It was a relatively good year for the Oscars' red carpet. At least it's safe to say that even the dresses that landed these stars on the worst dressed list aren't bad on their own, just in the context of, you know, the biggest awards show of the year. Click on any image for more pictures and a full rundown on what, exactly, makes these looks so bad.
Bai Ling vs Lolene: Who Wore It Worse?


text:
package com.svbio.workflow.samples.ckbundle;

import java.math.BigDecimal;
import java.math.RoundingMode;

import static java.math.BigDecimal.ONE;
import static java.math.BigDecimal.TEN;
import static java.math.BigDecimal.ZERO;

final class PiComputer {
    private static final BigDecimal TWO = BigDecimal.valueOf(2);
    private static final BigDecimal THREE = BigDecimal.valueOf(3);
    private static final BigDecimal FOUR = BigDecimal.valueOf(4);

    private PiComputer() {
        throw new AssertionError(String.format("No %s instances for you!", getClass().getName()));
    }

    /**
     * Linear fractional transformation that maps a {@link BigDecimal} number {@code x} to {@code (a * x + b) / c}.
     *
     *
A fractional transformation (also called a Möbius transformation) of the above form can be represented as a
     * upper-triangular 2 x 2 matrix where the left column contains {@code a} and 0, and the right column contains
     * {@code b} and {@code c}. Using this representation, the composition of linear fractional transformations is
     * equivalent to Matrix multiplication. Formally, let {@code H} be the mapping from linear functional transformation
     * to the matrices as defined before. That is, if {@code f(x) = (a * x + b) / c}, then
     * {@code H(f) = ((a, 0) (b, c))}. It is easy to verify that {@code H(f o g) = H(f) H(g)}. Here, {@code f o g}
     * denotes the function that maps {@code x} to {@code f(g(x))}.
     */
    private static final class MoebiusTransformation {
        private final BigDecimal topLeft;
        private final BigDecimal topRight;
        private final BigDecimal bottomRight;

        private MoebiusTransformation(BigDecimal topLeft, BigDecimal topRight, BigDecimal bottomRight) {
            this.topLeft = topLeft;
            this.topRight = topRight;
            this.bottomRight = bottomRight;
        }

        @Override
        public String toString() {
            return String.format("x -> (%s * x + %s) / %s", topLeft, topRight, bottomRight);
        }

        private static MoebiusTransformation unit() {
            return new MoebiusTransformation(ONE, ZERO, ONE);
        }

        private BigDecimal evaluate(BigDecimal x, RoundingMode roundingMode) {
            return topLeft.multiply(x).add(topRight).divide(bottomRight, roundingMode);
        }

        /**
         * Returns a new Möbius transformation that is the composition of the current transformation after the
         * given transformation.
         *
         * @param other other Möbius transformation
         * @return the new Möbius transformation
         */
        private MoebiusTransformation concatenate(MoebiusTransformation other) {
            return new MoebiusTransformation(
                topLeft.multiply(other.topLeft),
                topLeft.multiply(other.topRight).add(topRight.multiply(other.bottomRight)),
                bottomRight.multiply(other.bottomRight)
            );
        }
    }

    private static MoebiusTransformation sequence(BigDecimal index) {
        return new MoebiusTransformation(
            index,
            FOUR.multiply(index).add(TWO),
            TWO.multiply(index).add(ONE)
        );
    }

    private static MoebiusTransformation digitExtractionTransformation(int digit) {
        return new MoebiusTransformation(
            TEN,
            TEN.negate().multiply(BigDecimal.valueOf(digit)),
            ONE
        );
    }

    /**
     * Returns a {@link String} containing the given number of digits of the decimal representation of π.
     *
     *
This method implements a streaming algorithm (also called spigot algorithm in the literature) based
     * on the paper Unbounded Spigot Algorithms for the Digits of Pi by
     * Jeremy Gibbons (2006).
     *
     * @param numDigits number of digits in the decimal representation to include
     */
    static String computePi(int numDigits) {
        StringBuilder stringBuilder = new StringBuilder(numDigits + 1);
        MoebiusTransformation state = MoebiusTransformation.unit();
        int digitCount = 0;
        BigDecimal index = ONE;
        while (digitCount < numDigits) {
            state = state.concatenate(sequence(index));
            index = index.add(ONE);
            int digit = state.evaluate(THREE, RoundingMode.FLOOR).intValue();
            if (digit == state.evaluate(FOUR, RoundingMode.FLOOR).intValue()) {
                ++digitCount;
                if (digitCount == 2) {
                    stringBuilder.append('.');
                }
                stringBuilder.append(digit);
                state = digitExtractionTransformation(digit).concatenate(state);
            }
        }
        return stringBuilder.toString();
    }
}

redpajama_set_name: RedPajamaGithub


ATHENS (Reuters) - The euro zone will consider measures to support Greece as long as it implements its latest bailout but has no plans to review the situation before spring next year, European Central Bank executive board member Joerg Asmussen said on Wednesday.
Asked to comment on the German finance minister's statement that Greece would need a new bailout package, Asmussen said the issue was not discussed at talks with Greek Finance Minister Yannis Stournaras during his visit to Athens.
"We have not discussed this, we have focused on making the current program a success, more growth and jobs," he said.
He declined to be drawn into discussion of a third package, saying he had nothing to add to a Eurogroup decision in November last year that pledged additional assistance to Greece until it regained access to financial markets, as long as it implemented its bailout and generated a primary surplus.


text:
Leasingham is the name of a hamlet on the southern edge of the Clare Valley, in the Mid North of South Australia. It is approximately 117 kilometres from the state's capital, Adelaide and 17 kilometres south of the town of Clare.

Although the surrounding country is elevated, the township is located in lower country at the intersection of Main North Road and the connecting roads to Mintaro (to the East) and Hoyleton (to the West). Leasingham is bisected by the Eyre Creek as it makes its way to the Wakefield River. The Riesling Trail is on the western fringe of the township, following the former railway route between Auburn and Clare.

History
The Leasingham district was the ancestral home of the Ngadjuri people. The first European explorer, John Hill, passed through the district in April 1839. Surveys soon followed, the township being named after the village and civil parish of Leasingham in Lincolnshire, England and the site was once owned by E. Campbell, who held approximately 120 hectares (60 acres) of freehold land, which he sold in 1853. By 1866 Leasingham had about 130 residents and, although ribbon development occurred around the Main North Road crossroad, it never prospered into a town.

Governance
Leasingham is in the District Council of Clare and Gilbert Valleys, the state electoral district of Frome, and the Australian House of Representatives Division of Grey.

Land use
Leasingham is an agricultural district in which dry grain farming predominates. Additionally, being centrally located in quite a picturesque district, it caters for tourism and has a restaurant, caravan park, bed and breakfast accommodation and a number of tasting/cellar door sales operations of local wineries.

References

Towns in South Australia
Mid North (South Australia)

redpajama_set_name: RedPajamaWikipedia


text:
UK Artists
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CMW – Country Music Week
Cornbury Festival
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The Long Road Festival
Bluesfest
Quick Fired Questions
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Scotty McCreery – Seasons Change – Album Review ⭐⭐⭐⭐⭐
By hannahjessica31 on March 17, 2018 • ( 8 Comments )
One door closed and an even bigger one opened for Scotty McCreery last year when he stood by his truth and his belief in his music and his single Five More Minutes despite the fact there were doubters. Letting fuel add to the fire and pushing him to succeed even further, Scotty did one of the bravest things a musician can do and released his most beloved work without a label backing.
Proof that what is in your heart and in your gut should determine which path you choose rather than the direction someone tells you to take, Scotty's bold move and heartfelt lyrics soon got him turned up loud on country radio and soon got the industry to come knockin'.
One label closed and another one opened and almost in the blink of an eye Scotty has released his best and most honest work yet in the form of an album titled Seasons Change.
Seasons Change may be Scotty's fourth album but it's certainly his first album where he was given a lot more artistic freedom to write about what he knew, his experiences and where he, at 24 is in life right now and where he is is a dang good place but it wasn't exactly a smooth road to get there.
Driving over a few bumps and pot holes, Scotty has learned a lot along the driveway to success and Seasons Change is pretty much the map to that very journey.
Starting with the title track – Seasons Change, Scotty has kept his strong traditional yet modern style and sound that the mainstream country scene lacks of slightly . This is the perfect kick off to the album, it has an uplifting, addictive, upbeat tempo and Scotty's baritone vocals reel you in within the first line. The song sets a good flow for the rest of the album and you know it is going to be a good'un. Seasons Change is the perfect anthem lyrically to what Scotty faced changing labels and it is inspirational in the sense of there is always a silver lining.
Wherever You Are is another upbeat track that is instantly likeable. I can imagine this song blasting all over country radio and can hear everyone singing along loudly in their cars and at concerts. It pretty much makes you want to sing along loudly to it wherever you are.
Boys Back From Home is an anthemic tune about Scotty's real life friends and the things they got up to back in the day. It is nice to have a little bit of an invite into Scotty's world, after all, is that not what country music is all about?
Five More Minutes is just about as authentic as you can get and just about as country as country does. A song that deservedly took the number one slot, Scotty plays tribute to his grandfather in this moving song where he wishes for just five more minutes with him. Somewhat of a ballad, the painfully beautiful and relatable lyrics pull at the heartstrings. Sonically, the song is stunning and emotive in itself.
In Between is a wonderfully catchy tune with an infectious chorus that you can imagine being sung by the likes of Keith Urban or Tim McGraw so watch this space guys and make room for Scotty. The twangy sound of the traditional instruments on this track bring this song to an even higher level and every instrumental detail is crafted so perfectly.
Any hardcore fan of Scotty will likely feel highly emotional listening to This Is It or anyone in general really. This ballad is absolutely gorgeous and is such a lovely story and as Scotty describes "is the blueprint to his proposal to his fiancé" (Gabi.) This Is It is without a doubt going to be the next big wedding song for newlyweds – its stunning emotive melody brings happy tears and the lyrics are incredibly romantic. Hearing this song live will be one of those moments you won't forget.
Wrong Again has a fun, slick, rock start to it and has an exciting chorus which is good fun to sing along to. The song gives a slightly different edge to Scotty showing that he can be a lot more diverse. The same goes for the song Move It On Out which holds more of a pop country style to it but not in a "is it country or is it pop?" kind of way as it is clearly country music. I can imagine this song being a big summer anthem and a huge radio hit that you hear everywhere especially in places all around CMA Festival. It will be interesting to see if it gets released as a single and around that time.
Barefootin' has a nice soulful, Jazzy edge to it and has a really great brass segment to it. This upbeat tune should get you dancing in no time and has a Chris Young/ Josh Turner feel to it.
Still has a smooth, gentle melody to it and tells a story of young love right at the beginning and how that same love grows still. This is such a beautiful love letter to, I should imagine Gabi. If this song doesn't get you slow dancing with your partner then I don't know what will. I love the chorus, it is simple yet incredibly effective, even the vocals that seem to stay on one level are just really infectious.
Home In My Mind is one the fans have been raving about for a while. I keep mentioning the word beautiful but that's it exactly what this album is. This song is so sweet and tells of real places of Scotty's home. Life on the road is hard and Scotty translates those feelings perfectly in this song because no matter how big you get or how busy you are, home is always where the heart is and is always on your mind.
Every track from this album is single worthy and a perfect fit for country radio, so much so that I would be pleased with any of them being released as a single. Scotty deserves the success coming his way and I am excited to see how far this album travels as I am positive it will reach great heights.
Scotty has a voice made for country music and his talents as a songwriter should be championed after this well crafted album.
A soundtrack to his life, Scotty has set the bar high this year and I challenge any artist to deliver a more honest, authentic album that hits quite like Seasons Change does. Scotty is the one to watch this year and has worked hard to get here. As a winner of a contest show, some would say he has a lot to prove but this album is more than proof, Scotty is the real deal.
Listen to Seasons Change here
Purchase Seasons Change here
Categories: Album Reviews, Favourites, Latest
Tagged as: Scotty McCreery
Interview: Ashley Campbell talks C2C Festival, playing tribute to her father, The Lonely One and playing the first ever Black Deer Festival
CURTIS GRIMES announces first ever UK tour dates in June 2018
Roseann D Schrichte says:
Thank you for a comprehensive, complete and utterly AWESOME review of Scotty McCreerys' album "Seasons Change"!!!!!
Joan Schroeder says:
Thank you for your fantastic review of Scotty McCreery's Season's Change. You hit the nail on the head and we are all excited to hear Scotty's new music on the radio.
KN says:
It's rare when you hear an album where every song is something special. That's what you get with this album. Country music story telling at its finest with a smooth, deep delivery from one of country music's best male voices. Scotty Mccreery, at only 24 years old, is country music gold. Truly, a legend in the making.
Excellent review of Seasons Change. The artistry of the album shows through on every aspect. From the personal song writing, the arrangements on the songs to the flowing placement of the songs on the album, and of course, Scotty's voice shows this to be the project that will launch Scotty to the top in Country music.
Rosebug says:
As I listened to the album for the first time, I expected to find at least one or two songs I might fast forward through but I was wrong. All eleven songs are relatable & relevant. Personal favorites besides the already # 1 hit, Five More Minutes are Still, Barefootin, In Between and Home in My Mind. Thank you for your honest review and acknowledgement of the incredible talent Scotty McCreery has displayed in this latest album.
Thank you for this AWESOME review. We have always known he's special.
Judy Hicks says:
A legend is born and, oh how he can sing. I love this album and there is none better.
Kathleen L. Stevens says:
This is an accurate review of Scotty's album this is what he deserves. This album is so clean, so good, vocals are fantastic, lyrics so touching. He feels every song and it touches because he feels what he is singing. This young man is unique, the best that I have heard in years. He has a golden voice like none other.
Rachel Reinert – Ocean (Acoustic Live)
https://youtu.be/1EC-K36PZmo
Jenny Toleman – Rock & Roll to my Country Soul (Acoustic Live)
https://youtu.be/If_8gHn_M4s
Michael Ray – Think A Little Less – Live
https://youtu.be/oUxj8jaCkdg
Calibama – Sierra (Acoustic Live)
https://youtu.be/ealMF7BCgdQ
Christie Huff – Love-Ish (Acoustic Live)
https://youtu.be/fpXRT9g2H3Q
Kelly Archer – Love Like Mine (From Nashville TV show – Acoustic Live)
https://youtu.be/Qgtjm13-p64
David Mhyre – Cowgirl (Acoustic Live)
https://youtu.be/ccuakF4Shxk
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willow - Taylor Swift
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Musicians In The Spotlight: An Interview with drummer Elijah Wood on being an integral part of the Shania Twain Tour, the crazy and exciting stage production, advice to aspiring drummers and more
THE LUMINEERS - release a trilogy of music videos exploring alcohol and drug addiction
CRYSTAL GAYLE INDUCTED INTO GRAND OLE OPRY BY SISTER AND OPRY LEGEND LORETTA LYNN
Country Music Week -- Duane Eddy twangs the London Palladium -- ****
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redpajama_set_name: RedPajamaCommonCrawl


