# 0.16 Appendix: optimization theory

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Optimization theory is the branch of applied mathematics whose purpose is to consider a mathematical expression in order to find a set of parameters that either maximize or minimize it. Being an applied discipline, problems usually arise from real-life situations including areas like science, engineering and finance (among many other). This section presents some basic concepts for completeness and is not meant to replace a treaty on the subject. The reader is encouraged to consult further references for more information.

## Solution of linear weighted least squares problems

$\underset{h}{\text{min}}\phantom{\rule{0.277778em}{0ex}}{\parallel d-\mathbf{C}h\parallel }_{2}$

which can be written as

$\underset{h}{\text{min}}\phantom{\rule{0.277778em}{0ex}}{\left(d,-,\mathbf{C},h\right)}^{T}\left(d,-,\mathbf{C},h\right)$

omitting the square root since this problem is a strictly convex one. Therefore its unique (and thus global) solution is found at the point where the partial derivatives with respect to the optimization variable are equal to zero. That is,

$\begin{array}{cc}\hfill \frac{\partial }{\partial h}\left\{{\left(d,-,\mathbf{C},h\right)}^{T},\left(d,-,\mathbf{C},h\right)\right\}& =\frac{\partial }{\partial h}\left\{{d}^{T},d,-,2,{d}^{T},\mathbf{C},h,+,{\left(\mathbf{C},h\right)}^{T},\mathbf{C},h\right\}\hfill \\ & =-2{\mathbf{C}}^{T}d+2{\mathbf{C}}^{T}\mathbf{C}h=0\hfill \\ & ⇒\phantom{\rule{0.277778em}{0ex}}{\mathbf{C}}^{T}\mathbf{C}h={\mathbf{C}}^{T}d\hfill \end{array}$

The solution of [link] is given by

$h={\left({\mathbf{C}}^{T},\mathbf{C}\right)}^{-1}{\mathbf{C}}^{T}d$

where the inverted term is referred [link] , [link] as the Moore-Pentrose pseudoinverse of ${\mathbf{C}}^{T}\mathbf{C}$ .

In the case of a weighted version of [link] ,

$\underset{h}{\text{min}}\phantom{\rule{0.277778em}{0ex}}{\parallel \sqrt{w}\left(d,-,\mathbf{C},h\right)\parallel }_{2}^{2}=\sum _{k}{w}_{k}{|{d}_{k}-{C}_{k}h|}^{2}$

where ${C}_{k}$ is the $k$ -th row of $\mathbf{C}$ , one can write [link] as

$\underset{h}{\text{min}}\phantom{\rule{0.277778em}{0ex}}\left(\mathbf{W},\left(,d,-,\mathbf{C},h\right){\right)}^{T}\left(\mathbf{W},\left(,d,-,\mathbf{C},h\right)\right)$

where $\mathbf{W}=\text{diag}\left(\sqrt{w}\right)$ contains the weighting vector $w$ . The solution is therefore given by

$h={\left({\mathbf{C}}^{T},{\mathbf{W}}^{T},\mathbf{W},\mathbf{C}\right)}^{-1}{\mathbf{C}}^{T}{\mathbf{W}}^{T}\mathbf{W}d$

## Newton's method and ${l}_{p}$ Linear phase systems

Consider the problem

$\underset{a}{\text{min}}\phantom{\rule{0.277778em}{0ex}}g\left(a\right)={\parallel A\left(\omega ;a\right)-D\left(\omega \right)\parallel }_{p}$

for $a\in {\mathbb{R}}^{M+1}$ . Problem [link] is equivalent to the better posed problem

$\begin{array}{ccc}\hfill \underset{a}{\text{min}}\phantom{\rule{0.277778em}{0ex}}f\left(a\right)=g{\left(a\right)}^{p}& =& {\parallel A\left(\omega ;a\right)-D\left(\omega \right)\parallel }_{p}^{p}\hfill \\ & =& \sum _{i=0}^{L}\mid {C}_{i}a-{D}_{i}{\mid }^{p}\hfill \end{array}$

where ${D}_{i}=D\left({\omega }_{i}\right)$ , ${\omega }_{i}\in \left[0,\pi \right]$ , ${C}_{i}=\left[{C}_{i,0},...,{C}_{i,M}\right]$ , and

$\mathbf{C}=\left[\begin{array}{c}{C}_{0}\\ ⋮\\ {C}_{L}\end{array}\right]$

The $ij$ -th element of $\mathbf{C}$ is given by ${C}_{i,j}=\text{cos}\phantom{\rule{1mm}{0ex}}{\omega }_{i}\left(M-j\right)$ , where $0\le i\le L$ and $0\le j\le M$ . From [link] we have that

