"Do you think I am not scared?" she said. "I am terrified, honestly! But there are many other things I feel to. I allow myself to be scared, there's no use in fighting off fear, telling yourself you should not feel it and trying to tame it – firstly, it never works and it just eats up all your energy taming the fear, and secondly, if you are a normal human being, you will feel fear when a squad of riot cops descends upon a building you are staying in. But in the moments when I feel afraid, I try also to listen to other emotions that there are mixed with it and I try to remember what is important to me. I am really passionate about fighting for this house, the space here is hugely important and a temporary home for many. I am also very angry: how can people be 'illegal' and arresting them, beating them up, using tear gas on them is 'legal'? It is all of these other things that make me do what I do and, well, fear is just one thing among many I feel!"
The lesson I took from her words was that it is crucial to identify fear as fear, allow ourselves to feel it fully, but also to think carefully about whether acting out of fear is justified and to listen for other emotions that exist parallel to, or despite the fear. It is a lesson very helpful for me, especially in the recent months, with this sense of fear in the air...
Objects/Subjects in Exile. A conversation between Wayne Modest, Bonaventure Soh Bejeng Ndikung, and Margareta von Oswald
The Shame and Misery of Liberal Democracy: Europe and Migration Flows
The Mediterranean: A New Imaginary
The "Refugee Crisis" and the Current Predicament of the Liberal State
Škart Maps
Tagged jungle of calais,	migrants,	refugee influx into europe
Wayne Modest,	Margareta von Oswald,	Bonaventure Soh Bejeng Ndikung
Carlos Prieto del Campo
Adrian Lahound
Denise Ferreira da Silva
Đorđe Balmazović