This is a great deal if you're looking for something fun to do with your little one over the summer! I've been taking my kids to The Little Gym of Lancaster since before they could walk and we LOVE it! Today is actually the last day of my son's class for this session. We are already signed up to go back in the fall! He has so much fun there are we've made lots of great friends, both for him and for me!
Right now DoubleTake Deals is offering a great discount on summer classes at The Little Gym of Lancaster. $35 buys you four weeks of summer classes between June 13 and August 27, 2011. The regular price is $72, so this is over 50% off! Go here to read more about this deal!


What this article will cover: What are you doing in your "off" hours to improve your life?
You need to control your time at home rather than just haphazardly try to manage it. Setting priorities—not just at the office but at home—makes you boss of your own time. The busier you become, the more you have to manage, control and prioritize. Although I certainly don't have some scientific formula that will magically make this easier, I can tell you one thing: If you start with a commitment to success and then agree to control time, you will create an agenda that accommodates all you want.
You have to decide how you are going to use your time. You must command, control, and squeeze every second out of your weekends. If you don't do this, you will have people with different agendas pulling you in all sorts of directions. My schedule works for me because everyone in my life, from my wife to the people who work with me, knows what is most important to me and understands how I value time—even at home.
You're frequently encouraged to "slow down, relax, take it easy, find balance" and just "be happy" with where you are and what you have. Although this can sound great in theory, it can be very difficult for people who abandon every decision to be in control of their lives. Most people can't simply "relax and take it easy" because they never do enough to free themselves from the meager existence that comes as a result of mediocre actions.
If you work 9 to 5, you should still have a purpose, a mission, and a sense of accomplishment the other 128 hours a week you're not at your job. People who promote the new age, esoteric advice to "take it slow" are encouraging a mindset that isn't doing you any good. Consider the types of traits this thinking has created in you: laziness, procrastination, lack of urgency, sloth, tendency to blame others, irresponsibility, entitlement, expecting someone else to solve your problems.
Wake up! No one is going to save you.
No one is going to take care of your family or your retirement. No one is going to "make things" work out for you. The only way to do so is to utilize every moment of every day at 10X levels.
Happiness, security, confidence, and fulfillment come from utilizing your gifts and energy to achieve whatever you've decided success is for you. And it requires every bit of your time, which is yours to control.
If you want to turn your life around, think about what are some of your own time wasters? How much time do you spend on wasteful activities every day (e.g., watching television, smoking, drinking, oversleeping, video games)?
Control your time at home to control your life!
A great way to spend some time this upcoming month is going to be watching the Official Recording of 10XGrowthCon 2018. This is a 4-day event February 22-25th that will be the biggest business, entrepreneur, and lifestyle conference of the year.
You can watch the entire thing on demand from the comfort of your home.
Look, I'm just telling you like it is—if you want to improve your life it starts with how you spend your time at home. Get the official recording of this conference and you will have an investment that will pay you dividends for years to come.
What else are you doing on your weekends to improve your life?
Awesome advice thanks GC! I love that your content is practical and you actually give us the steps, not just the hocus pocus motivational fluff which doesn't actually have any substance!


Yesterday, a public consultation event on a new law to regulate the relationship between formal and informal Afghan justice sectors took place in Feyzabad, Badakhshan. The law shall increase legal certainty for Afghan citizens. In total, almost 40 people, including 15 women, attended the event. The Afghan Ministry of Justice (MoJ) implemented the event, supported by the Afghan-German Cooperation's project Promotion of the Rule of Law in Afghanistan (RoL).
Several legal systems exist in Afghanistan – traditional law, Islamic law and constitutional law. Instead of relying on formal courts, conflicting parties often refer their cases to informal local dispute resolution bodies, Jirgas and Shuras. Their decisions are based on traditional law and are not legally binding, causing legal uncertainties. Therefore, the MoJ is currently working on a law that shall enable the formal recognition of informal bodies' decisions if they meet basic constitutional standards. Particularly citizens living in rural areas where formal legal institutions are often absent will profit.
Current public consultations focus on procedural details regarding case referrals between formal and informal justice actors and the registration of decisions made by Jirgas and Shuras by formal courts. These details have major implications for the final law's efficiency in achieving its objective – increasing legal certainty for Afghan citizens, especially women and children. Further public consultations shall take place in Afghanistan's south in the weeks to come.
The programme Promotion of the Rule of Law in Afghanistan (RoL) is implemented by the Deutsche Gesellschaft fur Internationale Zusammenarbeit (GIZ) GmbH on behalf of the German Federal Ministry for Economic Cooperation and Development (BMZ). The programme aims at ensuring the rule of law, a well-functioning judicial system and legal security in Afghanistan. To ensure legal security for all citizens, RoL has been supporting so-called Huquq bureaus which handled more than 80,000 civil cases since 2013. In the same time, to support higher legal education, RoL provided more than 30,000 books to law and sharia faculties, implemented more than 150 courses on legal topics for students and lecturers and awarded more than 60 students with scholarships. The programme also emphasises promoting women's rights and strengthening their access to justice as well as the job market.


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without any burden. Take MIPOW Power Cube 6000 and start your relaxing journey.
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At the beginning of 2017 I had the opportunity to develop and design the graphics for Alien Covenant's bonus feature Phobos, as part of the MPC Design Team.
Here you can see a sneak peek trailer, some images from the film and the screen designs.
Full film available on Blu–Ray DVD.


Parents, educators, and practitioners have a united interest in identifying the qualities of youth associated with positive characters development. There is a growing academic interest in character as a key indicator of child development, and how it can benefit individuals and society as a whole.
The Society for Research in Child Development (SRCD) is hosting this conference to bring together developmental scientists, philosophers, educators, and practitioners to advance theory, research, and program practices pertinent to understanding character development as it occurs within and across the key settings of youth development. Youth and family professionals who work to promote character development will have much to add to the discussions at this special topic meeting.


text:
\section{Introduction}

If $(M,d s_1^2)$, $(N,d s_2^2)$ are Riemannian manifolds, then a warped product is the manifold $M \times N$ endowed with a Riemannian metric of the form
\begin{equation}
\label{warped prod riem}
    d s^2 = d s_1^2 + f^2 d s_2^2 \, ,
\end{equation}
where $f$ is a smooth function depending on the coordinates of $M$ only; said a warping function. This notion, called by \emph{warped product}, must be credited to Bishop and O'Neill \cite{bishop:oneill}. However, years earlier, metrics in the form of (\ref{warped prod riem}) were being studied with different names; in \cite{kruch}, for instance, they were called semi-reducible Riemannian spaces.

