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July  2016, 12(3): 833-850. doi: 10.3934/jimo.2016.12.833

## Auction games for coordination of large-scale elastic loads in deregulated electricity markets

 1 School of Automation, Beijing Institute of Technology, 5th South Street, Zhongguancun, Haidian District, Beijing 100081, China, China, China

Received  June 2014 Revised  April 2015 Published  September 2015

Auctions, e.g. market clearing price (MCP) auctions, have been widely adopted in electricity markets, and progressive second price (PSP) auctions are stated possessing promising properties of incentive compatibility and efficiency. In this work, we study the coordination of large-scale elastic loads in deregulated electricity markets under MCP and PSP auctions. To explore the performances of these auctions in the underlying problems, we focus on the key issues of the payment comparison, incentive compatibility and efficiency of Nash equilibrium (NE), and develop the following results: (i) The individual payment under MCP is always higher than that under PSP, and their difference vanishes asymptotically as the system scale increases; (ii) The incentive compatibility holds under PSP, and holds under MCP only with respect to others' efficient bid profile; (iii) The efficient bid profile under PSP is an NE, while that under MCP is an $\varepsilon$-NE which degenerates to an NE asymptotically as the system scale increases. With these analyses, we claim that it is pretty promising to apply both MCP and PSP auctions to the large-scale load coordination problems in deregulated electricity markets.
Citation: Suli Zou, Zhongjing Ma, Xiangdong Liu. Auction games for coordination of large-scale elastic loads in deregulated electricity markets. Journal of Industrial and Management Optimization, 2016, 12 (3) : 833-850. doi: 10.3934/jimo.2016.12.833
##### References:
 [1] H. Aalami, M. P. Moghaddam and G. Yousefi, Modeling and prioritizing demand response programs in power markets, Electric Power Systems Research, 80 (2010), 426-435. doi: 10.1016/j.epsr.2009.10.007. [2] K. Abbink, J. Brandts and P. Pezanis-Christou, Auctions for government securities: A laboratory comparison of uniform, discriminatory and spanish designs, Journal of Economic Behavior & Organization, 61 (2006), 284-303. [3] M. Albadi and E. El-Saadany, A summary of demand response in electricity markets, Electric Power Systems Research, 78 (2008), 1989-1996. doi: 10.1016/j.epsr.2008.04.002. [4] L. Ausubel and P. Cramton, Demand Reduction and Inefficiency in Multi-Unit Auctions, Working papers, University of Maryland, 2002. [5] F. E. Banks, Economics of electricity deregulation and privatization: an introductory survey, Energy, 21 (1996), 249-261. doi: 10.1016/0360-5442(95)00122-0. [6] E. Bompard, Y. Ma, R. Napoli and G. Abrate, The demand elasticity impacts on the strategic bidding behavior of the electricity producers, IEEE Transactions on Power Systems, 22 (2007), 188-197. doi: 10.1109/TPWRS.2006.889134. [7] D. Callaway and I. Hiskens, Achieving controllability of electric loads, Proceedings of the IEEE, 99 (2011), 184-199. doi: 10.1109/JPROC.2010.2081652. [8] S. Cha, T. Reen and N. Hah, Optimal charging strategies of electric vehicles in the UK power market, in 1st Conference on Innovative Smart Grid Technologies, Gaithersburg, Maryland, 2010, 1-8. [9] K. Dietrich, J. M. Latorre, L. Olmos and A. Ramos, Demand response and its sensitivity to participation rates and elasticities, in 8th International Conference on the European Energy Market (EEM), Zagreb, Croatia, 2011, 716-717. doi: 10.1109/EEM.2011.5953103. [10] N. Fabra, N. Von der Fehr and D. Harbord, Modeling electricity auctions, The Electricity Journal, 15 (2002), 72-81. doi: 10.1016/S1040-6190(02)00347-0. [11] T. Genc, Discriminatory versus uniform-price electricity auctions with supply function equilibrium, Journal of optimization theory and applications, 140 (2009), 9-31. doi: 10.1007/s10957-008-9437-8. [12] V. Gountis and A. Bakirtzis, Bidding strategies for electricity producers in a competitive electricity marketplace, IEEE Transactions on Power Systems, 19 (2004), 356-365. doi: 10.1109/TPWRS.2003.821474. [13] R. Jain and J. Walrand, An efficient Nash-implementation mechanism for network resource allocation, Automatica, 46 (2010), 1276-1283. doi: 10.1016/j.automatica.2010.05.013. [14] P. Jia and P. Caines, Analysis of quantized double auctions with application to competitive electricity markets, INFOR: Information Systems and Operational Research, 48 (2010), 239-250. doi: 10.3138/infor.48.4.239. [15] D. S. Kirschen, Demand-side view of electricity markets, IEEE Transactions on Power Systems, 18 (2003), 520-527. doi: 10.1109/TPWRS.2003.810692. [16] V. Krishna, Auction Theory, Academic Press, 2009. [17] A. Lazar and N. Semret, Design and analysis of the progressive second price auction for network bandwidth sharing, Telecommunication Systems, 13, (1999). [18] G. Li and J. Shi, Agent-based modeling for trading wind power with uncertainty in the day-ahead wholesale electricity markets of single-sided auctions, Applied Energy, 99 (2012), 13-22. doi: 10.1016/j.apenergy.2012.04.022. [19] Z. Ma, D. Callaway and I. Hiskens, Decentralized charging control of large populations of plug-in electric vehicles, IEEE Transactions on Control Systems Technology, 21 (2013), 67-78. [20] P. Maillé and B. Tuffin, The progressive second price mechanism in a stochastic environment, Netnomics, 5 (2003), 119-147. [21] O. Marce, H.-H. Tran and B. Tuffin, Double-sided auctions applied to vertical handover for mobility management in wireless networks, Journal of Network and Systems Management, 22 (2014), 658-681. doi: 10.1007/s10922-013-9269-1. [22] S. Nielsen, P. Sorknæs and P. A. Østergaard, Electricity market auction settings in a future danish electricity system with a high penetration of renewable energy sources: A comparison of marginal pricing and pay-as-bid, Energy, 36 (2011), 4434-4444. doi: 10.1016/j.energy.2011.03.079. [23] S. Sethi, H. Yan, J. Yan and H. Zhang, An analysis of staged purchases in deregulated time-sequential electricity markets, Journal of Industrial and Management Optimization, 1 (2005), 443-463. doi: 10.3934/jimo.2005.1.443. [24] A. Singh, Smart grid dynamic pricing, International Journal of Engineering Research and Applications (IJERA), 2 (2012), 705-742. [25] Y. R. Sood, N. P. Padhy and H. Gupta, Wheeling of power under deregulated environment of power system - a bibliographical survey, IEEE Transactions on Power Systems, 17 (2002), 870-878. doi: 10.1109/TPWRS.2002.800967. [26] K. Vitae and L. Lave, Demand response and electricity market efficiency, The Electricity Journal, 20 (2007), 69-85. [27] F. Wen and A. K. David, Optimal bidding strategies and modeling of imperfect information among competitive generators, IEEE Transactions on Power Systems, 16 (2001), 15-21. [28] J. H. Williams and F. Kahrl, Electricity reform and sustainable development in China, Environmental Research Letters, 3 (2008), 044009. doi: 10.1088/1748-9326/3/4/044009. [29] Z. Xu, W. Xu, W. Shao and Z. Zeng, Real-time pricing control on generation-side: Optimal demand-tracking model and information fusion estimation solver, IEEE Transactions on Power Systems, 29 (2014), 1522-1535. doi: 10.1109/TPWRS.2013.2296809. [30] X. Zhao and C. Ma, Deregulation, vertical unbundling and the performance of China's large coal-fired power plants, Energy Economics, 40 (2013), 474-483. doi: 10.1016/j.eneco.2013.08.003. [31] S. Zou, Z. Ma and X. Liu, Auction-based distributed efficient economic operations of microgrid systems, International Journal of Control, 87 (2014), 2446-2462. doi: 10.1080/00207179.2014.926395.