$\nabla f\left(a\right)=\left[\begin{array}{c}\frac{\partial }{\partial {a}_{0}}f\left(a\right)\\ ⋮\\ \frac{\partial }{\partial {a}_{M}}f\left(a\right)\end{array}\right]$

where ${a}_{j}$ is the $j$ -th element of $a\in {\mathbb{R}}^{M+1}$ and

$\begin{array}{ccc}\hfill \frac{\partial }{\partial {a}_{j}}f\left(a\right)& =& \frac{\partial }{\partial {a}_{j}}\sum _{i=0}^{L}\mid {C}_{i}a-{D}_{i}{\mid }^{p}\hfill \\ & =& \sum _{i=0}^{L}\frac{\partial }{\partial {a}_{j}}\mid {C}_{i}a-{D}_{i}{\mid }^{p}\hfill \\ & =& p\sum _{i=0}^{L}\mid {C}_{i}a-{D}_{i}{\mid }^{p-1}·\frac{\partial }{\partial {a}_{j}}\mid {C}_{i}a-{D}_{i}\mid \hfill \end{array}$

Now,

$\frac{\partial }{\partial {a}_{j}}\mid {C}_{i}a-{D}_{i}\mid =\text{sign}\left({C}_{i}a-{D}_{i}\right)·\frac{\partial }{\partial {a}_{j}}\left({C}_{i}a-{D}_{i}\right)={C}_{i,j}\phantom{\rule{0.277778em}{0ex}}\text{sign}\left({C}_{i}a-{D}_{i}\right)$

where Note that

$\underset{u\left(a\right)\to {0}^{+}}{lim}\frac{\partial }{\partial {a}_{j}}\mid u\left(a\right){\mid }^{p}=\underset{u\left(a\right)\to {0}^{-}}{lim}\frac{\partial }{\partial {a}_{j}}\mid u\left(a\right){\mid }^{p}=0$

$\text{sign}\left(x\right)=\left\{\begin{array}{cc}1& x>0\hfill \\ 0& x=0\hfill \\ -1& x<0\hfill \end{array}\right)$

Therefore the Jacobian of $f\left(a\right)$ is given by

$\nabla f\left(a\right)=\left[\begin{array}{c}p\sum _{i=0}^{L}{C}_{i,0}\phantom{\rule{0.277778em}{0ex}}\mid {C}_{i}a-{D}_{i}{\mid }^{p-1}\text{sign}\left({C}_{i}a-{D}_{i}\right)\\ ⋮\\ p\sum _{i=0}^{L}{C}_{i,M-1}\phantom{\rule{0.277778em}{0ex}}\mid {C}_{i}a-{D}_{i}{\mid }^{p-1}\text{sign}\left({C}_{i}a-{D}_{i}\right)\end{array}\right]$

The Hessian of $f\left(a\right)$ is the matrix ${\nabla }^{2}f\left(a\right)$ whose $jm$ -th element ( $0\le j,m\le M$ ) is given by

$\begin{array}{ccc}\hfill {\nabla }_{j,m}^{2}f\left(a\right)=\frac{\partial {a}^{2}}{\partial {a}_{j}\partial {a}_{m}}f\left(a\right)& =& \frac{\partial }{\partial {a}_{m}}\frac{\partial }{\partial {a}_{j}}f\left(a\right)\hfill \\ & =& \sum _{i=0}^{L}p\phantom{\rule{0.222222em}{0ex}}{C}_{i,j}\phantom{\rule{0.277778em}{0ex}}\frac{\partial }{\partial {a}_{m}}\mid {D}_{i}-{C}_{i}a{\mid }^{p-1}\text{sign}\left({D}_{i}-{C}_{i}a\right)\hfill \\ & =& \sum _{i=0}^{L}\alpha \frac{\partial }{\partial {a}_{m}}b\left(a\right)d\left(a\right)\hfill \end{array}$

where adequate substitutions have been made for the sake of simplicity. We have

$\begin{array}{ccc}\hfill \frac{\partial }{\partial {a}_{m}}b\left(a\right)& =& \frac{\partial }{\partial {a}_{m}}\mid {C}_{i}a-{D}_{i}{\mid }^{p-1}\hfill \\ & =& \left(p-1\right){C}_{i,m}\mid {C}_{i}a-{D}_{i}{\mid }^{p-2}\text{sign}\left({C}_{i}a-{D}_{i}\right)\hfill \\ \hfill \frac{\partial }{\partial {a}_{m}}d\left(a\right)& =& \frac{\partial }{\partial {a}_{m}}\text{sign}\left({D}_{i}-{C}_{i}a\right)=0\hfill \end{array}$