The class of warped product manifolds has shown itself to be rich, both wide and diverse, playing important roles in differential geometry as well as in physics. To illustrate, Bishop and O'Neill introduced warped products in \cite{bishop:oneill} as means to construct a large class of complete Riemannian manifolds with negative curvature. For this reason, it seems valuable to study notions of warped product metrics without the quadratic restriction, in the setting of Finsler geometry. Notably, progress in this direction has been stimulated by efforts to expand general relativity, such as the work of Asanov (e.g. \cite{asanov85}, \cite{asanov91}, \cite{asanov98}), which later motivated Kozma, Peter and Varga to study product manifolds $M \times N$ endowed with a Finsler metric
\begin{equation}
\label{warped prod 1}
    F = \sqrt{F_1^{2} + f^{2}F_{2}^{2}} \, ,
\end{equation}
called warped product, where $(M,F_1)$, $(N,F_2)$ are Finsler manifolds and $f$ is a smooth function on $M$ (see \cite{kozma:peter:varga}). Following the definition of Beem \cite{beem}, one may take $L = F^2$ to consider pseudo-Finsler metrics. For example, if $(M, F_1)$ is a $3$-dimensional Finsler manifold and $(\mathbb{R},F_2)$ is a Minkowski space, then
\begin{equation}
\label{static spacetime 1}
    L = f^2 F_2^2 - F_1^2
\end{equation}
is a Finsler metric with Lorentz signature, and $(\mathbb{R}\times M, L)$ may be regarded as a Finsler static spacetime. This is the case for \cite{li:chang}, where Li and Chang studied metrics in the form of (\ref{static spacetime 1}), given on coordinates $((t,r,\theta , \varphi),(y^t,y^r,y^\theta , y^\varphi))$ of the tangent bundle by
\begin{equation*}
    L = f^2 (y^t)^2 - \left[g^2(y^r)^2 + r^2\overline{F}^2\right] \, ,
\end{equation*}
with $\overline{F}$ a Finsler metric on coordinates $(\theta, \varphi,y^\theta , y^\varphi)$ and $f,g$ functions of $r$. They suggested the vacuum field equation for Finsler spacetime is equivalent to the vanishing of the Ricci scalar, and obtained a non-Riemannian exact solution similar to the Schwarzschild metric. Their results display viability to the explict construction of Ricci-flat Finsler metrics by warped product.

Recently, Chen, S. and Zhao have considered product manifolds $\mathbb{R}\times M$ with Finsler metrics arising from warped products in the following way: if $(M,\alpha^2)$, $(\mathbb{R},d t^2)$ are Riemannian manifolds, then $F^2 = d t^2 + f^2(t)\alpha^2$ is a warped product, which may be rewritten as $F = \alpha\sqrt{\left(\frac{d t}{\alpha}\right)^2 + f^2(t)}$. Letting $z = \frac{d t}{\alpha}$, they defined a class of Finsler metrics by
\begin{equation}
\label{warped prod 2}
    F = \alpha \sqrt{\phi(z,t)} \, ,
\end{equation}
also called warped product, where $\phi$ is a suitable function on $\mathbb{R}^2$, (see \cite{chen:shen:zhao}).

In the present work, we wish to consider Finsler metrics of similar type as (\ref{warped prod 2}), with another ``warping'', one that is consistent with the form of metrics modeling (Riemannian) static spacetimes and simplified by spherical symmetry over spatial coordinates, which emerged from the Schwarzschild metric in isotropic rectangular coordinates $(t,x^1,x^2,x^3)$:
\begin{equation}
\label{sch iso coord}
    d s^2 = \frac{\left(1 - \frac{m}{4\rho}\right)^2}{\left(1 + \frac{m}{4\rho}\right)^2} c^2 d  t^2 - \left(1 + \frac{m}{4\rho}\right)^4 \left[(d x^1)^2 + (d  x^2)^2 + (d x^3)^2\right] \, ,
\end{equation}
where $\rho=\sqrt{(x^1)^2+(x^2)^2+(x^3)^2}$ (see for example \cite{edd}, p. 93). Letting $z = \frac{d t}{\alpha}$ and $\alpha = \sqrt{(d x^1)^2 + (d x^2)^2 + (d x^3)^2}$, the Schwarzschild metric (\ref{sch iso coord}) is written as
$$d s = \alpha \sqrt{ \frac{\left(1 - \frac{m}{4\rho}\right)^2}{\left(1 + \frac{m}{4\rho}\right)^2} c^2 z^2 - \left(1 + \frac{m}{4\rho}\right)^4 } \, .$$

For $\mathbb{R}$, $\mathbb{R}^n$ with their Euclidean metrics $d t^2$, $\alpha^2$ (respectively), define a (positive-definite) Finsler metric on $\mathbb{R}\times\mathbb{R}^n$ by
\begin{equation*}
    F = \alpha \sqrt{\phi(z,\rho)} \, ,
\end{equation*}
where $z = \frac{d t}{\alpha}$, $\rho = \vert \overline{x}\vert$ for $\overline{x}\in \mathbb{R}^n$, and $\phi$ a suitable function on $\mathbb{R}^2$. We give the PDE characterization for the proposed metrics to be Ricci-flat:
\begin{thm}
\label{thm}
For $n\geq 2$, $F = \alpha\sqrt{\phi(z,\rho)}$ is Ricci-flat if and only if $P(z,\rho) = Q(z,\rho) = 0$. Furthermore, the Ricci-flat condition is weaker when $n=1$; namely, $P(z,\rho) + \rho^2Q(z,\rho) = 0$.
\end{thm}

Here $P$, $Q$ are functions of $\phi$ and its derivatives, described by equations (\ref{ricci comp}).

Next, we construct three examples. The first presents the Riemannian case to corroborate computations, the second are $m$-th root metrics, and the third consists of Randers norm. For $n\geq 3$, the non-Riemannian solutions are:
\begin{align*}
    \phi(z, \rho) &= (Az^m + B\rho^{-2m})^{\frac{2}{m}} \, , \; A,B > 0 &(\ref{m3sol})\\
    \phi(z, \rho) &= (\sqrt{A z^2 + B\rho^{-4}} + \varepsilon\sqrt{A}z)^2 \, , \; A,B > 0 \, , \; 0 < \vert \varepsilon \vert < 1 \, , \; C\in\mathbb{R}  &(\ref{r3sol})
\end{align*}
Whenever possible, we describe solutions with Lorentz signature, for which the $4$-dimensional metrics may also be studied as Finsler static spacetimes satisfying the vacuum field equation proposed in \cite{li:chang}.

\section{Geometric Quantities}

Set $M = \mathbb{R}\times\mathbb{R}^n$ with coordinates on $TM$
\begin{align*}
    x=(x^0,\overline{x}) &, \; \overline{x} = (x^1,\ldots , x^n) \, , \\
    y=(y^0,\overline{y}) &, \; \overline{y} = (y^1,\ldots , y^n) \, ;
\end{align*}
and consider a Finsler metric
\begin{equation}
\label{def}
    F = \alpha\sqrt{\phi(z,\rho)}\, ,
\end{equation}
where $\alpha= \vert\overline{y}\vert$, $z=\frac{y^0}{\vert\overline{y}\vert}$ and $\rho=\vert\overline{x}\vert$. Throughout our work, the following convention for indices is adopted: A, B, ... range from $0$ to $n$; i, j, ... range from $1$ to $n$.

This construction is the same as \cite{chen:shen:zhao} but for the ``warping''. Consequently, any calculations involving $F$ and its derivatives of any degree with respect to $y^A$ only will be similar in form to the calculations in \cite{chen:shen:zhao}, e.g. the fundamental form. The effects of the warping only appear when derivatives of $F$ with respect to $x^A$ are involved, e.g. spray coefficients. So the Hessian matrix, $g_{AB} = \frac{1}{2}[F^2]_{y^A y^B}$, is
\begin{equation}
\label{hessian matrix}
    (g_{AB}) = \left(
    \begin{array}{c|c}
    \frac{1}{2}\phi_{zz} & \frac{1}{2}\Omega_z\frac{y^j}{\alpha} \\
    \hline
    \frac{1}{2}\Omega_z\frac{y^i}{\alpha} & \frac{1}{2}\Omega\delta_{ij} - \frac{1}{2}z\Omega_z\frac{y^i y^j}{\alpha^2}
    \end{array}
    \right) \, ,
\end{equation}
where
\begin{equation}
    \Omega := 2\phi -z\phi_z\, ,
\end{equation}
and the same argument as \cite{chen:shen:zhao} to verify non-degeneracy of $F$ applies. It actually simplifies, because $\alpha$ is the Euclidean metric here. Thus,
$$\det (g_{AB}) = \frac{1}{2^{n+1}}\Omega^{n-1}\Lambda \, ,$$
where
\begin{equation}
    \Lambda:=\phi_{zz}(\Omega - z\Omega_z) - \Omega_z^2 = 2\phi\phi_{zz} - \phi_z^2\, ,
\end{equation}
and:

\begin{prop}[Prop.4.1, \cite{chen:shen:zhao}]
$F=\alpha\sqrt{\phi(z,\rho)}$ is strongly convex if and only if $\Omega, \Lambda > 0$.
\end{prop}

Moreover, letting $L = \alpha^2\phi(z,\rho)$, one has  a metric with Lorentz signature $(+,-,\ldots,-)$ if $\Omega,\Lambda < 0$, or $(-,+,\ldots,+)$ if $\Omega > 0$ and $\Lambda < 0$.

Henceforth, assume $F$ is non-degenerate. In this case, the inverse of $(g_{AB})$ is
\begin{equation}
\label{inverse matrix}
    (g^{AB}) = \left(
    \begin{array}{c|c}
    \frac{2}{\Lambda}(\Omega - z\Omega_z) & -\frac{2}{\Lambda}\Omega_z\frac{y^j}{\alpha} \\
    \hline
    -\frac{2}{\Lambda}\Omega_z\frac{y^i}{\alpha} & \frac{2}{\Omega}\delta^{ij} + \frac{2\phi_z\Omega_z}{\Omega\Lambda}\frac{y^i y^j}{\alpha^2}
    \end{array}
    \right) \, .
\end{equation}

The spray coefficients $G^C = \frac{1}{4}g^{CA}\left([F^2]_{y^A x^B}y^B - [F^2]_{x^A} \right)$ are:
\begin{subequations}
\begin{align}
    G^0 &= (U + z V)(x^m y^m)\alpha \, ,    \label{g0} \\
    G^i &= (V + W) y^i(x^m y^m) - W x^i\alpha^2 \, ,    \label{gi}
\end{align}
\end{subequations}
where
\begin{subequations}
\begin{align}
    U &:= \frac{1}{2\rho\Lambda}(2\phi\phi_{z\rho} - \phi_z\phi_{\rho})\, ,    \label{u}\\
    V &:= \frac{1}{2\rho\Lambda}(\phi_{\rho}\phi_{zz} - \phi_z\phi_{z\rho})\, ,    \label{v}\\
    W &:= \frac{1}{2\rho\Omega}\phi_{\rho}\, .    \label{w}
\end{align}
\end{subequations}