show all references

##### References:
 [1] H. Aalami, M. P. Moghaddam and G. Yousefi, Modeling and prioritizing demand response programs in power markets, Electric Power Systems Research, 80 (2010), 426-435. doi: 10.1016/j.epsr.2009.10.007. [2] K. Abbink, J. Brandts and P. Pezanis-Christou, Auctions for government securities: A laboratory comparison of uniform, discriminatory and spanish designs, Journal of Economic Behavior & Organization, 61 (2006), 284-303. [3] M. Albadi and E. El-Saadany, A summary of demand response in electricity markets, Electric Power Systems Research, 78 (2008), 1989-1996. doi: 10.1016/j.epsr.2008.04.002. [4] L. Ausubel and P. Cramton, Demand Reduction and Inefficiency in Multi-Unit Auctions, Working papers, University of Maryland, 2002. [5] F. E. Banks, Economics of electricity deregulation and privatization: an introductory survey, Energy, 21 (1996), 249-261. doi: 10.1016/0360-5442(95)00122-0. [6] E. Bompard, Y. Ma, R. Napoli and G. Abrate, The demand elasticity impacts on the strategic bidding behavior of the electricity producers, IEEE Transactions on Power Systems, 22 (2007), 188-197. doi: 10.1109/TPWRS.2006.889134. [7] D. Callaway and I. Hiskens, Achieving controllability of electric loads, Proceedings of the IEEE, 99 (2011), 184-199. doi: 10.1109/JPROC.2010.2081652. [8] S. Cha, T. Reen and N. Hah, Optimal charging strategies of electric vehicles in the UK power market, in 1st Conference on Innovative Smart Grid Technologies, Gaithersburg, Maryland, 2010, 1-8. [9] K. Dietrich, J. M. Latorre, L. Olmos and A. Ramos, Demand response and its sensitivity to participation rates and elasticities, in 8th International Conference on the European Energy Market (EEM), Zagreb, Croatia, 2011, 716-717. doi: 10.1109/EEM.2011.5953103. [10] N. Fabra, N. Von der Fehr and D. Harbord, Modeling electricity auctions, The Electricity Journal, 15 (2002), 72-81. doi: 10.1016/S1040-6190(02)00347-0. [11] T. Genc, Discriminatory versus uniform-price electricity auctions with supply function equilibrium, Journal of optimization theory and applications, 140 (2009), 9-31. doi: 10.1007/s10957-008-9437-8. [12] V. Gountis and A. Bakirtzis, Bidding strategies for electricity producers in a competitive electricity marketplace, IEEE Transactions on Power Systems, 19 (2004), 356-365. doi: 10.1109/TPWRS.2003.821474. [13] R. Jain and J. Walrand, An efficient Nash-implementation mechanism for network resource allocation, Automatica, 46 (2010), 1276-1283. doi: 10.1016/j.automatica.2010.05.013. [14] P. Jia and P. Caines, Analysis of quantized double auctions with application to competitive electricity markets, INFOR: Information Systems and Operational Research, 48 (2010), 239-250. doi: 10.3138/infor.48.4.239. [15] D. S. Kirschen, Demand-side view of electricity markets, IEEE Transactions on Power Systems, 18 (2003), 520-527. doi: 10.1109/TPWRS.2003.810692. [16] V. Krishna, Auction Theory, Academic Press, 2009. [17] A. Lazar and N. Semret, Design and analysis of the progressive second price auction for network bandwidth sharing, Telecommunication Systems, 13, (1999). [18] G. Li and J. Shi, Agent-based modeling for trading wind power with uncertainty in the day-ahead wholesale electricity markets of single-sided auctions, Applied Energy, 99 (2012), 13-22. doi: 10.1016/j.apenergy.2012.04.022. [19] Z. Ma, D. Callaway and I. Hiskens, Decentralized charging control of large populations of plug-in electric vehicles, IEEE Transactions on Control Systems Technology, 21 (2013), 67-78. [20] P. Maillé and B. Tuffin, The progressive second price mechanism in a stochastic environment, Netnomics, 5 (2003), 119-147. [21] O. Marce, H.-H. Tran and B. Tuffin, Double-sided auctions applied to vertical handover for mobility management in wireless networks, Journal of Network and Systems Management, 22 (2014), 658-681. doi: 10.1007/s10922-013-9269-1. [22] S. Nielsen, P. Sorknæs and P. A. Østergaard, Electricity market auction settings in a future danish electricity system with a high penetration of renewable energy sources: A comparison of marginal pricing and pay-as-bid, Energy, 36 (2011), 4434-4444. doi: 10.1016/j.energy.2011.03.079. [23] S. Sethi, H. Yan, J. Yan and H. Zhang, An analysis of staged purchases in deregulated time-sequential electricity markets, Journal of Industrial and Management Optimization, 1 (2005), 443-463. doi: 10.3934/jimo.2005.1.443. [24] A. Singh, Smart grid dynamic pricing, International Journal of Engineering Research and Applications (IJERA), 2 (2012), 705-742. [25] Y. R. Sood, N. P. Padhy and H. Gupta, Wheeling of power under deregulated environment of power system - a bibliographical survey, IEEE Transactions on Power Systems, 17 (2002), 870-878. doi: 10.1109/TPWRS.2002.800967. [26] K. Vitae and L. Lave, Demand response and electricity market efficiency, The Electricity Journal, 20 (2007), 69-85. [27] F. Wen and A. K. David, Optimal bidding strategies and modeling of imperfect information among competitive generators, IEEE Transactions on Power Systems, 16 (2001), 15-21. [28] J. H. Williams and F. Kahrl, Electricity reform and sustainable development in China, Environmental Research Letters, 3 (2008), 044009. doi: 10.1088/1748-9326/3/4/044009. [29] Z. Xu, W. Xu, W. Shao and Z. Zeng, Real-time pricing control on generation-side: Optimal demand-tracking model and information fusion estimation solver, IEEE Transactions on Power Systems, 29 (2014), 1522-1535. doi: 10.1109/TPWRS.2013.2296809. [30] X. Zhao and C. Ma, Deregulation, vertical unbundling and the performance of China's large coal-fired power plants, Energy Economics, 40 (2013), 474-483. doi: 10.1016/j.eneco.2013.08.003. [31] S. Zou, Z. Ma and X. Liu, Auction-based distributed efficient economic operations of microgrid systems, International Journal of Control, 87 (2014), 2446-2462. doi: 10.1080/00207179.2014.926395.
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Subject: Re: [virtio-comment] [PATCH v3 2/3] content: Document balloon feature page poison

• From: David Hildenbrand <david@redhat.com>
• To: Alexander Duyck <alexander.duyck@gmail.com>
• Date: Tue, 26 May 2020 10:24:39 +0200

On 20.05.20 18:25, Alexander Duyck wrote:
> On Wed, May 20, 2020 at 2:28 AM David Hildenbrand <david@redhat.com> wrote:
>>
>> On 20.05.20 04:02, Alexander Duyck wrote:
>>> From: Alexander Duyck <alexander.h.duyck@linux.intel.com>
>>>
>>> Page poison provides a way for the guest to notify the host of the content
>>> expected to be found in pages when they are added back to the guest after
>>> being discarded. The feature currently doesn't apply to the existing
>>> balloon features, however it will apply to an upcoming feature, free page
>>> reporting. Add documentation for the page poison feature describing the
>>> basic functionality and requirements.
>>>
>>
>> I would rephrase this, starting what it does *without* free page
>> reporting (which is not "provides a way for the guest to notify ..."),
>> and then eventually how this feature will also be used in the future as
>> well with free page reporting.
>
> Below is a rewrite on this description. I'm thinking that we can
> probably call out the advantage to free page reporting in a different
> way. Basically with the page poison feature we know a few things about
> the behavior and I have called them out in the new patch description:
>
> Page poison provides a way for the guest to notify the host that it is
> initializing or poisoning freed pages with some specific poison value. As a
> result of this we can infer a couple traits about the guest:
>
> 1. Free pages will contain a specific pattern within the guest.
> 2. Modifying free pages from this value may cause an error in the guest.
> 3. Pages will be immediately written to by the driver when deflated.
>
> There are currently no existing features that make use of this data. In the
> upcoming feature free page reporting we will need to make use of this to
> identify if we can evict pages from the guest without causing data
> corruption.
>
> Add documentation for the page poison feature describing the basic
> functionality and requirements.
>
> [...]
>
>>> diff --git a/content.tex b/content.tex
>>> index 816b6c1b052e..89e9948b7399 100644
>>> --- a/content.tex
>>> +++ b/content.tex
>>> @@ -5026,6 +5026,9 @@ \subsection{Feature bits}\label{sec:Device Types / Memory Balloon Device / Featu
>>>      page hinting. A virtqueue for providing hints as to what memory is
>>>      currently free is present. Configuration field \field{free_page_hint_cmd_id}
>>>      is valid.
>>> +\item[ VIRTIO_BALLOON_F_PAGE_POISON(4) ] The device has to be notified if
>>> +    the driver is expecting balloon pages to contain a certain value when
>>> +    returned. Configuration field poison_val is valid.
>>
>> That's not what it does in the context of this feature only, no?
>>
>> "A hint to the device, that the driver might immediately write
>> \field{poison_val} to pages after deflating them. Configuration field
>> \field{poison_val} is valid."
>
> I'll probably just use this wording with a few slight tweaks. Thinking
> about it though I will get rid of "might" and replace it with "will".

I think that's always guaranteed by Linux as of now, so "will" makes sense.

[...]

>>> +
>>> +If the guest is not initializing or poisoning freed pages it should reject
>>
>> Sometimes you use "write to pages after deflating", here you use "freed
>> pages"
>
> So when I am referencing "freed pages" I am talking about all free
> memory, while when I refer to "pages after deflating" I am talking
> about pages coming out of the balloon.
>
> My thought is that there maybe be additional uses for "poison_val" be
> to feed it into some future use other than just the balloon portion of
> the deflation. Basically what this is telling us is that we could look
> for a pattern of pages containing nothing but poison_val if we wanted
> to do some sort of same page merging, or maybe define something to
> optimize migration by defining a poison page similar to a zero page
> that could be used to reduce migration overhead in the future.
>
>>> +the VIRTIO_BALLOON_F_PAGE_POISON feature.
>>> +
>>> +If VIRTIO_BALLOON_F_PAGE_POISON feature has been negotiated, the guest
>>> +will place the expected poison value into the \field{poison_val}
>>
>> again, "expected" is misleading in the context of this patch only.
>
> I will rewrite this statement at follows:
>   If VIRTIO_BALLOON_F_PAGE_POISON feature has been negotiated, the driver
>   will place the initialization and/or poison value into the \field{poison_val}
>   configuration field data.
>
> I think I might strengthen things a bit as well. In the driver
> normative section I think I might add the following:
>   The driver MUST initialize and/or poison the deflated pages with
>   \field{poison_val} when they are reused by the driver.
>

Maybe simplify that whole "initialize and/or poison " handling across
this patch to "initialize with \field{poison_val}" - if the
initialization is used for poisoning or initialization doesn't matter
from spec POV.

In general looks good to me, I'll have another look at the full result.

Still wondering what to do with free page hinting ... in the meantime
I'll have a look at free page reporting :)

--
Thanks,

David / dhildenb

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# The random variable X is normally distributed with mean [u = 50 and standard devialion o = 7. Find the

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the random variable X is normally distributed with mean [u = 50 and standard devialion o = 7. Find the 97th percentile Assume The 97th percentile is (Round to two decimal places as needed: )

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637 views

The number $(25)_6$, in base $6$ is equivalent to _________ in binary number system.

1. $11001$
2. $10001$
3. $11000$
4. $10000$

### 1 comment

25 base 6 is represented 17 in base 10

So 10001 in base 2

Option B is correct

Option B

$(25)_{6}=2*6^{1}+5*6^{0}=(17)_{10}$

$(17)_{10}=(10001)_{2}$

option (b) is correct.