Note that the partial derivative of $d\left(a\right)$ at ${D}_{i}-{C}_{i}a=0$ is not defined. Therefore

$\begin{array}{ccc}\hfill \frac{\partial }{\partial {a}_{m}}b\left(a\right)d\left(a\right)& =& b\left(a\right)\frac{\partial }{\partial {a}_{m}}d\left(a\right)+d\left(a\right)\frac{\partial }{\partial {a}_{m}}b\left(a\right)\hfill \\ & =& \left(p-1\right){C}_{i,m}\mid {C}_{i}a-{D}_{i}{\mid }^{p-2}{\text{sign}}^{2}\left({C}_{i}a-{D}_{i}\right)\hfill \end{array}$

Note that ${\text{sign}}^{2}\left({C}_{i}a-{D}_{i}\right)=1$ for all ${D}_{i}-{C}_{i}a\ne 0$ where it is not defined. Then

${\nabla }_{j,m}^{2}f\left(a\right)=p\left(p-1\right)\sum _{i=0}^{L}{C}_{i,j}{C}_{i,m}\phantom{\rule{0.277778em}{0ex}}\mid {C}_{i}a-{D}_{i}{\mid }^{p-2}$

except at ${D}_{i}-{C}_{i}a=0$ where it is not defined.

• Given ${a}_{0}\in {\mathbb{R}}^{M+1}$ , $D\in {\mathbb{R}}^{L+1}$ , $\mathbf{C}\in {\mathbb{R}}^{L+1×M+1}$
• For $i=0,1,...$
1. Find $\nabla f\left({a}_{i}\right)$ .
2. Find ${\nabla }^{2}f\left({a}_{i}\right)$ .
3. Solve ${\nabla }^{2}f\left({a}_{i}\right)s=-\nabla f\left({a}_{i}\right)$ for $s$ .
4. Let ${a}_{+}={a}_{i}+s$ .
5. Check for convergence and iterate if necessary.

Note that for problem [link] the Jacobian of $f\left(a\right)$ can be written as

$\nabla f\left(a\right)=p{\mathbf{C}}^{T}y$

where

$y=\mid \mathbf{C}{a}_{i}-D{\mid }^{p-1}\text{sign}\left(\mathbf{C}{a}_{i}-D\right)=\mid \mathbf{C}{a}_{i}-D{\mid }^{p-2}\left(\mathbf{C}{a}_{i}-D\right)$

Also,

${\nabla }_{j,m}^{2}f\left(a\right)=p\left(p-1\right)\phantom{\rule{0.277778em}{0ex}}{C}_{j}^{T}\mathbf{Z}{C}_{m}$

where

$\mathbf{Z}=\text{diag}\left(\mid ,\mathbf{C},{a}_{i},-,D,{\mid }^{p-2}\right)$

and

${C}_{j}=\left[\begin{array}{c}{C}_{0,j}\\ ⋮\\ {C}_{L,j}\end{array}\right]$

Therefore

${\nabla }^{2}f\left(a\right)=\left({p}^{2}-p\right){\mathbf{C}}^{T}\mathbf{Z}\mathbf{C}$

From [link] , the Hessian ${\nabla }^{2}f\left(a\right)$ can be expressed as

${\nabla }^{2}f\left(a\right)=\left({p}^{2}-p\right){\mathbf{C}}^{T}{\mathbf{W}}^{T}\mathbf{W}\mathbf{C}$

where

$\mathbf{W}=\text{diag}\left(\mid ,\mathbf{C},{a}_{i},-,D,{\mid }^{\frac{p-2}{2}}\right)$

The matrix $\mathbf{C}\in {\mathbb{R}}^{\left(L+1\right)×\left(M+1\right)}$ is given by