The flag curvature tensor by Berwald's formula $$K_B^C = 2[G^C]_{x^B} - [G^C]_{x^A y^B} y^A + 2G^A[G^C]_{y^A y^B} - [G^C]_{y^A}[G^A]_{y^B}$$ gives
\begin{dgroup}[frame={0pt},framesep={4pt}]
\begin{dmath}
K_0^0 = \left[ \rho^2(U + z V)W_z - (2\rho^2W +1)(U_z + V + z V_z) \right] \alpha^2 + \left[ 2(V + W)(U_z + V + z V_z) - (V_z + W_z)(U + z V) + 2U(U_{zz} + 2V_z + z V_{zz}) - \frac{1}{\rho}(U_{z\rho} + V_{\rho} + z V_{z\rho}) - (U_z + V + z V_z)^2 - (U  - z U_z - z^2V_z)V_z \right](x^m y^m)^2
\end{dmath}

\begin{dmath}
K_j^i = - \left[ 2W + (2\rho^2W + 1)(V + W) \right]\alpha^2\delta^i_j +\left[ (V + W)^2 + 2U(V_z + W_z) - \frac{1}{\rho}(V_{\rho} + W_{\rho}) \right](x^m y^m)^2\delta^i_j + \left[ 2W(2W -z W_z) + W_z(U - zW) -\frac{2}{\rho}W_{\rho} \right]\alpha^2x^i x^j + \left[ (V + W) + z(V_z + W_z)(2\rho^2W + 1) + (\rho^2(V + W) + 1)(2W - z W_z) \right]y^i y^j - \left[ 2z U (V_{zz} + W_{zz}) + (3U - z U_z - z V + 5z W)(V_z + W_z) - \frac{z}{\rho}(V_{z\rho} + W_{z\rho}) \right] (x^m y^m)^2\frac{y^i y^j}{\alpha^2} + \left[ - (2W - z W_z)^2 -2U(W_z - z W_{zz}) + \frac{1}{\rho}(2W_{\rho} - z W_{z\rho}) + W_z(U - z U_z + z^2 W_z) \right](x^m y^m) x^i y^j + \left[ -(V + W)^2 + (V_z + W_z)(U + 3z W) + \frac{1}{\rho}(V_{\rho} + W_{\rho}) \right](x^m y^m) x^j y^i
\end{dmath}

\begin{dmath}
 K_j^0 = z\left[ (2\rho^2 W + 1)(V + U_z + z V_z) - \rho^2 W_z(U + z V) \right]\alpha y^j  + \left[ z(U + z V)(V_z + W_z) - 2z U(U_{zz} + 2V_z + z V_{zz}) + (U - z U_z - z^2 V_z)(5W - U_z) - \frac{1}{\rho}(U_{\rho} - z U_{z\rho} - z^2 V_{z\rho}) \right](x^m y^m)^2\frac{y^j}{\alpha} + \left[ (U + z V)(U_z - V + z V_z - 2W) + (V - 3W)(U - z U_z - z^2 V_z) + \frac{1}{\rho}(U_{\rho} + z V_{\rho}) \right](x^m y^m)\alpha x^j
\end{dmath}

\begin{dmath}
 K_0^i = \left[ \rho^2 W_z(V - W) - (2\rho^2W + 1)V_z \right]\alpha y^i + \left[ (2W - V - U_z)(V_z + W_z) + 2U(V_{zz} + W_{zz}) -\frac{1}{\rho}(V_{z\rho} + W_{z\rho}) \right](x^m y^m)^2 \frac{y^i}{\alpha} + \left[ (U_z - W)W_z - 2U W_{zz} + \frac{1}{\rho}W_{z\rho} \right](x^m y^m)\alpha x^i
\end{dmath}
\end{dgroup}

After simplification, the Ricci curvature is:
\begin{dmath*}
\ric = \sum K^A_A \\
        = \left[ - (2\rho^2 W + 1)(U_z + n V + (n-3) W) - 2(n W + \rho^2 W_z(U - z W)) \right]\alpha^2 + \left[ 2U(U_{zz} + n V_z + (n-2) W_z) -\frac{1}{\rho}(U_{z\rho} + n V_{\rho} + (n-3) W_{\rho}) + n V(V + 2W) + W((n-5) W + 2z W_z) + U_z(2W - U_z) \right](x^m y^m)^2
\end{dmath*}

Let the Ricci curvature components be
\begin{dgroup}[frame={0pt},framesep={4pt}]
\label{ricci comp}
\begin{dmath}
 P(z,\rho) \defeq - (2\rho^2 W + 1)(U_z + n V + (n-3) W) - 2(n W + \rho^2 W_z(U - z W))
\end{dmath}
\begin{dmath}
 Q(z, \rho) \defeq 2U(U_{zz} + n V_z + (n-2) W_z) -\frac{1}{\rho}(U_{z\rho} + n V_{\rho} + (n-3) W_{\rho}) + n V(V + 2W) + W((n-5) W + 2z W_z) + U_z(2W - U_z)
\end{dmath}
\end{dgroup}

So
\begin{dmath}
\label{ricci}
\ric = P\left(\frac{y^0}{\vert\overline{y}\vert},\vert\overline{x}\vert\right)\langle \overline{y} , \overline{y} \rangle + Q\left(\frac{y^0}{\vert\overline{y}\vert},\vert\overline{x}\vert\right)\langle \overline{x} , \overline{y} \rangle^2 \\
        = \left\langle P\left(\frac{y^0}{\vert\overline{y}\vert},\vert\overline{x}\vert\right)\overline{y} + Q\left(\frac{y^0}{\vert\overline{y}\vert},\vert\overline{x}\vert\right)\langle\overline{x},\overline{y}\rangle\overline{x} , \overline{y} \right\rangle
\end{dmath}

\begin{proof}[{\bf Proof of Theorem \ref{thm}}]
Suppose $\ric = 0$. Let $e_i$ denote the $n$-dimensional vector with $1$ in the $i^{th}$ entry and zeros elsewhere. Take $\overline{y} = e_i$ and $\overline{x} = \rho e_j$ for $\rho \geq 0$. By equation (\ref{ricci}),
$$P\left(y^0,\rho\right) + Q\left(y^0,\rho\right)\rho^2\delta^{ij} = 0 \, , \; \forall i,j \, .$$
For $n \geq 2$, pick $i\neq j$ to get $P(y^0,\rho) = 0$. Now set $i=j$ to conclude $Q(y^0,\rho) = 0$ for $\rho \neq 0$. Finally, $Q(y^0,0) = 0$ by continuity. The remaining assertions are clear.
\end{proof}

The above proof suggests metrics $F$ that are singular on $(x^0,0)$ or  metrics $F$ defined on $\mathbb{R}\times \mathbb{R}^n\setminus\{0\}$ should also be considered. This becomes evident on the examples bellow.

\section{Examples}

\begin{ex}[Riemannian metrics]
Suppose $\phi(z, \rho) = \e^{f(\rho)}z^2 + \e^{g(\rho)}$.

So $\Omega = 2\e^{g}$, $\Lambda = 4\e^{f + g}$ and $F = \alpha\sqrt{\phi}$ gives a positive-definite Riemannian metric.

The Ricci curvature components are:
\begin{align*}
    P &= -\frac{1}{4\rho}\left[ p_2\e^{f-g}z^2  + p_0 \right] \\
    Q &= -\frac{1}{4\rho^3}q_0
\end{align*}
where
\begin{dgroup*}
\begin{dmath*}
p_2 = 2\rho f^{\prime\prime} + \rho(f^{\prime})^2 + (n-2)\rho f^{\prime}g^{\prime} + 2(n-1)f^{\prime}
\end{dmath*}
\begin{dmath*}
p_0 = 2\rho g^{\prime\prime} + (n-2)\rho(g^{\prime})^2 + \rho f^{\prime}g^{\prime} + 2f^{\prime} + 2(2n-3)g^{\prime}
\end{dmath*}
\begin{dmath*}
q_0 = 2\rho f^{\prime\prime} + 2(n-2)\rho g^{\prime\prime} + \rho(f^{\prime})^2 - 2\rho f^{\prime}g^{\prime} - (n-2)\rho (g^{\prime})^2 - 2f^{\prime} - 2(n-2)g^{\prime}
\end{dmath*}
\end{dgroup*}

By independence of $z$ and $\rho$, the Ricci-flat equations for $n\geq 2$ become $p_2 = p_0 = q_0 = 0$. Taking $q_0 - p_2 + n p_0 = 0$ yields:
$$4(n-1)\rho g^{\prime\prime} + (n-2)(n-1)\rho(g^{\prime})^2 + 4(n-1)^2g^{\prime} = 0 $$

For $n\geq 3 :$
\begin{equation}
\label{Riemann edo}
    4\rho g^{\prime\prime} + (n-2)\rho(g^{\prime})^2 + 4(n-1)g^{\prime} = 0
\end{equation}
If $g^{\prime} = 0$, then (\ref{Riemann edo}) is trivially satisfied. So $g(\rho)=\overline{B}$ constant is a solution. Otherwise, (\ref{Riemann edo}) is a Bernoulli differential equation in $g^{\prime}$, which can be transformed to a linear ODE by letting $u:= (g^{\prime})^{-1}$. The equation reduces to:
$$4\rho u^{\prime} - (n-2)\rho - 4(n-1)u = 0$$
Its solution gives $g(\rho) = \ln(B\vert \rho^{2-n} + C\vert^{\frac{4}{n-2}})$, for $B,C\in \mathbb{R}$ constants with $B>0$.

To find $f$, substitute $g$ in $p_0 = 0$. When $g(\rho)=\overline{B}$, $f^{\prime} = 0$ and $f$ is constant also, say $f(\rho) = \overline{A}$. For $g(\rho) = \ln(B\vert \rho^{2-n} + C\vert^{\frac{4}{n-2}})$, equation $p_0= 0$ gives:
$$f^{\prime} = \frac{4(n-2)C\rho^{1-n}}{(C-\rho^{2-n})(C+\rho^{2-n})}$$
So $f(\rho) = \ln \left[A\left(\frac{C-\rho^{2-n}}{C+\rho^{2-n}}\right)^2\right]$, for some constant $A>0$.

Therefore, when $n\geq 3$, solutions are:
\begin{subequations}
\begin{align}
    \phi(z, \rho) &= Az^2 + B \, , \; A,B > 0 \\
    \phi(z, \rho) &= A\left(\frac{C-\rho^{2-n}}{C+\rho^{2-n}}\right)^2 z^2 + B\vert \rho^{2-n} + C\vert^{\frac{4}{n-2}} \, , \; A,B > 0 , \, C\in\mathbb{R}
\end{align}
\end{subequations}

For $n = 2$, equation (\ref{Riemann edo}) still holds, but it is already linear:
$$\rho g^{\prime\prime} + g^{\prime} = 0$$
So $g(\rho) = \ln (B\vert \rho\vert^C)$, for $B,C\in \mathbb{R}$ constants with $B>0$. Substitute $g$ in $p_0 = 0$ to get:
$$(C+2)f^{\prime} = 0$$
If $C\neq -2$, then $f^{\prime} = 0$. So $f(\rho) = \overline{A}$. When $C = -2$, equation $p_2 = 0$ yields:
$$2\rho f^{\prime\prime} + \rho(f^{\prime})^2 + 2f^{\prime} = 0$$
If $f^{\prime} = 0$, the above equation is trivially satisfied and then $f(\rho) = \overline{A}$. Else, it is a Bernoulli equation in $f^{\prime}$. As before, let $u:= (f^{\prime})^{-1}$ to get a linear ODE:
$$2\rho u^{\prime} - \rho - 2u = 0$$
It gives $f(\rho) = \ln (A_1 + A_2\ln\vert\rho\vert)^2$, for real constants $A_1 , A_2$.