1 vote


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# Another repertoire method example from Concrete Mathematics

This is my third post on the repertoire method. This method seems quite magical (or perhaps that means I don’t fully understand it 😅). In this case I am referring, as is often the case, to the book Concrete Mathematics by Graham, Patishnick, and Knuth - in particular the example 3, equation 2.40 under “General Methods” for solving a summation. There they present many ways to find the sum of squares $\unicode{9744}_n$ and what I am attempting to cement my knowledge in is the repertoire method approach in the aforementioned example. It’s frightfully clever.

To reproduce it here:

\begin{align} R_0 &= \alpha \\ R_n &= R_{n-1} + \beta + \gamma n + \delta n^2 \ \ \ \text{for} \ n > 0 \\ \end{align}

This will have a solution of the general form

$R_n = A(n)\alpha + B(n)\beta + C(n)\gamma + D(n)\delta$

Conveniently we have already determined $A(n)$, $B(n)$, and $C(n)$ which I wrote about in this blog post. More rigorously we can say that because if we set $\delta$ to $0$ then we end up with what we determined before.

The book then says, rather tersely for me anyway,

If we now plug in $R_n = n^3$, we find that $n^3$ is the solution when $\alpha = 0$, $\beta = 1$, $\gamma = –3$, $\delta = 3$. Hence

$3D(n) - 3C(n) + B(n) = n^3$

The process to arrive at that is pretty much the same as the $n^2$ as discussed here. Which goes something like this:

1 - We start with our recurrence form for which we want to find a closed form $$R_n = R_{n-1} + \beta + \gamma n + \delta n^2 \ \ \ \text{for} \ n > 0$$

As above the closed form will have the below general structure; we need to find $D(n)$ - we already know the rest from before: $$R_n = A(n)\alpha + B(n)\beta + C(n)\gamma + D(n)\delta$$

2a - This is the magic bit: we set the recurrence to $n^3$ and see what $\alpha$, $\beta$, $\gamma$, and $\delta$ need to be in order to make the RHS of the recurrence equal to $n^3$

\begin{align} n^3 &= (n-1)^3 + \alpha + \beta + \gamma n + \delta n^2 \\ &= n^3 + -3n^2 -1 + \alpha + \beta + \gamma n + \delta n^2 \\ \end{align}

2b - Re-arrange things to make the process clearer

\begin{align} n^3 &= n^3 + \delta n^2 - 3n^2 + \gamma n + 3n + \beta -1 + \alpha \\ &= n^3 + n^2(\delta - 3) + n(\gamma + 3) + \beta -1 + \alpha \end{align}

2c - Now we can see what we need to set $\alpha$, $\beta$, $\gamma$, and $\delta$ to in order to make that equation true i.e. $n^3 = n^3$. Specifically $\alpha = 0$, $\beta = 1$, $\gamma = -3$, and $\delta = 3$

3 - Then we substitute (linearly apply?) the found values for $\alpha$, $\beta$, $\gamma$, and $\delta$ into the general form set to n^3.

\begin{align} n^3 = A(n)\cdot0 + B(n)\cdot1 + C(n)\cdot-3 + D(n)\cdot3 \\ n^3 = B(n) - 3C(n) + 3D(n) \\ 3D(n) - 3C(n) + B(n) = n^3 \\ \end{align}

Last re-ordering to match what is in the book at the top of page 45. This will tell us what $D(n)$ needs to be linearly combined with - namely $3$ - for that general form to be $n^3$.

3 - Express in terms of our target - sum of squares $\unicode{9744}_n$. Now that we have built a repertoire let’s see what things have to be set to in order to find $\unicode{9744}_n$. So $\unicode{9744}_n$ equals $\unicode{9744} _{n-1} + n^2$ thus we’ll get $\unicode{9744}_n = R_n$ by setting $\alpha = \beta = \gamma = 0$ and $\delta = 1$. In other words $R_n = A(n)\cdot0 + B(n)\cdot0 + C(n)\cdot0 + D(n)\cdot1$.

So this shows us that $D(n)$ is what we are after - namely - $\unicode{9744}_n$.

4 - Finally substitute in the already known values for $A(n)$, $B(n)$, $C(n)$ and finally solving for $D(n)$.

As mentioned above, surprisingly, solving for $D(n)$ gives us our closed form! Let’s take a look:

\begin{align} 3D(n) &= n^3 + 3C(n) - B(n) \\ 3D(n) &= n^3 - \frac{3(n+1)n}{2} - n \\ 3D(n) &= n(n + 1/2)(n + 1) \\ D(n) &= \frac{n(n + 1/2)(n + 1)}{3} \\ \end{align}

So sometimes, in the process of building a repertoire, one of the items actually ends up being The Solution!

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# Trigonometric summation

Geometry Level 5

$\displaystyle{f^{ n }\left( \theta \right) =\sum _{ k=0 }^{ n }{ { 4 }^{ k }\sec ^{ 2 }{ \left( { 2 }^{ k }\theta \right) } } }$ Define a function $$f$$ as above and if $\displaystyle{f^{ 504 }\left( \cfrac { \pi }{ { 2 }^{ 507 } } \right) \equiv { 2 }^{ a }-\csc ^{ 2 }{ \left( \cfrac { \pi }{ { 2 }^{ b } } \right) } }$ for positive integers $$a,b$$ . Evaluate $$a-b$$.

×


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# Let $a$, $b$, and $c$ be positive real numbers. What is the smallest possible value of…

Let $$a$$, $$b$$, and $$c$$ be positive real numbers. What is the smallest possible value of $$(a+b+c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)$$?

I don't know how to approach this problem, though I think it might use the AM-GM inequality. Can someone please help?

• Hint: The inequality is homogeneous so we may assume $a+b+c=1.$ – Display name May 20 '20 at 0:44

Hint: Do you know why $$( a + b+ c) ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ) \geq 9$$?

Arithmetic Mean - Harmonic Mean.

Hence, conclude that $$2 ( a + b + c) ( \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} ) \geq 9$$

When does equality occur?

• Where does the factor 2 comes from. – hamam_Abdallah May 20 '20 at 0:52
• Because you require the same numbers as there are in the denominator, $(a+b)+(b+c)+(c+a)= 2(a+b+c)$. – MathAnimal May 20 '20 at 1:02
• Alternatively, let $x = a+b, y = b+c, z = c+a$ to see the transformation. – Calvin Lin May 20 '20 at 1:04

AM-GM inequality is a good idea: \begin{align} & (a + b + c)\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c}\right) \\ = & \frac{1}{2}((a + b) + (a + c) + (b + c))\left(\frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c}\right) \\ \geq & \frac{1}{2} \times 3\sqrt[3]{(a + b)(a + c)(b + c)} \times 3\sqrt[3]{\frac{1}{(a + b)(a + c)(b + c)}} \\ = & \frac{9}{2} \end{align}

The equality holds when $$a + b = a + c = b + c$$.

• You can simplify the criterion for equality: $a + b = a + c = b + c \iff a = b = c$. – Clement Yung May 20 '20 at 1:05
• Sure, I intended to leave as it is. – Zhanxiong May 20 '20 at 1:15

Well, by Cauchy-Schwarz, $$((a+b)+(a+c)+(b+c))\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right) \ge (1+1+1)^2=9$$ $$\iff (a+b+c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right) \ge \frac{9}{2}$$


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# Different series than Taylor series for a function

## Main Question or Discussion Point

I am trying to solve an integral that has $\frac{1}{1+x^2}$ as a factor in the integrand. In my book it is claimed that if we use $\displaystyle \frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n \frac{1}{x^{2n+2}}$ the problem can be solved immediately. But, I am confused as to where this series representation comes from... It certainly has nothing to do with Taylor series it seems.

haushofer
That doesn't seem right to me.

jasonRF
Gold Member
The expression the OP gave is valid for $|x|>1$. It is the generalized binomial as mathman said, although I think the equation he posted isn't quite correct.

mathwonk
Homework Helper
have you ever done any complex analysis? in that setting we often find it convenient to complete the complex plane by adding in a point at infinity, and then one can make a taylor expansion centered at infinity.

in your case set w = 1/x and then 1/(1+x^2) becomes w^2/(1+w^2) which has a taylor expansion centered at w= 0, or x = infinity:

namely 1/(1+x^2) = w^2/(1+w^2) = (geometric series) w^2 (1 -w^2 + w^4 ±.....) = (1/x^2)(1 - 1/x^2 + 1/x^4 ± .....) = 1/x^2 -1/x^4 + 1/x^6 ±.......

(in particular mathman's equation seems off by a factor of 1/x^2.)

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haushofer
The expression the OP gave is valid for $|x|>1$. It is the generalized binomial as mathman said, although I think the equation he posted isn't quite correct.
Ah, yes, I implicitly assumed the integration interval is between 0 and 1, but the exercise is probably an integral which has an integration interval outside of this. My bad.

FactChecker
Gold Member
One can obtain this series from the Taylor series $\frac 1 {1-r} = 1+r+r^2+r^3+\cdots$, which is valid for $|r|<1$.
$$\frac 1 {1+x^2} = x^{-2} \frac 1 {x^{-2}+1} = x^{-2} \frac 1 {1 - (-x^{-2})}$$
$$= x^{-2} (1+(-x^{-2})+(-x^{-2})^2+(-x^{-2})^3+(-x^{-2})^4+\cdots)$$ which is valid for $|x^{-2}|<1$.

If you study power series and analytic functions, the series $\frac 1 {1-r} = 1+r+r^2+r^3+\cdots$, $|r|<1$ will be fundamental.

mathwonk
Homework Helper
that is of course exactly what i said above in #5, but maybe looks more elementary by not mentioning the point at infinity where x^(-2) = 0. note too that the series is also valid at x = infinity since both sides equal zero. indeed this is the reason the expansion works.