$\mathbf{C}=\left[\begin{array}{ccccccc}\text{cos}M{\omega }_{0}& \text{cos}\left(M-1\right){\omega }_{0}& \cdots & \text{cos}\left(M-j\right){\omega }_{0}& \cdots & \text{cos}{\omega }_{0}& 1\\ \text{cos}M{\omega }_{1}& \text{cos}\left(M-1\right){\omega }_{1}& \cdots & \text{cos}\left(M-j\right){\omega }_{1}& \cdots & \text{cos}{\omega }_{1}& 1\\ ⋮& ⋮& \ddots & ⋮& & ⋮& ⋮\\ \text{cos}M{\omega }_{i}& \text{cos}\left(M-1\right){\omega }_{i}& \cdots & \text{cos}\left(M-j\right){\omega }_{i}& \cdots & \text{cos}{\omega }_{i}& 1\\ ⋮& ⋮& & ⋮& \ddots & ⋮& ⋮\\ \text{cos}M{\omega }_{L-1}& \text{cos}\left(M-1\right){\omega }_{L-1}& \cdots & \text{cos}\left(M-j\right){\omega }_{L-1}& \cdots & \text{cos}{\omega }_{L-1}& 1\\ \text{cos}M{\omega }_{L}& \text{cos}\left(M-1\right){\omega }_{L}& \cdots & \text{cos}\left(M-j\right){\omega }_{L}& \cdots & \text{cos}{\omega }_{L}& 1\end{array}\right]$

The matrix $\mathbf{H}={\nabla }^{2}f\left(a\right)$ is positive definite (for $p>1$ ). To see this, consider $\mathbf{H}={\mathbf{K}}^{T}\mathbf{K}$ where $\mathbf{K}=\mathbf{W}\mathbf{C}$ . Let $z\in {\mathbb{R}}^{M+1}$ , $z\ne 0$ . Then

${z}^{T}\mathbf{H}z={z}^{T}{\mathbf{K}}^{T}\mathbf{K}z={\parallel \mathbf{K}z\parallel }_{2}^{2}>0$

unless $z\in N\left(\mathbf{K}\right)$ . But since $\mathbf{W}$ is diagonal and $\mathbf{C}$ is full column rank, $N\left(\mathbf{K}\right)=0$ . Thus ${z}^{T}\mathbf{H}z\ge 0$ (identity only if $z=0$ ) and so $\mathbf{H}$ is positive definite.

## Newton's method and ${l}_{p}$ Complex linear systems

Consider the problem

$\underset{x}{\text{min}}\phantom{\rule{0.277778em}{0ex}}e\left(x\right)={\parallel \mathbf{A}x-b\parallel }_{p}^{p}$

where $\mathbf{A}\in {\mathbb{C}}^{m×n}$ , $x\in {\mathbb{R}}^{n}$ and $b\in {\mathbb{C}}^{m}$ . One can write [link] in terms of the real and imaginary parts of $\mathbf{A}$ and $b$ ,

$\begin{array}{ccc}\hfill e\left(x\right)& =& \sum _{i=1}^{m}{|{A}_{i}x-{b}_{i}|}^{p}\hfill \\ & =& \sum _{i=1}^{m}{|\text{Re}\left\{{A}_{i}x-{b}_{i}\right\}+jIm\left\{{A}_{i}x-{b}_{i}\right\}|}^{p}\hfill \\ & =& \sum _{i=1}^{m}{|\left({R}_{i}x-{\alpha }_{i}\right)+\left({Z}_{i}x-{\gamma }_{i}\right)|}^{p}\hfill \\ & =& \sum _{i=1}^{m}{\left(\sqrt{{\left({R}_{i}x-{\alpha }_{i}\right)}^{2}+{\left({Z}_{i}x-{\gamma }_{i}\right)}^{2}}\right)}^{p}\hfill \\ & =& \sum _{i=1}^{m}{g}_{i}{\left(x\right)}^{p/2}\hfill \end{array}$

where $\mathbf{A}=\mathbf{R}+j\mathbf{Z}$ and $b=\alpha +j\gamma$ . The gradient $\nabla e\left(x\right)$ is the vector whose $k$ -th element is given by

$\frac{\partial }{\partial {x}_{k}}e\left(x\right)=\frac{p}{2}\sum _{i=1}^{m}\left[\frac{\partial }{\partial {x}_{k}},{g}_{i},\left(x\right)\right]{g}_{i}{\left(x\right)}^{\frac{p-2}{2}}=\frac{p}{2}{q}_{k}\left(x\right)\stackrel{^}{g}\left(x\right)$

where ${q}_{k}$ is the row vector whose $i$ -th element is

$\begin{array}{cc}\hfill {q}_{k,i}\left(x\right)=\frac{\partial }{\partial {x}_{k}}{g}_{i}\left(x\right)& =2\left({R}_{i}x-\alpha {\alpha }_{i}\right){R}_{ik}+2\left({Z}_{i}x-\gamma {\gamma }_{i}\right){Z}_{ik}\hfill \\ & =2{R}_{ik}{R}_{i}x+2{Z}_{ik}{Z}_{i}x-\left[2{\alpha }_{i}{R}_{ik}+2{\gamma }_{i}{Z}_{ik}\right]\hfill \end{array}$