Thus, for $n=2$, solutions are:
\begin{subequations}
\begin{align}
    \phi(z, \rho) &= Az^2 + B\vert\rho\vert^C \, , \; A,B > 0 , \, C\in\mathbb{R}\setminus\{-2\} \\
    \phi(z, \rho) &= (A_1 + A_2\ln\vert\rho\vert)^2z^2 + B\rho^{-2} \, , \; A_1,A_2\in\mathbb{R}, \, B > 0
\end{align}
\end{subequations}

For $n = 1$, the Ricci-flat condition gives $p_2 = p_0 + q_0 = 0$, by independence of $z$ and $\rho$. This gives:
$$2 f^{\prime\prime} + (f^{\prime})^2 - f^{\prime}g^{\prime} = 0$$
So either $f(\rho) = \overline{A}$ and $g$ is an arbitrary smooth function of $\rho$, or $g = \ln(f^{\prime})^2 + f + \overline{B}$ for any smooth function $f$ of $\rho$.

Hence, solutions for $n=1$ are:
\begin{subequations}
\begin{align}
    \phi(z, \rho) &= Az^2 + \e^{g(\rho)} \, , \; A > 0 \, , \; g\in C^{\infty} \\
    \phi(z, \rho) &= \e^{f(\rho)}\left(z^2 + B[f^{\prime}(\rho)]^2 \right) , \; f\in C^{\infty} , \, B > 0
\end{align}
\end{subequations}

Finally, if $\phi(z, \rho) = \e^{f(\rho)}z^2 - \e^{g(\rho)}$, then $\Omega = - 2\e^{g}$ and $\Lambda = - 4\e^{f + g}$. So the associated metric $L = \alpha^2\phi$ has Lorentz signature $(+,-,\ldots,-)$. In this case, the Ricci curvature components are:
\begin{align*}
    P &= \frac{1}{4\rho}\left[ p_2\e^{f-g}z^2  - p_0 \right] \\
    Q &= -\frac{1}{4\rho^3}q_0
\end{align*}
where $p_2$, $p_0$ and $q_0$ are as before. Thus, by independence of $z$ and $\rho$, the Ricci-flat equations reduce to the same system as the positive-definite case.
\end{ex}

\begin{ex}[$m^{th}$-root metrics]
If $\phi(z,\rho) = \left(\e^{f(\rho)}z^{m} + \e^{g(\rho)}\right)^{\frac{2}{m}}$ for an even integer $m > 2$,  then $\Omega = \frac{2\e^g}{(\e^f z^m + \e^g)^{1-\frac{2}{m}}}$ and $\Lambda = \frac{4(m-1)\e^{f+g}z^{m-2}}{(\e^f z^m + \e^g)^{2(1-\frac{2}{m})}}$. So $F = \alpha\sqrt{\phi}$ is a positive-definite $m^{th}$-root metric.

The Ricci curvature components are:
\begin{align*}
    P &= -\frac{1}{2m^2(m-1)\rho}\left[ p_{2m}\e^{2(f-g)}z^{2m} + p_m\e^{f-g}z^m  + p_0 \right] \\
    Q &= \frac{1}{4m^2(m-1)^2\rho^3}\left[ q_{2m}\e^{2(f-g)}z^{2m} - q_m\e^{f-g}z^m  - q_0 \right]
\end{align*}
where
\begin{dgroup*}
\begin{dmath*}
p_{2m} = (m-2)(m+n-2)\rho(f^{\prime})^2
\end{dmath*}
\begin{dmath*}
p_m = 2m(m-1)\rho f^{\prime\prime} + m(m-1)\rho(f^{\prime})^2 + (n-2)(3m-4)\rho f^{\prime}g^{\prime} + m[(n-2)(3m-4)+2(m-1)]f^{\prime}
\end{dmath*}
\begin{dmath*}
p_0 = 2m(m-1)\rho g^{\prime\prime} + 2(m-1)(n-2)\rho(g^{\prime})^2 + m\rho f^{\prime}g^{\prime} + m^2f^{\prime} + 2m(m-1)(2n-3)g^{\prime}
\end{dmath*}
\begin{dmath*}
q_{2m} = (m-2)[2m^2 + (n-2)(3m-2)]\rho(f^{\prime})^2
\end{dmath*}
\begin{dmath*}
q_m = 2(m-2)[ m(m-1)(n-2)\rho f^{\prime\prime} - m(m+n-1)\rho(f^{\prime})^2 + 2(n-2)(m-1)\rho f^{\prime}g^{\prime} + m(m-1)(n-2)f^{\prime}]
\end{dmath*}
\begin{dmath*}
q_0 = 2m^2(m-1)\rho f^{\prime\prime} + 4m(m-1)^2(n-2)\rho g^{\prime\prime} + m^2\rho(f^{\prime})^2 - 4m(m-1)\rho f^{\prime}g^{\prime} - 4(m-1)^2(n-2)\rho (g^{\prime})^2 - 2m^2(m-1)f^{\prime} - 4m(m-1)^2(n-2)g^{\prime}
\end{dmath*}
\end{dgroup*}

By independence of $z$ and $\rho$, the Ricci-flat equations for $n\geq 2$ are $p_{2m} = p_m = p_0 = q_{2m} = q_m = q_0 = 0$. Since $m > 2$, $p_{2m} = q_{2m} = 0$ imply $f^{\prime} = 0$, and equations $p_m = q_m = 0$ are automatically satisfied. The remaining equations reduce to:
\begin{equation}
\label{m p eq}
    m\rho g^{\prime\prime} + (n-2)\rho (g^{\prime})^2 + (2n-3)mg^{\prime} = 0
\end{equation}
\begin{equation}
\label{m q eq}
      (n-2)[m\rho g^{\prime\prime} - \rho(g^{\prime})^2 - mg^{\prime}] = 0
\end{equation}
So $f(\rho) = \overline{A}$ and $g(\rho)$ must be determined from the above equations.

For $n \geq 3$, combine equations (\ref{m p eq}) and (\ref{m q eq}) to eliminate $g^{\prime\prime}$. This gives:
$$g^{\prime}(\rho g^{\prime} + 2m) = 0$$
If $g^{\prime} = 0$, then $g(\rho) = \overline{B}$. Otherwise, $\rho g^{\prime} + 2m = 0$ and so $g(\rho) = \ln(B\rho^{-2m})$ for some constant $B > 0$.

Therefore, when $n\geq 3$, solutions are:
\begin{subequations}
\begin{align}
    \phi(z, \rho) &= (Az^m + B)^{\frac{2}{m}} \, , \; A,B > 0  \\
    \phi(z, \rho) &= (Az^m + B\rho^{-2m})^{\frac{2}{m}} \, , \; A,B > 0  \label{m3sol}
\end{align}
\end{subequations}

For $n = 2$, (\ref{m q eq}) is vacuous and (\ref{m p eq}) gives a linear ODE:
$$\rho g^{\prime\prime} + g^{\prime} = 0$$
So $g(\rho) = \ln(B\vert\rho\vert^C)$, for constants $B > 0$ and $C\in\mathbb{R}$.

Hence, solutions for $n=2$ are:
\begin{equation}
    \phi(z,\rho) =  (Az^m + B\vert\rho\vert^C)^{\frac{2}{m}} \, , \; A,B > 0 \, , \; C\in\mathbb{R}
\end{equation}

For $n = 1$, the Ricci-flat equations are $2(m-1)p_{2m} - q_{2m} = 2(m-1)p_m + q_m = p_0 + q_0 = 0$, by independence of $z$ and $\rho$. As before, since $m > 2$, $2(m-1)p_{2m} - q_{2m} = 0$ implies $f^{\prime} = 0$, and equation $2(m-1)p_m + q_m = 0$ is automatically satisfied. The remaining equation gives:
$$m\rho g^{\prime\prime} - \rho (g^{\prime})^2 - m g^{\prime} = 0$$
If $g^{\prime} = 0$, the above equation is trivially satisfied; then $g(\rho) = \overline{B}$. Otherwise, this is yet again a Bernoulli equation in $g^{\prime}$. Let $u := (g^{\prime})^{-1}$ to obtain a linear ODE:
$$m\rho u^{\prime} + \rho + m u = 0$$
Its solution gives $g(\rho) = \ln(B\vert \rho^2 + C \vert^{-m})$ for constants $B > 0$ and $C\in\mathbb{R}$.

So, for $n=1$, solution are:
\begin{subequations}
\begin{align}
    \phi(z, \rho) &= (Az^m + B)^{\frac{2}{m}} \, , \; A,B > 0  \\
    \phi(z, \rho) &= (Az^m + B\vert \rho^{2} + C \vert^{-m})^{\frac{2}{m}} \, , \; A,B > 0 \, , \; C\in\mathbb{R}
\end{align}
\end{subequations}

For an odd integer $m>2$, all formulas still hold, but the metric generated changes signature according to the sign of $z$, because it determines the sign of $\Lambda$. For $m=2$, $\phi$ simplifies to give a Riemannian metric; in this case, the non-trivial Ricci-flat equations are multiples of the previously found equations for Riemannian metrics.

Finally, taking $\phi(z,\rho) = \left(\e^{f(\rho)}z^{m} - \e^{g(\rho)}\right)^{\frac{2}{m}}$ for some integer $m > 2$ with $m \equiv 2 (\text{mod } 4)$ gives a well-defined metric $L = \alpha^2\phi$ with Lorentz signature $(+,-,\ldots,-)$, since $\Omega = - \frac{2\e^g}{(\e^f z^m - \e^g)^{1-\frac{2}{m}}}$ and $\Lambda = - \frac{4(m-1)\e^{f+g}z^{m-2}}{(\e^f z^m - \e^g)^{2(1-\frac{2}{m})}}$.

In this setting, the Ricci components are:
\begin{align*}
    P &= -\frac{1}{2m^2(m-1)\rho}\left[ p_{2m}\e^{2(f-g)}z^{2m} - p_2\e^{f-g}z^2  + p_0 \right] \\
    Q &= \frac{1}{4m^2(m-1)^2\rho^3}\left[q_{2m}\e^{2(f-g)}z^{2m} + q_m\e^{f-g}z^m - q_0 \right]
\end{align*}
where $p_i$, $q_j$ are as before. Thus, the Ricci-flat equations coincide with the positive-definite case.

With some thought, one might consider these equations for other values of $m$. When $m > 2$ is divisible by $4$, one may take $L = \alpha^m\left(\e^{f(\rho)}z^{m} - \e^{g(\rho)}\right)$ to consider Finsler spacetimes in the sense of Pfeifer and Wohlfarth \cite{pfeifer:wohlfarth}. When $m > 2$ is odd, $F = \alpha\left(\e^{f(\rho)}z^{m} - \e^{g(\rho)}\right)^{\frac{1}{m}}$ already makes sense. However, in both cases, one needs to become concerned with the domain of $z$ and $\rho$ to ensure $\Omega$, $\Lambda$ are defined and their sign give the appropriate signature.
\end{ex}

\begin{ex}[Randers metrics]
Assume $\phi(z,\rho) = (\sqrt{\e^{f(\rho)}z^2 + \e^{g(\rho)}} + \varepsilon\e^{\frac{f(\rho)}{2}}z)^2$ with $0 < \vert \varepsilon \vert < 1$, so $F = \alpha\sqrt{\phi}$ gives a positive-definite Randers metric.