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have you ever done any complex analysis? in that setting we often find it convenient to complete the complex plane by adding in a point at infinity, and then one can make a taylor expansion centered at infinity.

in your case set w = 1/x and then 1/(1+x^2) becomes w^2/(1+w^2) which has a taylor expansion centered at w= 0, or x = infinity:

namely 1/(1+x^2) = w^2/(1+w^2) = (geometric series) w^2 (1 -w^2 + w^4 ±.....) = (1/x^2)(1 - 1/x^2 + 1/x^4 ± .....) = 1/x^2 -1/x^4 + 1/x^6 ±.......

(in particular mathman's equation seems off by a factor of 1/x^2.)
I have not studied complex analysis yet. But, does this have anything to do with what's called a Laurent series? I'm asking because on WolframAlpha, when I input $\frac{1}{1+x^2}$ it gives the Maclaurin series expansion at 0 and also gives $\displaystyle \sum_{n=0}^{\infty} (-1)^n \frac{1}{x^{2n+2}}$ and calls it the Laurent series expansion at $\infty$.

FactChecker
Yes. I hadn't noticed that you had already given the same answer till later. When I realized that, I decided not to delete my post because I think that the expansion of $\frac 1 {1-z}$ is so important and I wanted to state that.


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# Softmax Regression

(Difference between revisions)
 Revision as of 23:24, 7 May 2011 (view source)Zellyn (Talk | contribs)m (→Parameterization)← Older edit Revision as of 23:26, 7 May 2011 (view source)Zellyn (Talk | contribs) m (→Binary Logistic Regression)Newer edit → Line 190: Line 190: &= &= - \frac{e^{ \Theta_1^T x^{(1)} } }{ 1 + e^{ \Theta_1^T x^{(i)} } } + \frac{e^{ \Theta_1^T x^{(i)} } }{ 1 + e^{ \Theta_1^T x^{(i)} } } \cdot \cdot - \frac{1}{e^{ \Theta_1^T x^{(1)} } } + \frac{1}{e^{ \Theta_1^T x^{(i)} } } \begin{bmatrix} \begin{bmatrix} e^{ \Theta_1^T x^{(i)} } \\ e^{ \Theta_1^T x^{(i)} } \\

## Introduction

Softmax regression, also known as multinomial logistic regression, is a generalisation of logistic regression to problems where there are more than 2 class labels.

Recall that in logistic regression, our hypothesis was of the form:

\begin{align} h_\theta(x) = \frac{1}{1+\exp(-\theta^Tx)}, \end{align}

where we trained the logistic regression weights to optimize the (conditional) log-likelihood of the dataset using p(y | x) = hθ(x). In softmax regression, we are interested in multi-class problems where each example (input image) is assigned to one of k labels. One example of a multi-class classification problem would be classifying digits on the MNIST dataset where each example has label 1 of 10 possible labels (i.e., where it is the digit 0, 1, ... or 9).

To extend the logistic regression framework which only outputs a single probability value, we consider a hypothesis that outputs K values (summing to 1) that represent the predicted probability distribution. Formally, let us consider the classification problem where we have m k-dimensional inputs $x^{(1)}, x^{(2)}, \ldots, x^{(m)}$ with corresponding class labels $y^{(1)}, y^{(2)}, \ldots, y^{(m)}$, where $y^{(i)} \in \{1, 2, \ldots, n\}$, with n being the number of classes.

Our hypothesis hθ(x), returns a vector of probabilities, such that

\begin{align} h(x^{(i)}) = \begin{bmatrix} P(y^{(i)} = 1 | x^{(i)}) \\ P(y^{(i)} = 2 | x^{(i)}) \\ \vdots \\ P(y^{(i)} = n | x^{(i)}) \end{bmatrix} = \frac{1}{ \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} } \begin{bmatrix} e^{ \theta_1^T x^{(i)} } \\ e^{ \theta_2^T x^{(i)} } \\ \vdots \\ e^{ \theta_n^T x^{(i)} } \\ \end{bmatrix} \end{align}

where $\theta_1, \theta_2, \ldots, \theta_n$ are each k-dimensional column vectors that constitute the parameters of our hypothesis. Notice that $\frac{1}{ \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} }$ normalizes the distribution so that it sums to one.

Strictly speaking, we only need n − 1 parameters for n classes, but for convenience, we use n parameters in our derivation.

Now, this hypothesis defines a predicted probability distribution given some x, P(y | x(i)) = h(x(i)). Thus to train the model, a natural choice is to maximize the (conditional) log-likelihood of the data, $l(\theta; x, y) = \sum_{i=1}^{m} \ln { P(y^{(i)} | x^{(i)}) }$.

Motivation: One reason for selecting this form of hypotheses comes from connections to linear discriminant analysis. In particular, if one assumes a generative model for the data in the form $p(x,y) = p(y) \times p(x | y)$ and selects for p(x | y) a member of the exponential family (which includes Gaussians, Poissons, etc.) it is possible to show that the conditional probability p(y | x) has the same form as our chosen hypotheses h(x). For more details, see the CS 229 Lecture 2 Notes.

## Optimizing Softmax Regression

Expanding the log-likelihood expression, we find that:

\begin{align} \ell(\theta) &= \ln L(\theta; x, y) \\ &= \ln \prod_{i=1}^{m}{ P(y^{(i)} | x^{(i)}) } \\ &= \sum_{i=1}^{m}{ \ln \frac{ e^{ \theta^T_{y^{(i)}} x^{(i)} } }{ \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} } } \\ &= \theta^T_{y^{(i)}} x^{(i)} - \ln \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} \end{align}

Unfortunately, there is no closed form solution to this optimization problem (although it is concave), and we usually use an off-the-shelf optimization method (e.g., L-BFGS, stochastic gradient descent) to find the optimal parameters. Using these optimization methods require computing the gradient ($\ell(\theta)$ w.r.t. θk), which can can be derived as follows:

\begin{align} \frac{\partial \ell(\theta)}{\partial \theta_k} &= \frac{\partial}{\partial \theta_k} \theta^T_{y^{(i)}} x^{(i)} - \ln \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} \\ &= I_{ \{ y^{(i)} = k\} } x^{(i)} - \frac{1}{ \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} } \cdot e^{ \theta_k^T x^{(i)} } \cdot x^{(i)} \qquad \text{(where } I_{ \{ y^{(i)} = k\} } \text{is 1 when } y^{(i)} = k \text{ and 0 otherwise) } \\ &= x^{(i)} ( I_{ \{ y^{(i)} = k\} } - P(y^{(i)} = k | x^{(i)}) ) \end{align}

With this, we can now find a set of parameters that maximizes $\ell(\theta)$, for instance by using L-BFGS with minFunc.

### Weight Regularization

When using softmax regression in practice, it is important to use weight regularization. In particular, if there exists a linear separator that perfectly classifies all the data points, then the softmax-objective is unbounded (given any θ that separates the data perfectly, one can always scale θ to be larger and obtain a better objective value). With weight regularization, one penalizes the weights for being large and thus avoids these degenerate situations.

Weight regularization is also important as it often results in models that generalize better. In particular, one can view weight regularization as placing a (Gaussian) prior on θ so as to prefer θ with smaller values.

In practice, we often use a L2 weight regularization on the weights where we penalize the squared value of each element of θ. Formally, we use:

\begin{align} w(\theta) = \frac{\lambda}{2} \sum_{i}{ \sum_{j}{ \theta_{ij}^2 } } \end{align}

This regularization term is added together with the log-likelihood function to give a cost function, J(θ), which we want to minimize (note that we want to minimize the negative log-likelihood, which corresponds to maximizing the log-likelihood):

\begin{align} J(\theta) = -\ell(\theta) + \frac{\lambda}{2} \sum_{i}{ \sum_{j}{ \theta_{ij}^2 } } \end{align}

The gradients with respect to the cost function must then be adjusted to account for the weight decay term:

\begin{align} \frac{\partial J(\theta)}{\partial \theta_k} &= x^{(i)} ( I_{ \{ y^{(i)} = k\} } - P(y^{(i)} = k | x^{(i)}) ) + \lambda \theta_k \end{align}

Minimizing J(θ) now performs regularized softmax regression.

## Parameterization

We noted earlier that we actually only need n − 1 parameters to model n classes. To see why this is so, consider our hypothesis again:

\begin{align} h(x^{(i)}) &= \frac{1}{ \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} } \begin{bmatrix} e^{ \theta_1^T x^{(i)} } \\ e^{ \theta_2^T x^{(i)} } \\ \vdots \\ e^{ \theta_n^T x^{(i)} } \\ \end{bmatrix} \\ &= \frac{e^{ \theta_n^T x^{(i)} } }{ \sum_{j=1}^{n}{e^{ \theta_j^T x^{(i)} }} } \cdot \frac{1}{e^{ \theta_n^T x^{(i)} } } \begin{bmatrix} e^{ \theta_1^T x^{(i)} } \\ e^{ \theta_2^T x^{(i)} } \\ \vdots \\ e^{ \theta_n^T x^{(i)} } \\ \end{bmatrix} \\ &= \frac{1}{ \sum_{j=1}^{n}{e^{ (\theta_j^T - \theta_n^T) x^{(i)} }} } \begin{bmatrix} e^{ (\theta_1^T - \theta_n^T) x^{(i)} } \\ e^{ (\theta_2^T - \theta_n^T) x^{(i)} } \\ \vdots \\ e^{ (\theta_n^T - \theta_n^T) x^{(i)} } \\ \end{bmatrix} \\ \end{align}

Letting Θj = θj − θn for $j = 1, 2 \ldots n - 1$ gives

\begin{align} h(x^{(i)}) &= \frac{1}{ 1 + \sum_{j=1}^{n-1}{e^{ \Theta_j^T x^{(i)} }} } \begin{bmatrix} e^{ \Theta_1^T x^{(i)} } \\ e^{ \Theta_2^T x^{(i)} } \\ \vdots \\ 1 \\ \end{bmatrix} \\ \end{align}

Showing that only n − 1 parameters are required.

In practice, however, it is often easier to implement the version which is over-parametrized although both methods will lead to the same classifier.

### Binary Logistic Regression

In the special case where n = 2, one can also show that softmax regression reduces to logistic regression:

\begin{align} h(x^{(i)}) &= \frac{1}{ 1 + e^{ \Theta_1^T x^{(i)} } } \begin{bmatrix} e^{ \Theta_1^T x^{(i)} } \\ 1 \\ \end{bmatrix} \\ &= \frac{e^{ \Theta_1^T x^{(i)} } }{ 1 + e^{ \Theta_1^T x^{(i)} } } \cdot \frac{1}{e^{ \Theta_1^T x^{(i)} } } \begin{bmatrix} e^{ \Theta_1^T x^{(i)} } \\ 1 \\ \end{bmatrix} \\ &= \frac{1}{ e^{ -\Theta_1^T x^{(i)} } + 1 } \begin{bmatrix} 1 \\ e^{ -\Theta_1^T x^{(i)} } \\ \end{bmatrix} \\ \end{align}


text:
gaussmixt {condmixt} R Documentation

## Mixture of Gaussians

### Description

Density and distribution function for a mixture of Gaussians with m components.

### Usage

dgaussmixt(params, x, log = FALSE, trunc = TRUE)
pgaussmixt(params, x, trunc = TRUE)

### Arguments

 params matrix of mixture parameters of dimension 3 x m, where m is the number of components, so that each column contains the mixture parameters (pi, mu, sigma) related to one component x vector of sample data log logical, if TRUE, probabilities p are given as log(p). trunc logical, if TRUE (default), the Gaussian density is truncated below zero.

### Value

dgaussmixt gives the density and pgaussmixt gives the distribution function

Julie Carreau

### See Also

Normal

[Package condmixt version 1.1 Index]


text:
Physics Help Forum mass-spring prpblem

 General Physics General Physics Help Forum

 Aug 30th 2017, 09:09 AM #1 Junior Member   Join Date: Aug 2017 Posts: 8 mass-spring prpblem the problem: Suppose at time zero, the bob was drawn upward four units from the equilibrium position, let C=2, K=2, m=1 lbm, initial speed=2 unit/sec find an expression for body's position. and in the solution it says: y''+6y'+5y=0 my question is: from where does the numbers (6) and (5) that is in te above DE came from ? 2. Relevant equations spring equation: my''+cy'+ky=f(t) consider f(t) =0 3. The attempt at a solution stopped at first because of the first step Last edited by aows61; Aug 30th 2017 at 01:31 PM. Reason: correction
 Aug 30th 2017, 12:25 PM #2 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 552 Have you got some kind of friction in the problem? A spring ODE doesn't usually have a velocity term unless some kind of dampening is involved. aows61 likes this. __________________ Burn those raisin muffins. Burn 'em all I say.
Aug 30th 2017, 12:28 PM   #3
Junior Member

Join Date: Aug 2017
Posts: 8
spring-mass

 Originally Posted by kiwiheretic Have you got some kind of friction in the problem? A spring ODE doesn't usually have a velocity term unless some kind of dampening is involved.
this is all the information that i have about the problem:
and here is a link of you want to check:
>>>https://i.imgur.com/xjqAOUY.jpg

Aug 30th 2017, 01:18 PM   #4
Senior Member

Join Date: Apr 2015
Location: Somerset, England
Posts: 1,035
 Originally Posted by aows61 this is all the information that i have about the problem: and here is a link of you want to check: >>>https://i.imgur.com/xjqAOUY.jpg

y' is the velocity and you say that at t=0 the velocity = +2.

This is a differential equation you can solve for the boundary condition.

But what I don't know is what your sign convention is.

Which way is + and which way is -?

 Aug 30th 2017, 01:29 PM #5 Junior Member   Join Date: Aug 2017 Posts: 8 here is the attachment, just press on this link>>>>> https://imgur.com/xjqAOUY
 Aug 30th 2017, 01:30 PM #6 Junior Member   Join Date: Aug 2017 Posts: 8 for the sign convention: upward is negative and downward is positive
 Aug 30th 2017, 01:32 PM #7 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 552 I think he meant like this (reattached) Attached Thumbnails   aows61 likes this. __________________ Burn those raisin muffins. Burn 'em all I say.
 Aug 30th 2017, 01:33 PM #8 Junior Member   Join Date: Aug 2017 Posts: 8 THanks indeed @KIWIHERETIC , appreciate your help,
 Aug 30th 2017, 01:46 PM #9 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 I think the issue is either (a) the answer is incorrect in that it doesn't agree with the values of c and k that are given, or (b) there are units at play here that have not been properly defined. For example, they say m = 1 lb_m, so the first term of the DE must be divided by 32.2 lbm-ft/lbf-s^2 in order to get units of force in lbf for a given acceleration in ft/s^2. Unfortunately we are not told what the units of c and k are. If we assume that c is in lbf/(ft/s) and k is in lbf/ft, then the DE would be: $\displaystyle \frac 1 {32.2} y'' + 2 y' +2y = 0$ But the problem states that the initial displacement is "4 units," not 4 ft, and initial velocity is "2 units per second," not 2 ft per second, so I'm guessing that somewhere in the original problem statement they define what they mean by a distance of 1 "unit." They should also be clear about what the units are for both C and K. aows61 likes this. Last edited by ChipB; Aug 30th 2017 at 02:14 PM.
Aug 30th 2017, 01:54 PM   #10
Junior Member

Join Date: Aug 2017
Posts: 8
thanks

 Originally Posted by ChipB I think the issue is either (a) the answer is incorrect in that it doesn't agree with the values of c and k that are given, or (b) there are units at play here that have not been properly defined. For example, they say m = 1 lb_m, so the first term of the DE must be divided by 32.2 lbm-ft/lbf-s^2 in order to get units of force in lbf for a given acceleration in ft/s^2. Unfortunately we are not told what the units of c and k are. If we assume that c is in lbf/(ft/s) and k is in lbf/ft, then the DE would be: $\displaystyle \frac 1 {32.2} y'' + 2 y' +2y = 0$ But the problem states that the initial displacement is "4 units," not 4 ft, and initial velocity is "2 units per second," not 2 ft per second, so I'm guessing that somewhere in the original problem statement there were units defined for m, c and k that we don't know about.
i also guess that there is some mistake within the solution.
even though there is some unit change it will not produce any 5 or 6, right ?
i solved it and the roots of the characteristic equation were complex (r1=-1-i, and r2=-1+i)

 Tags massspring, prpblem

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text:
## Calculus: Early Transcendentals 8th Edition

$V=\frac{64 \pi }{15}$
{Step 1 of 5}: Find, find the points of intersection of $y=2x$ and $y=x^{2}$ $x^{2}=2x$ $x^{2}-2x=0$ $x \left( x-2 \right) =0$ Or $x=0$ or $x=2$ The points of Intersection are (0, 0) and (2, 4). {Step 2 of 5}: Sketch the curves. {Step 3 of 5}: Rotate the shaded region about the x-axis. Sketch the solid obtained by rotating the shaded region about the x-axis and the washer. {Step 4 of 5}: The outer radius of the washer is 2x The inner radius of the washer is x2 The cross-sectional area of the washer $A \left( x \right) = \pi \left[ \left( 2x \right) ^{2}- \left( x^{2} \right) ^{2} \right]$ $= \pi \left[ 4x^{2}-x^{4} \right]$ {Step 5 of 5}: The volume of the solid obtained by rotating the shaded region about the x-axis $V= \int _{0}^{2}A \left( x \right) dx$ $= \int _{0}^{2} \pi \left[ 4x^{2}-x^{4} \right] dx$ $= \pi \int _{0}^{2} \left( 4x^{2}-x^{4} \right) dx$ $= \pi \left[ \frac{4}{3}x^{3}-\frac{1}{5}x^{5} \right] _{0}^{2}$ [By the fundamental theorem of calculus, part 2] $= \pi \left[ \frac{32}{3}-\frac{32}{5} \right]$ $V=\frac{64 \pi }{15}$


text:
Puppy Linux Discussion Forum
"THE" alternative forum : puppylinux.info

 The time now is Wed 21 Feb 2018, 21:36 All times are UTC - 4
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Author Message
wert

Joined: 31 Mar 2015
Posts: 207
Location: USA

 Posted: Fri 29 Sep 2017, 12:03    Post subject:  How do i succesfully install an individual .deb package?