Indeed, $\Omega = 2\left(\frac{\sqrt{\e^f z^2 + \e^g} + \varepsilon\e^{\frac{f}{2}}z}{\sqrt{\e^f z^2 + \e^g}}\right)\e^g$ and $\Lambda = 4\left(\frac{\sqrt{\e^f z^2 + \e^g} + \varepsilon\e^{\frac{f}{2}}z}{\sqrt{\e^f z^2 + \e^g}}\right)^3\e^{f+g}$.

The Ricci curvature components are:
\begin{dgroup*}
\begin{dmath*}
P = -\frac{1}{4\rho\sqrt{\e^f z^2+\e^g}(\sqrt{\e^f z^2+\e^g} + \varepsilon\e^{\frac{f}{2}}z)}\left[ p_4\e^{2f-g}z^4 + 2\varepsilon p_3\e^{\frac{3f}{2}-g}\sqrt{\e^f z^2+\e^g}z^3 + p_2\e^f z^2 + \varepsilon p_1\e^{\frac{f}{2}}\sqrt{\e^f z^2+\e^g}z + p_0\e^g \right]
\end{dmath*}
\begin{dmath*}
Q = \frac{1}{4\rho^3(\e^f z^2+\e^g)^2(\sqrt{\e^f z^2+\e^g} + \varepsilon\e^{\frac{f}{2}}z)^2}\left[ q_6\e^{3f}z^6 + 2\varepsilon q_5\e^{\frac{5f}{2}}\sqrt{\e^f z^2+\e^g}z^5 + 2q_4\e^{2f+g}z^4 + 4\varepsilon q_3e^{\frac{3f}{2}+g}\sqrt{\e^f z^2+\e^g}z^3 + q_2\e^{f+2g}z^2 + 2\varepsilon q_1\e^{\frac{f}{2}+2g}\sqrt{\e^f z^2+\e^g}z + q_0\e^{3g} \right]
\end{dmath*}
\end{dgroup*}
where $p_i$, $q_j$ are functions of $\rho$, $f$, $g$ and its derivatives of order up to two. Particularly,
\begin{dgroup*}
\begin{dmath*}
p_4 = 2(\varepsilon^2 + 1)\rho f^{\prime\prime} - ((n+1)\varepsilon^2 - 1)\rho(f^{\prime})^2 + (n-2)(\varepsilon^2 + 1)\rho f^{\prime}g^{\prime} + 2(n-1)(\varepsilon^2 + 1)f^{\prime}
\end{dmath*}
\begin{dmath*}
p_3 = 2\rho f^{\prime\prime} - \frac{1}{4}((n+2)\varepsilon^2 + (n-2))\rho (f^{\prime})^2 + (n-2)\rho f^{\prime}g^{\prime} + 2(n-1)f^{\prime}
\end{dmath*}
\end{dgroup*}

For $n\geq 2$, the Ricci-flat equations reduce to $p_i = q_j = 0$, by independence of $z$ and $\rho$. Taking $p_4 - (\varepsilon^2 + 1)p_3 = 0$ reads
$$ \frac{(n+2)}{4}(\varepsilon^2 - 1)^2\rho (f^{\prime})^2 = 0 \, .$$
So $f^{\prime} = 0$, and the remaining equations simplify to:
\begin{equation}
\label{r p eq}
    2\rho g^{\prime\prime} + (n-2)\rho (g^{\prime})^2 + 2(2n-3)g^{\prime} = 0
\end{equation}
\begin{equation}
\label{r q eq}
    (n-2)[2\rho g^{\prime\prime} - \rho(g^{\prime})^2 - 2g^{\prime}] = 0
\end{equation}
Hence, $f(\rho) = \overline{A}$ and $g(\rho)$ must be determined from the above equations.

When $n\geq 3$, one may combine equations (\ref{r p eq}) and (\ref{r q eq}) to eliminate $g^{\prime\prime}$, obtaining:
$$(n-1)(\rho g^{\prime} + 4)g^{\prime} = 0$$
If $g^{\prime} = 0$, then $g(\rho) = \overline{B}$. Otherwise, $\rho g^{\prime} + 4 = 0$ and so $g(\rho) = \ln (B\rho^{-4})$ for some constant $B>0$.

Thus, solutions for $n \geq 3$ are:
\begin{subequations}
\begin{align}
    \phi(z, \rho) &= (\sqrt{A z^2 + B} + \varepsilon\sqrt{A}z)^2 \, , \; A,B > 0 \,  , \; 0 < \vert \varepsilon \vert < 1 \\
    \phi(z, \rho) &= (\sqrt{A z^2 + B\rho^{-4}} + \varepsilon\sqrt{A}z)^2 \, , \; A,B > 0 \, , \; 0 < \vert \varepsilon \vert < 1 \, , \; C\in\mathbb{R} \label{r3sol}
\end{align}
\end{subequations}

For $n = 2$, (\ref{r q eq}) is vacuous and (\ref{r p eq}) becomes a linear ODE:
$$\rho g^{\prime\prime} + g^{\prime} = 0$$
So $g(\rho) = \ln(B\vert\rho\vert^C)$, for constants $B > 0$ and $C\in\mathbb{R}$.

Hence, when $n=2$, solutions are:
\begin{equation}
    \phi(z,\rho) =  \left( \sqrt{A z^2 + B\vert \rho \vert ^{C}} + \varepsilon\sqrt{A}z \right)^2 \, , \; A,B > 0 \, , \; C\in\mathbb{R} , \; 0 < \vert \varepsilon \vert < 1
\end{equation}

Finally, for $n = 1$, the Ricci-flat condition once again implies $f^{\prime} = 0$, although the computation is lengthier and will be omitted. All remaining equations are automatically satisfied.

Therefore, for $n=1$, solutions are:
\begin{equation}
    \phi(z,\rho) = \left( \sqrt{A z^2 + \e^{g(\rho)}} + \varepsilon\sqrt{A}z \right)^2 \, , \; A > 0 \, , \; g\in C^{\infty} \, , \; 0 < \vert \varepsilon \vert < 1
\end{equation}

Clearly, one may rewrite solutions as $\phi(z,\rho) = \left( \sqrt{A z^2 + \e^{g(\rho)}} + D z \right)^2$ for any constant $D$ satisfying $D^2A^{-1} < 1$. More generally, it is possible to look for solutions in the form $\phi(z,\rho) = \left( \sqrt{\e^{f(\rho)} z^2 + \e^{g(\rho)}} \pm h(\rho) z \right)^2$ with $h^2(\rho) < \e^{f(\rho)}$, but the calculations quickly become cumbersome. In either case, it is uncertain how to consider Lorentz signature (if possible).
\end{ex}

\section{Discussion}

The Hessian of the Ricci curvature
$$\ric_{AB} = \frac{1}{2}\left[\ric \right]_{y^A y^B}$$
was the first notion for Ricci curvature tensor of Finsler metrics introduced by Akbar-Zadeh in 1988. Evidently, $\ric_{AB} = 0$ if and only if $\ric = 0$, and they imply the vanishing of the scalar curvature $R = g^{AB}\ric_{AB}$. By defining the modified Einstein tensor
$$ G_{AB} = \ric_{AB} - \frac{1}{2}g_{AB}R $$
in \cite{li:chang}, Li an Chang established the equivalence between the vacuum field equation for Finsler spacetime and the vanishing of the Ricci curvature. However, the notion of Ricci curvature tensor for Finsler metrics is not unique. If $R^{\;A}_{B\;CD}$ is the Riemann curvature tensor for Finsler metrics introduced by Berwald in 1926, then
$$\widetilde{\ric}_{AB} = \frac{1}{2}\left( R^{\;C}_{A\;CB} + R^{\;C}_{B\;CA} \right)$$
is another notion of Ricci curvature tensor introduced by Li and S. in \cite{li:shen}. Moreover, these Ricci tensors differ by a non-Riemannian quantity; namely,
$$ \widetilde{\ric}_{AB} - \ric_{AB} = H_{AB} = \frac{1}{2}\left( [\chi_B]_{y^A} + [\chi_A]_{y^B} \right) \, , $$
where the $\chi$-curvature tensor is given by
$$\chi_A = \frac{1}{2}\left[ \Pi_{x^B y^A}y^B - \Pi_{x^A} - 2\Pi_{y^A y^B}G^B \right]$$
with $\Pi = \frac{\partial G^C }{\partial y^C}$. So $\widetilde{\ric}_{AB} = 0$ if and only if $\ric_{AB} = 0$ and $H_{AB} = 0$; in words, the vanishing of $\widetilde{\ric}_{AB}$ is a stronger condition than the vanishing of $\ric_{AB}$. In particular, if $\ric = 0$ and $\chi_A = 0$, then $\widetilde{\ric}_{AB} = 0$.

For the proposed metrics $F = \alpha\sqrt{\phi(z,\rho)}$, we have $\Pi = \Psi(x^m y^m)$, where
\begin{equation}
\label{Psi}
    \Psi := U_z + (n+2)V + (n-1)W \, ,
\end{equation}
and the $\chi$-curvature is
\begin{dgroup}
\begin{dmath}
\chi_0 = \left[ \frac{1}{2\rho}\Psi_{z\rho} - U\Psi_{zz} - W\Psi_{z} \right]\frac{(x^my^m)^2}{\alpha} + \frac{1}{2}(2\rho^2 W + 1)\Psi_z\alpha
\end{dmath}
\begin{dmath}
\chi_i = \left[ zU\Psi_{zz} - \frac{z}{2\rho}\Psi_{z\rho} + (U + 2zW)\Psi_z \right] \frac{(x^m y^m)^2}{\alpha^2}y^i -\frac{z}{2}(2\rho^2 W + 1)\Psi_z y^i - (U + zW)\Psi_z(x^m y^m)x^i
\end{dmath}
\end{dgroup}
Clearly, $\Psi_z = 0$ is a sufficient condition for the vanishing of the $\chi$-curvature. By direct verification, all solutions in previous section satisfy $\Psi_z = 0$. Thus, they are strongly Ricci-flat metrics: $\widetilde{\ric}_{AB} = \ric_{AB} = 0$.