Hi! i'm not very experienced with linux. I have tried installing the following app but it doesn't load
Overgrive. The package is overgrive_3.2.3_all.deb but when I try opening the app, its returns the following
 Code: root# ./overgrive ./overgrive: line 1: $'\003\363\r': command not found ./overgrive: line 2:$'^\277VWc\b@s\356\031dd\001lZdd\001l\001Z\001dd\001l\002Z\002xKe\003e\004e\001j\005\203\001\203\001D]4Z\006e\001j\005e\006\031d\002d\003d\004d\005f\004k\006rhd\006Z\an\006d\aZ\aq:We\ad\ak\002r\r\001dd\001l\bZ\be\bj': command not found ./overgrive: line 3: $'d\v\203\002\001dd\fl': command not found ./overgrive: line 4:$'m\vZ\v\001dd\rl': command not found ./overgrive: line 5: $'m\fZ\f\001dd\016l': command not found ./overgrive: line 6:$'m\rZ\r\001dd\017l': command not found ./overgrive: line 7: $'m\016Z\017\001dd\020l': command not found ./overgrive: line 8:$'m\020Z\020\001dd\021l': command not found ./overgrive: line 9: syntax error near unexpected token Z' ./overgrive: line 9: mZnddlZddlZddlmZddlmZddlmZddlmZddlmddlmZddlZddlm Z ddlm!Z!ddl"Z"ddl#m#Z#dd#m$Z$ddl%Z%ddl&Z&ddl'Z'ddl(Z(ddl)Z)ddl*Z*ddl+Z+ddl,Z,ddl-Z-ddl.Z.ddl/Z/ddl0Z0ddl1Z1ddl2Z2ddl3Z3ddl4Z4ddl5Z5ddl6Z6ddl7Z7ddl8Z8ddl9Z9dZ:dZ;de+je?d!Z@d"ZAd#ZBd\$ZCd%ZDd&ZEd'ZFd(ZGd)ZHd*ZId+ZJe?d,ZKd-ZLe?e?d.ZMd/ZNd0ZOd1ZPd2ZQd3ZRd4ZSd5ZTd6ZUd7ZVd8ZWd9ZXd:ZYd;ZZdZ]d?Z^d@Z_dAZdBZadCZbdDZcdEZddFZedGZfe?dHZgdIZhdJZidKZjdLZkdMZldNZmdOZndPZoe?dQZpdRZqdSZrdTZsdUZtdVZudWZvdXZwdYZxdZZyd[Zzd\Z{d]Z|d^Z}d_Z~dZdaZdbZdcZddZdededfZdgefdhYZdiefdjYZdkefdlYZdmefdnYZdoZdpZdqZdrZdsZdtZduZdvZe?dwZdxZdyZdzZd{Ze?d|Zd}Zd~ZdefdYZedZdZdZe?dZ e?dZ¡e?dZ¢dZ£e?e?dZ¤ee?dZ¥dZ¦dZ§dZ¨dZ©dZªdZ«dZ¬dZdZ®dZ¯dZ°dZ±dZ²ee?dZ³ee?dZ´ee?d^[[?1;2cZµdZ¶da·dZ¸dddZ¹d Zºd¡Z»d¢Z¼d£Z½d¤Z¾d¥Z¿d¦ZÀd§ZÁd¨ZÂd©eÒZÑnXeÑrejÓd±neÍjÔd²ejՃeÍjփejËjÌeÆd³dd´Z×daØeÒaÙeÒaÚgaÛgaÜiaÝeÒaÞe#j߃ZàeàjádµZâejËjÌeÆd¶Zãe½eãeRd·eRd¸eRd¹eRdºeRd»eRd¼eRd½eÇeRdeRd¾eRd¿eRdejZäejåjædÀZçeRdÁeäeRdÂeèeçedkrô' root# 1;2c
. I've found that installing things via ppm is better since the software ultimately works when installed through ppm even though some may be outdated. This is because PPM detects all dependancies and installs them automatically, But however overgrive isn't in ppm. Can I add it's repository in ppm? If so how? Thanks in advance for any help.

My specs:
-Version-
Kernel : Linux 3.14.20 (i686)
Version : #1 SMP Thu Jan 8 13:19:06 GMT 2015
C Library : GNU C Library version 2.19 (stable)
Distribution : tahrpup - 6.0.2

Desktop Environment : Unknown (Window Manager: JWM)
dancytron

Joined: 18 Jul 2012
Posts: 1015

 Posted: Fri 29 Sep 2017, 15:17    Post subject:

It seems to be a python application.

The .desktop file located in usr/share/applications has this as it's exec line.

 Quote: Exec=python2 /opt/thefanclub/overgrive/overgrive

So the first thing to try would be typing "python2 /opt/thefanclub/overgrive/overgrive" in the console and see what you get.

The second thing to try would be to install python from the PPM and then try again.

Python applications can be problematic in Puppy, sometimes they work and sometimes they don't.

The README.debian file in /usr/share/docs seems to indicate the same thing.

anikin

Joined: 10 May 2012
Posts: 982

 Posted: Fri 29 Sep 2017, 16:09    Post subject:

 Quote: How do i succesfully install an individual .deb package?

Have a look here:
http://murga-linux.com/puppy/viewtopic.php?p=956596#956596
bigpup

Joined: 11 Oct 2009
Posts: 10054
Location: Charleston S.C. USA

 Posted: Fri 29 Sep 2017, 16:49    Post subject:

Also, you need to understand that Puppy is not 100% a standard Linux operating system.
It does have it's own way of doing things and it's own specific internal core programs.
Those may not be compatible with just any deb package.

Linux operating systems do require programs to be compiled specifically for that system.
Even to the point of what version of the OPS.

PPM(Puppy Package Manager) keeps this all sorted out.

Using something else, requires some added work.

Sometimes it just works.
Other times, you see what happens and adjust.

Always, always run the update database to PPM and then search for programs.
PPM>Configure>Update Database

_________________
I have found, in trying to help people, that the things they do not tell you, are usually the clue to solving the problem.
When I was a kid I wanted to be older.... This is not what I expected
bigpup

Joined: 11 Oct 2009
Posts: 10054
Location: Charleston S.C. USA

 Posted: Fri 29 Sep 2017, 16:54    Post subject:

 Quote: Distribution : tahrpup - 6.0.2

I hope you have used the Quickpet>Tahrpup updates.

Tahrpup 6.0.2 was a bug fix testing version and Tahrpup is now up to V6.0.5.

_________________
I have found, in trying to help people, that the things they do not tell you, are usually the clue to solving the problem.
When I was a kid I wanted to be older.... This is not what I expected
wert

Joined: 31 Mar 2015
Posts: 207
Location: USA

 Posted: Fri 29 Sep 2017, 17:31    Post subject:

bigpup wrote:
 Quote: Distribution : tahrpup - 6.0.2

I hope you have used the Quickpet>Tahrpup updates.

Tahrpup 6.0.2 was a bug fix testing version and Tahrpup is now up to V6.0.5.
Yes I have dude. And thanks alot for ur details. BTW why doesn't the Quickpet>Tahrpup updates also update the ppm database? Or is ppm not part of tahrpup project?
puppyluvr

Joined: 06 Jan 2008
Posts: 3460
Location: Chickasha Oklahoma

 Posted: Fri 29 Sep 2017, 18:03    Post subject:

Hello,
The PPM points to a static repo, which isn't updated, to prevent conflict.
Quickpet is an evolving repo updated by devs when new things are approved...
The Ubuntu repos the PPM uses are updatable, and you can point it at new repos as well, but it requires editing a text file manually...

_________________
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wert

Joined: 31 Mar 2015
Posts: 207
Location: USA

 Posted: Sat 14 Oct 2017, 18:04    Post subject:

 puppyluvr wrote: Hello, The PPM points to a static repo, which isn't updated, to prevent conflict. Quickpet is an evolving repo updated by devs when new things are approved... The Ubuntu repos the PPM uses are updatable, and you can point it at new repos as well, but it requires editing a text file manually...

Nice info man. Thanks
wert

Joined: 31 Mar 2015
Posts: 207
Location: USA

 Posted: Sat 14 Oct 2017, 18:28    Post subject:

 bigpup wrote: Always, always run the update database to PPM and then search for programs. PPM>Configure>Update Database

I'm afraid updating database to PPM is not gonna work with me. It's drudgery, Plus I can't have that time. How does someone keep clicking the dialogue boxes constantly like do you actually do that? Or is there a tweak to make it update all at once without bringing an awfully long series of alert boxes that I'm to point and click serially?

BTW would updating the database to PPM show me Ubuntu 16.04 LTS programs or do newer ubuntu releases post-14.0 require me to edit the text file you mentioned earlier first. Thanks
rockedge

Joined: 11 Apr 2012
Posts: 464
Location: Connecticut, United States

 Posted: Sat 14 Oct 2017, 20:28    Post subject:

you only need to update the ppm occasionally. It takes about 5 minutes total time. If 5 minutes and hitting enter about 7 times is drudgery your time must be extremely valuable. Crack open some documents and teach yourself something about Linux...I don't share hard won knowledge to save some punk ass kid some time....dude.
wert

Joined: 31 Mar 2015
Posts: 207
Location: USA

 Posted: Sat 11 Nov 2017, 05:32    Post subject:

 rockedge wrote: you only need to update the ppm occasionally. It takes about 5 minutes total time. If 5 minutes and hitting enter about 7 times is drudgery your time must be extremely valuable. Crack open some documents and teach yourself something about Linux...I don't share hard won knowledge to save some punk ass kid some time....dude.

Why not just automate the thing instead of making someone press enter. Infact why not integrate in into tahr updates? And why not make tahr updates run on schedule? It seems pretty easy to preconfigure the OS with those things.

It's 2017 not the 90's. A linux OS of today needs to be automated sufficiently before releasing it. I mean, nowadays you want people to be doing stuff like that with their time?

I can't afford time to teach myself an OS. I already spent years learning windows in school. I'm a student teaching myself further things. All I need is to start my pc and get to business. Speaking of which I'm business man too. You can't expect me to do the maintenance work and not compromise my agenda. That's why there are handy men. I don't aim to be one
musher0

Joined: 04 Jan 2009
Posts: 11694

 Posted: Sat 11 Nov 2017, 06:11    Post subject:

Hi gang.

Meanwhile, musher0 left the grounds where the controversy was flaming to do
some useful research...

Here: https://www.thefanclub.co.za/overgrive/installation-instructions-ubuntu
 Quote: Ubuntu 12.04 (32bit / 64bit) or later installed. The software should work on most Debian based systems if they support Gtk3.
(Bold, etc., by me.)

"Voilà pourquoi votre fille est muette" (famous sentence by Molière) !
Transl.: "That is why your daughter is a mute." Ok. Not funny. Anyway...

Puppy at this time has only partial support for GTK3, namely for gnumeric. Or in a
few cases "compiled in our kennels" , the dev has compiled the needed GTK3
components along with the app.

Also, as dancytron has mentioned, Puppy has only partial support for python. I tried
the overgrive on my machine and a message said that python2 was missing.

Puppy has:
 Quote: [/usr/lib]>lg -d pyth* drwxr-xr-x 37 2048 nov 5 22:20 python2.7/ drwxr-xr-x 32 3648 oct 20 19:47 python3.5/
but not the true python 2, it seems.

Sorry wert, you're out of luck. Any other app can do what overgrive does? Maybe
a GTK2 one, based on perl?