In addition to the examples presented here, it seems to be feasible (although lengthy) to construct other types of (strongly) Ricci-flat metrics in the proposed form; particularly, one may look for series expansions. The same type of construction also seems to work well for Ricci-isotropic metrics, $\ric = [(n+1)-1] k(x)F^2$. At the very least the PDE characterization is similar to describe; namely, for $n \geq 2$, $F=\alpha\sqrt{\phi(z,\rho)}$ is Ricci isotropic if and only if $P = n k \phi$ and $Q=0$. It might be wise, however, to spend such efforts with a wider class of warped product Finsler metrics, which may allow for global solutions on $\mathbb{R}\times M$; for instance, a class of Finsler metrics defined by
\begin{equation}
    F = \alpha \sqrt{\phi(z,\overline{x})} \, ,
\end{equation}
for $\alpha$ any Riemannian metric on $M$, $z$ as before and $\phi$ some appropriate function on $\mathbb{R}\times M$.

redpajama_set_name: RedPajamaArXiv


I have an unbelievable story to share with you.
I had the chance to talk with the world-famous Samantha Hess, the affection-starved woman who divorced her husband to cuddle everyone she possibly could. She turned cuddling into her job, seriously. And then she got famous for it.
She claims to have spent nearly 75,000 minutes cuddling in the last 2.5 years and has been featured on the cover of the Wall Street Journal and America's Got Talent. Through her greatest struggle, a touch-deprived marriage, she was able to find her purpose and offer her unique gift to the world.
It's such a fun interview. She's one of the most lively humans I've ever talked to.
What about the "sexual" part that is bound to come up?
How do you scale the cuddling business?
Samantha Hess, owner of Cuddle Up To Me has established herself as the industry expert. In the last 2 years she has professionally cuddled more than 50,000 minutes, written a book, opened a 3,500 sq ft retail space, hired 3 employees, designed and ran the world's first cuddling convention, and now she brings the only hands on certification program that will standardize the profession and legitimize the industry. Samantha Hess is on a mission to take over the world, one hug at a time!
Samantha recently released Touch: The Power of Human Connection. The book explores the science and psychology behind cuddling and offers detailed advice for improving intimacy. The book is available at http://cuddleuptome.com or Amazon.
Samantha can be reached by email at samantha@cuddleuptome.com, at the studio: 503-719-6203 (or 1-844-2CUDDLE), by cell phone: 503-703-6459 or on Skype under cuddleuptome.


text:
package com.choodon.rpc.framework.cluster.support;

import com.choodon.rpc.base.RPCCallback;
import com.choodon.rpc.base.RPCFuture;
import com.choodon.rpc.base.common.RPCConstants;
import com.choodon.rpc.base.common.URL;
import com.choodon.rpc.base.common.URLParamType;
import com.choodon.rpc.base.extension.ExtensionLoader;
import com.choodon.rpc.base.extension.SpiMeta;
import com.choodon.rpc.base.log.LoggerUtil;
import com.choodon.rpc.base.protocol.RPCRequest;
import com.choodon.rpc.base.protocol.RPCResponse;
import com.choodon.rpc.base.util.URLTools;
import com.choodon.rpc.framework.cluster.ha.HaStrategy;
import com.choodon.rpc.framework.cluster.loadbalance.LoadBalance;
import com.choodon.rpc.framework.referer.Referer;
import com.choodon.rpc.framework.referer.RefererManager;
import com.choodon.rpc.registry.api.EventTypeEnum;
import com.choodon.rpc.registry.api.RegistryFactory;
import com.choodon.rpc.registry.api.RegistryService;

@SpiMeta(name = RPCConstants.DEFAULT)
public class DefaultCluster extends AbstractCluster {

    @Override
    public RPCResponse syncCall(RPCRequest request) throws Exception {
        if (referers.size() == 0) {
            LoggerUtil.warn(mergerURL.getParameter(RPCConstants.CLUSTER_KEY) + " has no providers.");
            return null;
        }
        return haStrategy.syncCall(request, referers, loadBalance);
    }

    @Override
    public RPCFuture asyncCall(RPCRequest request) throws Exception {
        if (referers.size() == 0) {
            LoggerUtil.warn(mergerURL.getParameter(RPCConstants.CLUSTER_KEY) + " has no providers.");
            return null;
        }
        return haStrategy.asyncCall(request, referers, loadBalance);
    }

    @Override
    public void callback(RPCRequest request, RPCCallback callBack) throws Exception {
        if (referers.size() == 0) {
            LoggerUtil.warn(mergerURL.getParameter(RPCConstants.CLUSTER_KEY) + " has no providers.");
            return;
        }
        haStrategy.callback(request, callBack, referers, loadBalance);
    }

    @Override
    public void notify(URL url, EventTypeEnum envetType) {
        switch (envetType) {
            case ADD:
                addNotify(url);
                break;
            case UPDATE:
                updateNotify(url);
                break;
            case REMOVE:
                removeNotify(url);
                break;
        }

    }

    private void addNotify(URL url) {
        String catagray = url.getParameter(RPCConstants.CATAGRAY);
        switch (catagray) {
            case RPCConstants.PROVIDER_CATAGRAY:
                Referer referer = RefererManager.create(url);
                referers.add(referer);
                break;
            default:
                LoggerUtil.error("unknown catagray,Notify URL: " + url);
                break;
        }
    }

    private void updateNotify(URL url) {
        String catagray = url.getParameter(RPCConstants.CATAGRAY);
        switch (catagray) {
            case RPCConstants.PROVIDER_CATAGRAY:
            default:
                LoggerUtil.error("unknown catagray,Notify URL: " + url);
        }
    }

    private void removeNotify(URL url) {
        String catagray = url.getParameter(RPCConstants.CATAGRAY);
        switch (catagray) {
            case RPCConstants.PROVIDER_CATAGRAY:
                Referer referer = RefererManager.create(url);
                referers.remove(referer);
            default:
                LoggerUtil.error("unknown catagray,Notify URL: " + url);
        }
    }

    @Override
    public void init(URL interfaceURL, URL protocolURL, URL registryURL) {
        mergerURL = URL.copy(interfaceURL);
        mergerURL.addParameter(RPCConstants.CLUSTER_KEY, URLTools.getClusterKey(interfaceURL, protocolURL, registryURL));
        mergerURL.addParameters(protocolURL.getParameters());
        String haStrategyName = interfaceURL.getParameter(URLParamType.haStrategy.getName(), URLParamType.haStrategy.getValue());
        haStrategy = ExtensionLoader.getExtensionLoader(HaStrategy.class).getExtension(haStrategyName);
        String loadBalanceName = interfaceURL.getParameter(URLParamType.loadBalance.getName(), URLParamType.loadBalance.getValue());
        loadBalance = ExtensionLoader.getExtensionLoader(LoadBalance.class).getExtension(loadBalanceName);
        RegistryFactory registryFactory = ExtensionLoader.getExtensionLoader(RegistryFactory.class).getExtension(registryURL.getProtocol());
        RegistryService registryService = registryFactory.getRegistryService(registryURL);
        registryService.subscribe(URLTools.buildRegistryConsumerURL(protocolURL, interfaceURL), this);
    }
}