IHTH

_________________
musher0
~~~~~~~~~~
"Logical entities must not be multiplied beyond necessity." | |
« Il ne faut pas multiplier les entités logiques sans nécessité. » (Ockham)
bigpup

Joined: 11 Oct 2009
Posts: 10054
Location: Charleston S.C. USA

 Posted: Sat 11 Nov 2017, 06:18    Post subject:

 Quote: I can't afford time to teach myself an OS. I already spent years learning windows in school.

So, how did you learn Windows 10?
No way are you going to tell me Windows 10 is just like Windows XP, 7 or even 8.

Puppy does some things that are common, but others are Puppy way.
Also, the Linux way.

Next time you buy a new car, ask them why it is not setup to work the same way the last car you had.
Why did you put that switch, there, when it was over here, in my last car?

In PPM(Puppy Package Manager)
configure>Options
un-check Use verbose method during update of package databases.
That not being checked should make PPM auto update everything, when you select update database>update now.

_________________
I have found, in trying to help people, that the things they do not tell you, are usually the clue to solving the problem.
When I was a kid I wanted to be older.... This is not what I expected
musher0

Joined: 04 Jan 2009
Posts: 11694

 Posted: Sat 11 Nov 2017, 06:56    Post subject:

Option A: use the PPM to locate and install the file.

Option B:
-- download the deb archive from pkgs.org or an ubuntu repo to a convenient and
____ safe place
-- open a ROX window in that "place"
-- double-click on the deb archive with your mouse and it will install itself.

Option C:
-- open a terminal in that directory (aka "place")
-- type
 Code: petget FileNameOf.deb < Enter >
and it will install itself.

Do we have another question related to the subject at hand?

BFN.

_________________
musher0
~~~~~~~~~~
"Logical entities must not be multiplied beyond necessity." | |
« Il ne faut pas multiplier les entités logiques sans nécessité. » (Ockham)

Last edited by musher0 on Sat 11 Nov 2017, 07:11; edited 1 time in total
musher0

Joined: 04 Jan 2009
Posts: 11694

 Posted: Sat 11 Nov 2017, 07:09    Post subject:

wert wrote:
 rockedge wrote: you only need to update the ppm occasionally. It takes about 5 minutes total time. If 5 minutes and hitting enter about 7 times is drudgery your time must be extremely valuable. Crack open some documents and teach yourself something about Linux...I don't share hard won knowledge to save some punk ass kid some time....dude.
(...)
Speaking of which I'm [a] business man too. You can't expect me to do the maintenance work and not compromise my agenda. That's why there are handy men. I don't aim to be one

With respect, Sir Not-Aiming-to-Be-a-Handyman,

if you don't have time to learn new stuff, why don't you stay with the devil you
know -- and not bother us?

Best regards.

~~~~~~~~
PS. As for me, I am proud to have become a Linux / Puppy "handyman".
It was worth 100 times the effort.

_________________
musher0
~~~~~~~~~~
"Logical entities must not be multiplied beyond necessity." | |
« Il ne faut pas multiplier les entités logiques sans nécessité. » (Ockham)
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text:
Retrieval Practice with Kindle: Feel the Learn

I use Amazon’s free Kindle Reader for all of my (online and offline) book reading, except for any book that I really want that just can’t be had digitally. Besides notes and highlights, the Reader has a nifty little Flashcards feature that works really well for retrieval practice. Here’s how I do retrieval practice with Kindle.

Step 1: Construct the Empty Flashcard Decks

Currently I’m working through Sarah Guido and Andreas Müller’s book Introduction to Machine Learning with Python. I skimmed the chapters before starting and decided that the authors’ breakdown by chapter was pretty good—not too long and not too short. So, I made a flashcard deck for each chapter in the book, as shown at the right. On your Kindle Reader, click on the stacked cards icon. Then click on the large + sign next to “Flashcards” to create and name each new deck.

Depending on your situation, you may not have a choice in how you break things down. But I think it’s good advice to set up the decks—however far in advance you want—before you start reading.

So, if I were assigned to read the first half of Chapter 2 for a class, I would create a flashcard deck for the first half of Chapter 2 before I started reading. And, although I didn’t set titles in this example, it’s probably a good idea to give the flashcard deck a title related to what it’s about (e.g., Supervised Learning).

You still need to read and comprehend the content. Retrieval practice adds, it doesn’t replace. So, I read and highlight and write notes like I normally would. I don’t worry at this point about the flashcards, about what is important or not. I just read for the pleasure of finding things out. I highlight things that strike me as especially interesting and write notes with questions, or comments I want to make on the text.

Read a section of the content represented by one flashcard deck. Since I divided my decks by chapter, I read the first chapter straight through, highlighting and making notes as I went.

The reading doesn’t have to be done in one sitting. The important thing is to just focus on reading one section before moving on to the next step.

Step 3: Create the Fronts for the Flashcards

Now, go through the content of your first section of reading and identify important concepts, items worth remembering, things you want to be able to produce. You’ll want to add these as prompts on your flashcards. You don’t necessarily have to write these all down in a list. You can enter a prompt on a flashcard, return to the text for another prompt, enter a prompt on another flashcard, and on and on.

Screenshot 1

Screenshot 2

Screenshot 3

When you have at least one prompt, click on the flashcard deck and then click on Add a Card (Screenshot 1) and enter the prompt.

Enter the prompt at the top. (Screenshot 2) This will be the front of the flashcard you will see when testing yourself. Leave the back blank for the moment. Click Save and Add Another Card at the bottom right to repeat this with more prompts.

When you are finished entering one card or all the cards, click on Save at the top right. This will automatically take you to the testing mode (Screenshot 3), which you’ll want to ignore for a while. Click on the stacked cards icon to return to the text for more prompts. When you come back to the flashcards, your decks may have shifted, since the most recently edited deck will be at the top.

Importantly, though, Screenshot 3 is the screen you will see when you return and click on a deck. To add more cards from this screen, click on the + sign at the bottom right. When you are done entering the cards for a section, get ready for the retrieval practice challenge! This is where it gets good (for learning).

Step 4: Create the Backs for the Flashcards

Rather than simply enter the backs of the flashcards from the information in the book, I first fill out the backs by simply trying to retrieve what I can remember. For example, for the prompt, “Write the code for the Iris model, using K Nearest Neighbors,” I wrote something like this on the back of the card:

import numpy as np
from sklearn.datasets import load_iris
from sklearn.model_selection import train_test_split
iris_dataset = load_iris()
X_train, X_test, y_train, y_test = train_test_split(iris_dataset.data, iris_dataset.target)

There are a lot of omissions here and some errors, and I moved things around after I wrote them down, but I tried as hard as I could to remember the code. To make the back of the card right, I filled in the omissions and corrected the errors. As I went through this process with all the cards in a section, I edited the fronts and backs of the cards and even added new cards as the importance of some material presented itself more clearly.

Create the backs of the flashcards for a section by first trying as hard as you can to retrieve the information asked for in the prompt. Then, correct the information and fill in omissions. Repeat this for each card in the deck.

Step 5: Test Yourself and Feel the Learn

One thing you should notice when you do this is that it hurts. And it should. In my view, the prompts should not be easy to answer. Another prompt I have for a different chapter is “Explain how k-neighbors regression works for both 1 neighbor and multiple neighbors.” My expectations for my response are high—I want to answer completely with several details from the text, not just a mooshy general answer. I keep the number of cards per chapter fairly low (about 5 to 10 cards per 100 pages). But your goals for retaining information may be different.

But once you have a set of cards for a section, come back to them occasionally and complete a round of testing for the section. To test yourself, click on the deck and respond to the first prompt you see without looking at the answer. Try to be as complete (correct) as possible before looking at the correct response.

To view the correct response, click on the card. Then, click on the checkmark if you completely nailed the response. Anything short of that, I click on the red X.

For large decks, you may want to restudy those items you got incorrect. In that case, you can click on Study Incorrect to go back over just those cards you got wrong. There is also an option to shuffle the deck (at the bottom left), which you should make use of if the content of the cards build on each other, making them too predictable.

Contiguity Effective for Deductive Inference

research post

The discourse that surrounds the technicalities in this paper contains an agenda: to convince readers that the benefits of retrieval practice extend beyond the boring old “helps you remember stuff” caricature to something more “higher order” like deductive inference. But I’m not convinced that the experiments show this. Rather, what they demonstrate fairly convincingly is that informational contiguity, not retrieval practice, benefits inference-making. A related result from the research, on the benefits of text coherence, is explained here.

The Setup

The arch-nemesis of this research is a paper by Tran, et al. last year, which appeared to show some domain limitations on retrieval practice:

They found that retrieval practice failed to benefit participants’ later ability to make accurate deductions from previously retrieved information. In their study, participants were presented sentences one at a time to learn . . . The sentences could be related to one another to derive inferences about particular scenarios. Although retrieval practice was shown to improve memory of the sentences relative to a restudy control condition, there was no benefit on a final inference test that required integration of information from across multiple sentences.

So, in this study, researchers replicated Tran et al.’s methods, except in one important way: they did not present the sentences to be learned one at a time but together instead.

Participants were each presented with four scenarios (two of which are outlined at right) consisting of seven to nine premises in the form of sentences. In each scenario, deductions to specific conclusions were possible. For two of the four scenarios, subjects used retrieval practice. They were given a chance to read the sentences in a scenario at their own pace and then were shown the sentences again for five minutes—in cycles where the order of the sentences was randomized. During this five-minute session, subjects were asked to type in the missing words in each premise (between one and three missing words). The complete sentences were then shown as feedback. Each participant used restudy for the other two scenarios. During the restudy five-minute session, subjects simply reread the premises again, in cycles again, with the order of the premises randomized for each cycle.

The Results and Discussion

Two days later, participants were given a 32-item multiple choice test which “assessed participants’ ability to draw logical conclusions derived from at least two premises within each scenario.” And consistent with the researchers’ hypothesis, the retrieval practice conditions yielded significantly better results on a test of deductive inference than did the restudy conditions.