text:
Charizma & Peanut Butter Wolf :: Big Shots
LPReview
This album is as unlikely as they come. Charles Hicks, the rapping half of this duo, has been dead for over a decade. There's no intention to shock or offend with that statement, it's just something to bear in mind as you read this review. "Big Shots" was originally recorded with the intention of being released on Hollywood BASIC, but Charizma's untimely death in Dcember of 1993 shocked his producer and friend Wolf so much that the already recorded material was put on the shelf while Wolf went on hiatus.
Like KRS-One after Scott LaRock's death, Wolf returned to the hip-hop world stung by the needless tragedy but more determined to make an impact and a statement with his life. After finding success as a producer on his own, the enterprising Wolf would go on to form Stones Throw Records, today one of the best recognized and respected of labels in indie hip-hop. Clearly though, Wolf was still haunted by what happened to his close friend Hicks, because the first single Stones Throw released was their track "My World Premiere." Seven years after the rap world got their first taste of Charizma, Wolf has at long last polished their unreleased album to a shine he feels worthy of sharing with the public.
If you're old enough, think back to what you remember was fresh in 1993. Think about the artists and albums that excited you. Cypress Hill. Digable Planets. Wu-Tang Clan. A year in which Snoop Dogg was not an industry veteran, but Dr. Dre's newest protege with a much anticipated solo album in the wings. To evaluate this "new" album, "Big Shots" must (as best as is possible) be considered in this context. As such, Charizma is a good MC for his era. He has the Project Blowed sound of Aceyalone vocally, the straightforward no-nonsense flow of Nas, and a style that lives up to his nom de plume by grabbing your attention. On the production side, these early works of Wolf are on par with any of the best beat maestros of the day. On "Devotion" he slings notes like a young Diamond D. "Ice Cream Truck" is some smoothed out "Peachfuzz" K.M.D. type shit. "Soon to be Large" is the type of bomb Pete Rock could have dropped if he worked with the aspiring rapper, sans horny horns. "Methods" is a spacy sounding suave song which would have fit into "Bizarre Ride II the Pharcyde" perfectly. On and on and on, to the break of dawn, the music on "Big Shots" is type strong like that.
Peanut Butter Wolf was clearly coming into his own working on this album, and would eventually go on to create "Peanut Butter Breaks" that would bring him acclaim in the underground the world over. As for Charizma himself, he clearly had promise as an MC. Promise enough to make him listenable on this album. For those who have heard "World Ultimate" by The Nonce, one may be struck by how similar he is to that group's producer/rapper Nouka Base. Their song "Mix Tapes" was a friendly, upbeat, likeable jam, but it also showed Nouka to be a simplistic rapper whose words could not grab your attention like his voice or beats did. Far be it from one to speak ill of the dead, but for better or worse Charizma is this type of MC. If there's such a thing as being too nice, Charles Hicks may very well have been guilty of it. "Tell You Something" has an intimidating title, but Charizma is still as sweet as a sugar cookie:
"You, the one with the big yeahhh
And you, ohhh you look so good
And you, you shoplift every day
And you and you and you come here, yeahhh
High school girls you need to learn some things
College girls you need to use your brains
Cause I'm a creator, and a dominator
Then I'm down for hittin skins later"
Listening to Charizma rap, the closest he comes to being foul-mouthed is the sentence "I like hip-hop, you like R&B crap." Intimidating? No. Charizma's personality may have won him friends and fans, but in a battle over who is more hardcore he'd probably lose to Will Smith. While this may sound like a fatal indictment of "Big Shots," it's really not. In some ways, it's refreshing to hear an MC this sincerely nice, one who clearly believed he didn't have to follow the gangster rap trends that were starting to dominate hip-hop on the West coast in '93. At the same time, he avoids the trap of sounding like a polka dot joke like Kwame or a quickly forgotten rapper-of-the-minute like Redhead Kingpin. No pun intended, but on the strength of his charisma, Hicks could have had a long career in hip-hop. With Peanut Butter Wolf at his side providing beats, the two would no doubt have gone far; and given time to develop his raps the C in Charizma might eventually have stood for Complex too. Like Dooley-O's "Watch My Moves 1990" though, this album is an artifact of a bygone era unearthed like a fossil and put on display for all hip-hop to see. While it would have sold well back in the day, today's hip-hop fans may be more interested in Wolf's music than Charizma's lyrics. That's a shame, but no one's to blame. We'll never know what the charismatic MC would have done if he was still alive today. With this release, Charles Hicks, rest in peace.
6Overall Score
CharizmaPeanut Butter Wolf
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text:
Home » Laws of the Federation » GOVERNMENT PROMISSORY NOTES ACT
GOVERNMENT PROMISSORY NOTES ACT
Short title.
Interpretation.
Issue of promissory notes for the purpose of raising authorised loans.
Loans to be charged upon revenue.
Particulars to be published in the Federal Gazette.
Issue and effect of promissory notes.
Transferability.
Appropriation of revenue for payment of principal sum and interest.
Payment of principal sum and interest.
Cessation of liability to pay interest.
Place of payment.
Sinking funds
Establishment of sinking funds.
Appropriation of revenue for sinking fund.
Investment of sinking fund.
Cessation of contributions to sinking fund.
Expenses to be paid out of sinking fund.
Deficiency in sinking fund to be a charge upon revenue.
Issue of duplicates and renewals.
Liability after renewal.
Discharge of liability.
Exemption from stamp duties.
Power to make regulations.
23. Saving
An Act to make provision for the creation and issue of Government Promissory
Notes for the purpose of raising loans.
[12th May, 1960]
This Act may be cited as the Government Promissory Notes Act.
"Central Bank" means the Central Bank of Nigeria;
"Government promissory note" means a promissory note issued under section 3 of
this Act;
"Minister" means the Minister for the time being charged with responsibility for
matters relating to finance.
Issue of promissory notes for the purpose of raising authorised loans
(1) Whenever by any Act whether enacted before or after the coming into operation
of this Act, authority has been or is given to raise any sum of money by way of loan for
any purpose mentioned in that Act, or whenever it is necessary to raise any sum of money
for the purpose of repaying any loan raised by the Federal Government under this or any
other Act, the Minister may, from time to time, raise such sum or part thereof under the
provisions of this Act from Governments of other countries and bodies corporate by the
issue of securities in the form of Government promissory notes.
(2) Subject to subsection (1) of this section, the Governor of the Central Bank may
issue naira denominated notes or other forms of securities or make any arrangement in
respect of outstanding foreign obligations of the Federal Government in the course of
rescheduling or restructuring such obligations.
4. Loans to be charged upon revenue
The principal sums and interest represented or secured by any Government promis-
sory notes are hereby charged upon and shall be payable out of the general revenue and assets of the Federation.
5. Particulars to be published in the Federal Gazette
The Minister shall, in respect of each loan raised by the issue of Government promis-
sory notes, cause the following particulars to be published in the Federal Gazette-
(a) the face value of the notes;
(b) the rate of interest payable on the loan;
(c) the date of redemption of the notes and, if the Minister has reserved an option to redeem the notes at an earlier date, the fact that he has reserved such an option and the terms and conditions on which the option may be exercised.
Issue and effect of promissory notes
(1) Every Government promissory note shall bear the signature of the Minister and of the Accountant-General of the Federation for and on behalf of the Government of the
Federation and shall, when issued, bind the Government of the Federation to pay the
principal sum mentioned in that note and the interest thereon in accordance with the provisions of this Act or in pursuance of an option to redeem that note reserved by the Minister.
(2) Naira denominated notes issued by the Central Bank in accordance with sec-
tion 3 (2) of this Act shall bear the signature of the Governor of the Central Bank and
shall, when issued, bind the Federal Government to pay the principal sum mentioned in
that note and the interest thereon in accordance with the provisions of this Act.
7. Transferability
(1) Every Government promissory note shall specify whether or not it is transferable,
and, if transferable, the extent to which and the manner in which it may be transferred.
(2) Notwithstanding anything contained in the Bills of Exchange Act, a Government
promissory note shall not be transferable except to the extent and in the manner specified
in the note.
[Cap. B8.]
8. Appropriation of revenue for payment of principal sum and interest
So long as any principal sum or interest is payable under this Act in respect of any
Government promissory notes, the Minister shall appropriate out of the general revenue
and assets of Nigeria a sum sufficient to meet all principal sums and interest so payable,
and the Central Bank shall thereupon pay such principal sum or interest out of the sum so
appropriated.
Payment of principal sum and interest
(1) The principal sum and interest due on any Government promissory note shall be
payable on such dates and, in the case of interest, at such rate as shall be specified in the
(2) Where any amount has become payable on any date as interest due on any Gov-
ernment promissory note, no interest on that amount shall, after that date, be paid or pay-
able by or on behalf of the Federal Government to any person in any circumstances.
10. Cessation of liability to pay interest
No person shall be entitled to claim interest on any Government promissory note in
respect of any period which has elapsed after the earliest date on which demand could
lawfully have been made for the payment of the principal sum due on the note.
11. Place of payment
All payments of interest and all payments of the principal sum due on any Govern-
ment promissory note shall be made at the Central Bank in Lagos:
Provided that the Central Bank or any person authorised in writing by the Central
Bank in that behalf may pay any such interest or principal sum at any other place,
whether within or without Nigeria, in pursuance of any arrangement which the Central
Bank may make for that purpose.
12. Establishment of sinking funds
(1) The Minister may, after consultation with the Governor of the Central Bank, es-
tablish sinking funds for the purpose of redeeming loans raised by the issue of Govern-
ment promissory notes:
Provided that no sinking fund so established shall be in respect of more than one
such loan.
(2) The Minister shall, in respect of any sinking fund established under subsection (1)
of this section, direct by notice in the Federal Gazette, the rate and intervals at which appropriation out of the general revenue and assets of the Federation shall be made as a
contribution to the sinking fund and the date from which such contributions shall com-
mence.
13. Appropriation of revenue for sinking fund
After the date directed by the Minister under section 12 of this Act as the date from
which contributions to a sinking fund established under that section shall commence, the
Minister shall, at the intervals so directed for the payment of contributions to the fund,
appropriate out of the general revenue and assets of the Federation a sum determined in
accordance with the rate so directed.
14. Investment of sinking fund
(1) All moneys appropriated under section 13 of this Act as contributions to a sinking
fund established under section 12 of this Act; shall be paid to the Central Bank and may
then be invested in such stock or securities or in such other investments or classes of in-
vestments as may be approved by the Minister.
(2) The Central Bank may, from time to time, with the approval of the Minister, vary
any investment made under subsection (1) of this section or may realise and re-invest any
moneys invested under that subsection.
(3) The dividends, interest, bonus and other profits of any investment of any part of
any such sinking fund shall be invested by the Central Bank so as to form part of that
sinking fund in like manner as moneys appropriated under section 13 of this Act as con-
tributions to that sinking fund.
15. Cessation of contributions to sinking fund
Notwithstanding anything to the contrary contained in this Act, if at any time the
Governor of the Central Bank is satisfied that a sinking fund established under section 12
of this Act will be sufficient with further accumulations of interest, but without further
payments of contributions, to enable the loan in respect of which it was established to be
redeemed at the time fixed for its redemption, he shall inform the Minister accordingly,
and the Minister may thereupon suspend further payments of contributions to that sinking
fund:
Provided, however, that the contributions to that sinking fund shall be recom-
menced if the Governor of the Central Bank at any time thereafter informs the Minister
that he is no longer satisfied that the sinking fund with further accumulations of interest,
will be sufficient for the redemption of that loan.
16. Expenses to be paid out of sinking fund
There shall be paid out of any sinking fund established under section 12 of this Act,
expenses specifically incurred in, or incidental to, the investment and management of that
fund and the repayment of the loan in respect of which it was established.
17. Deficiency in sinking fund to be a charge upon revenue
In the event of a sinking fund established under section 12 of this Act being found, at
the time fixed for the repayment of the loan in respect of which it was established, to be
insufficient for such redemption, the deficiency shall be made good out of the general
revenue and assets of the Federation.
18. Issue of duplicates and renewals
(1) The Accountant-General of the Federation may issue duplicate Government
promissory notes in such circumstances as may be prescribed.
(2) The Accountant-General of the Federation may issue renewals of Government
19. Liability after renewal
(1) Where a renewed Government promissory note has been issued under section 18
of this Act, the note so issued shall be deemed to constitute a new contract between the
Federal Government and the holder and any other Government or body corporate, deriv-
ing title thereafter through the holder.
(2) No such renewal shall affect the rights as against the Federal Government or any
other Government or body corporate to the promissory note so renewed.
20. Discharge of liability
The liability of the Government of the Federation shall-
(a) in respect of any Government promissory note redeemed on or after the date on which payment of the principal sum becomes due, be discharged after the lapse of six years from that date;
(b) in respect of any Government promissory note in place of which a duplicate is
issued under section 18 of this Act, be discharged after the lapse of six years
from the date of the issue of such duplicate or from the date of the last pay-
ment of interest on such note, whichever date is the later;
(c) in respect of any Government promissory note for which a renewed note is
from the date of the issue of the renewed note.
21. Exemption from stamp duties
All Government promissory notes and all renewals and duplicates thereof shall be in
such form as may be prescribed and shall be free from stamp duty, anything in any other
Act to the contrary notwithstanding.
22. Power to make regulations
(1) The Minister may, after consultation with the Governor of the Central Bank,
make regulations for the purpose of giving effect to the provisions of this Act.
(2) In particular and without prejudice to the generality of the provisions of subsec-
tion (2) of this section, the regulations may provide for all or any of the following mat-
ters-
(a) the manner in which payment of interest in respect of Government promissory notes is to be made and acknowledged;
(b) the circumstances in which Government promissory notes must be renewed
before further payment of interest thereon may be claimed;
(c) the issue of duplicate Government promissory notes;
(d) the renewal of Government promissory notes;
(e) the manner of payment of interest to joint holders of Government promissory
(f) the disposal of unclaimed interest;
(g) the payment of principal sums and interest and transfer of Government promissory notes in the case of bodies corporate under a legal disability;
(h) all matters required or permitted by this Act to be prescribed and all matters incidental to or connected with the matters herein before enumerated.
23. Savings
Nothing contained in this Act shall affect the provisions of any other enactment re-
lating to the manner in which or the terms and conditions on which loans may be raised
by the Federal Government.
NATIONAL BOUNDARY COMMISSION (ESTABLISHMENT) ACT
NIGERIAN EXPORT-IMPORT BANK ACT
MOTOR SPIRIT (RETURNS) ACT


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Daisy Bates House
Home to Civil Rights activist Daisy Bates (1914-1999), the Daisy Bates House is a National Historical Landmark for its role in the Civil Rights Movement and the Little Rock Central High School desegregation crisis of 1957-1958. Along with her husband, Daisy Bates assisted in the efforts for desegregation, opening her home to be a safe haven for the nine students involved. The Central High School desegregation crisis was "the first time a President used federal powers to uphold and implement a federal court decision regarding school desegregation" ("Daisy Bates House"). The house became the central headquarters, pick up, and drop off locations for the Little Rock Nine, the nine African American students whose goal was to desegregate the school, which lead to the home becoming a frequent target for damage and violence from supporters of segrigation. However, the strength of Bates and the Little Rock Nine sent a message to South: desegrigtaion was happening and Jim Crow laws were coming to an end and no longer would be tolerated.
Daisy Bates
Daisy Bates and the Little Rock Nine
Daisy Bates was president of the Arkansas state conference of the National Association for the Advancement of Colored People (NAACP) branches and was co-editor of the Arkansas State Press, in addition to serving as a mentor to the Little Rock Nine, who ultimately defeated segregated at Little Rock Central High School. As a symbol for the legal fight to desegregate the public school system, Bates played a significant role in both the Civil Rights Movement and the Little Rock Central High School desegregation crisis.
http://www.nps.gov/nr/travel/civilrights/ar2.htm
National Parks and Services
Daisy Bates Bio
This property is privately owned and is not open to the public.
Created by Katie Olszeski on December 13th 2013, 6:02:48 pm.
Last updated by Alexa Antill on July 19th 2015, 5:21:38 pm.