Yet, it’s not at all clear that retrieval practice was the cause of the better performance with respect to inference-making. There was another cause preceding it: the improved contiguity of the presented information, as compared with Tran et al.’s one-at-a-time procedure. It’s possible that the effectiveness of retrieval practice is limited to recall of already-integrated information, and the contiguity of the premises in this study allowed for such integration, which, in turn, allowed retrieval practice to outperform restudy. It is a possibility the researchers raise in the paper and one that, in my view, the current research has not effectively answered:

However, other recent studies have failed to find a benefit of retrieval practice for learning educational materials (Leahy et al. 2015; Tran et al. 2015; Van Gog and Sweller 2015). These studies all used learning materials that required learners to simultaneously relate multiple elements of the materials during study and/or test. Such materials that are high in element interactivity need constituent elements to be related to one another in order for successful learning or task completion (element interactivity may also be considered as a measure of the complexity of materials, see (Sweller 2010)).

What we can say, with some confidence, is that even if the benefits of retrieval practice were limited to improvements in recall (as prior research has demonstrated), such improvements do not stand in the way of improvements to higher-order reasoning, such as inference-making. (And shaping the path for students, such as improving informational contiguity can have a positive effect too.)

Eglington, L., & Kang, S. (2016). Retrieval Practice Benefits Deductive Inference Educational Psychology Review DOI: 10.1007/s10648-016-9386-y


text:
D4 branes ending on NS5 branes (Witten's construction): query about harmonic functions

This query stems from Witten's paper, ''Solutions of Four-Dimensional Field Theories via M Theory'' (hep-th/9703166). Specifically, consider Type IIA string theory, and suppose one has a stack of NS5 branes (''fivebranes'') of Type IIA with D4 branes (''fourbranes'') suspended between them.

The fourbrane worldvolume is parametrized by the coordinates $(x^0, x^1, x^2, x^3, x^6)$. The fivebrane worldvolume is parametrized by $(x^0, x^1, x^2, x^3, x^4, x^5)$. Let $v \equiv x^4 + i x^5$.

Quoting the reference (cf. page 4 of the above paper)

...note that $x^6$ is determined as a function of $v$ by minimizing the total fivebrane worldvolume. For large $v$ the equation for $x^6$ reduces to a Laplace equation,

$$\nabla^2 x^6 = 0$$

Here $\nabla^2$ is the Laplacian on the fivebrane worldvolume. $x^6$ is a function only of the directions normal to the fourbrane ends, that is only of $v$ and $\bar{v}$.

Is it obvious that the coordinate $x^6$ should be harmonic?

For branes, one usually writes a metric and the fact that the supersymmetry variation of the gravitino is zero for this background metric gives rise to an integrability condition, which in turn yields a differential equation which tells us that some function of the coefficients in the metric is harmonic. But that function is usually a function of the coordinates transverse to the brane (which of course $v$ and $\bar{v}$ are in this case). But why is it $x^6$ which must satisfy such an equation?

Does this have to do with the fact that when one compactifies on a circle, $x^6$ is like a scalar field?


text:
# line passing through a point and extending

What if I only want to extend the line 1cm in one direction but not the other? I found this on another post here from a year ago.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}

\begin{document}
\begin{tikzpicture}[dot/.style={circle,inner sep=1pt,fill,label={#1},name=#1},
extended line/.style={shorten >=-#1,shorten <=-#1},
extended line/.default=1cm]

\node [dot=A] at (0,0) {};
\node [dot=B] at (3,1) {};
\node [dot=P] at (1.9,-1.6) {};

\draw [extended line=0.5cm] (A) -- (B);
\draw [extended line] ($(A)!(P)!(B)$) -- (P);

\fill [red] ($(A)!(P)!(B)$) circle [radius=2pt];

\end{tikzpicture}
\end{document}

-

Your figure extends on both ends since extended line style has shorten <=-#1. If you remove that, then the line will not be extended on the other end. But I suggest that you add another style.

one end extended/.style={shorten >=-#1},
one end extended/.default=1cm,

Here is the complete code and usage.

%\documentclass{article}
\documentclass[preview,border=5]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}

\begin{document}
\begin{tikzpicture}[dot/.style={circle,inner sep=1pt,fill,label={#1},name=#1},
extended line/.style={shorten >=-#1,shorten <=-#1},
extended line/.default=1cm,
one end extended/.style={shorten >=-#1},
one end extended/.default=1cm,
]

\node [dot=A] at (0,0) {};
\node [dot=B] at (3,1) {};
\node [dot=P] at (1.9,-1.6) {};

\draw [extended line=0.5cm] (A) -- (B);
\coordinate (P') at ($(A)!(P)!(B)$);
\draw [one end extended] (P') -- (P);

\end{tikzpicture}
\end{document}

## Edit

I added P' as name of the coordinate ($(A)!(P)!(B)$) as suggested in comment to speed up compilation time.

-
Saving the projected point as a named node might speed up the compilation a bit. :-) –  kiss my armpit Mar 9 '13 at 0:33
@GarbageCollector Done. I haven't timed this. But I wasn't able to see a noticeable difference in compilation time. :) –  hpesoj626 Mar 9 '13 at 0:42

In PSTricks, we can use nodesepA (only for A end), nodesepB (only for B end), or nodesep (for both A and B ends) to shorten or extend the line ends.

\documentclass[pstricks]{standalone}
\usepackage{pst-eucl}
\begin{document}
\begin{pspicture}(8,6)
\pstGeonode(2,3){A}(6,5){B}(6,2){P}
\pstProjection[PointName=none]{A}{B}{P}
\psset{nodesep=-1}
\pstLineAB{A}{B}
\pstLineAB[nodesepB=0]{P}{P'}
\end{pspicture}
\end{document}

-

This is possible with tkz-euclide or tkz-base but here a solution with only tikz

\documentclass{scrartcl}
\usepackage{tikz}
\usetikzlibrary{calc}

\tikzset{%
($(\tikztostart)!-#1!(\tikztotarget)$)--($(\tikztotarget)!-#2!(\tikztostart)$)%
\tikztonodes}}
}

\begin{document}
\begin{tikzpicture}[dot/.style={circle,inner sep=1pt,fill,name=#1}]

\node [dot=A,label={below:$A$}] at (0,0) {};
\node [dot=B,label={below:$B$}] at (3,1) {};
\node [dot=P,label={right:$P$}] at (1.9,-1.6) {};

\draw [add=.5 and .5] (A) to (B);
\coordinate (P') at ($(A)!(P)!(B)$);
\draw [add=0 and .5] (P') to (P);


text:
# Is my Peano Axiom proof correct?

I have recently begun reading Terry Tao's Real Analysis 1. The Peano Axiom proofs in the book are very new to me. Because of this, I have little intuition as to whether my proofs are correct. In addition to this, solutions to the exercises are not available in the text and I don't know anyone who is well-versed in this. This is the motivation behind my question. Any feedback is welcome! Thank you very much for taking the time, I truly appreciate it.

Here are the axioms as stated in his book:

Axiom 2.1: $$0$$ is a natural number.

Axiom 2.2: If $$n$$ is a natural number, then $$n++$$ is also a natural number.

Axiom 2.3: $$0$$ is not the successor of any natural number; i.e., we have $$n++\not= 0$$ for every natural number $$n.$$

Axiom 2.4: Different natural numbers must have different successors; i.e., if $$n,m$$ are natural numbers and $$n\not=m$$, then $$n++\not=m++$$. Equivalently, if $$n++ =m++$$, then we must have $$n=m.$$

Axiom 2.5: Let $$P(n)$$ be any property pertaining to a natural number $$n$$. Suppose that $$P(0)$$ is true,and suppose that whenever $$P(n)$$ is true, $$P(n++)$$ is also true. Then $$P(n)$$ is true for every natural number $$n.$$

We can also assume that recursive definitions are well defined.

Definition of addition: Let $$m$$ be a natural number. To add zero to $$m$$, we define $$0 +m:=m$$. Now suppose inductively that we have defined how to add $$n$$ to $$m$$. Then we can add $$n++$$ to $$m$$ by defining $$(n++ ) +m:= (n+m)++.$$

Additionally, the following have been proven:

Lemma 2.2.2: $$n+0=n$$ for all natural numbers $$n$$.

Lemma 2.2.3: For any natural numbers $$n$$ and $$m, n+(m++ ) = (n+m)++.$$

Proposition 2.2.4: $$n+m=m+n$$ for all natural numbers $$n,m.$$

Given this, is the following proof right?

Theorem 1: For any natural numbers $$a,b,c$$ we have $$(a+b)+c=a+(b+c).$$

My proof: We induct on $$c.$$ First, we verify the base case: $$(a+b)+0$$ The above equals $$a+b$$ by lemma 2.2.2. Additionally, this can be written as $$a+(b+0)$$ by lemma 2.2.2. This concludes the base case. Next, assume that there exists a natural number $$c$$ such that $$(a+b)+c=a+(b+c)$$ for all natural numbers $$a,b.$$ Then, for $$c++$$ we have $$(a+b)+(c++)=((a+b)+c)++$$ by lemma 2.2.3. By the inductive hypothesis, this becomes $$(a+(b+c))++.$$ Using lemma 2.2.3, this becomes $$a+(b+c)++.$$ Applying lemma 2.2.3 once more yields $$a+(b+(c++))$$ as desired.

$$\square$$

2. Instead of "assume that there exists a natural number $$c$$ such that...", you should write "take any natural number $$c$$, and assume that for all $$a,b$$, $$(a + b) + c = a + (b + c)$$". It is under these assumptions that you are going to show that the statement "for all $$a,b$$, $$(a + b) + c\!+\!\!+ = a + (b + c\!+\!\!+)$$" holds for the same $$c$$.
edit: One more comment about style. Instead of the phrase "this becomes", you can always write chains of equalities. For example, you can write "by Lemma 2.2.3, $$(a + (b + c))\!+\!\!+ = a + (b + c)\!+\!\!+$$". This way the reader doesn't have to guess which side/subexpression of the original expression you mean, which may improve readability in more complicated arguments.