Therefore one can express the gradient of $e\left(x\right)$ by $\nabla e\left(x\right)=\frac{p}{2}\mathbf{Q}\stackrel{^}{g}$ , where $\mathbf{Q}=\left[{q}_{k,i}\right]$ as above. Note that one can also write the gradient in vector form as follows

$\nabla e\left(x\right)=p\phantom{\rule{0.277778em}{0ex}}\left[{\mathbf{R}}^{T},\text{diag},\left(\mathbf{R}x-\alpha \right),+,{\mathbf{Z}}^{T},\text{diag},\left(\mathbf{Z}x-\gamma \right)\right]·\left[{\left({\left(\mathbf{R}x-\alpha \right)}^{2},+,{\left(\mathbf{Z}x-\gamma \right)}^{2}\right)}^{\frac{p-2}{2}}\right]$

The Hessian $\mathbf{H}\left(x\right)$ is the matrix of second derivatives whose $kl$ -th entry is given by

$\begin{array}{cc}\hfill {\mathbf{H}}_{k,l}\left(x\right)& =\frac{{\partial }^{2}}{\partial {x}_{k}\partial {x}_{l}}e\left(x\right)\hfill \\ \hfill & =\frac{\partial }{\partial {x}_{l}}\frac{p}{2}\sum _{i=1}^{m}{q}_{k,i}\left(x\right){g}_{i}{\left(x\right)}^{\frac{p-2}{2}}\hfill \\ & =\frac{p}{2}\sum _{i=1}^{m}\left[{q}_{k,i},\left(x\right),\frac{\partial }{\partial {x}_{l}},{g}_{i},{\left(x\right)}^{\frac{p-2}{2}},+,{g}_{i},{\left(x\right)}^{\frac{p-2}{2}},\frac{\partial }{\partial {x}_{l}},{q}_{k,i},\left(x\right)\right]\hfill \end{array}$

Now,

$\begin{array}{cc}\hfill \frac{\partial }{\partial {x}_{l}}{g}_{i}{\left(x\right)}^{\frac{p-2}{p}}& =\frac{p-2}{2}\left[\frac{\partial }{\partial {x}_{l}},{g}_{i},\left(x\right)\right]{g}_{i}{\left(x\right)}^{\frac{p-4}{2}}\hfill \\ & =\frac{p-2}{2}{q}_{l,i}\left(x\right){g}_{i}{\left(x\right)}^{\frac{p-4}{2}}\hfill \\ \hfill \frac{\partial }{\partial {x}_{l}}{q}_{k,i}\left(x\right)& =2{R}_{ik}{R}_{il}+2{Z}_{ik}{Z}_{il}\hfill \end{array}$

${\mathbf{H}}_{k,l}\left(x\right)=\frac{p\left(p-2\right)}{4}\sum _{i=1}^{m}{q}_{k,i}\left(x\right){q}_{l,i}\left(x\right){g}_{i}{\left(x\right)}^{\frac{p-4}{4}}+p\sum _{i=1}^{m}\left({R}_{ik}{R}_{il}+{Z}_{ik}{Z}_{il}\right){g}_{i}{\left(x\right)}^{\frac{p-2}{2}}$

Note that $\mathbf{H}\left(x\right)$ can be written in matrix form as

$\begin{array}{cc}\hfill \mathbf{H}\left(x\right)=& \frac{p\left(p-2\right)}{4}\left(\mathbf{Q},\phantom{\rule{0.277778em}{0ex}},\text{diag},\left(g,{\left(x\right)}^{\frac{p-4}{2}}\right),{\mathbf{Q}}^{T}\right)+\hfill \\ & p\left({\mathbf{R}}^{T},\text{diag},\left(g,{\left(x\right)}^{\frac{p-2}{2}}\right),\mathbf{R},+,{\mathbf{Z}}^{T},\text{diag},\left(g,{\left(x\right)}^{\frac{p-2}{2}}\right),\mathbf{Z}\right)\hfill \end{array}$

Therefore to solve [link] one can use Newton's method as follows: given an initial point ${x}_{0}$ , each iteration gives a new estimate ${x}^{+}$ according to the formulas

$\begin{array}{ccc}\mathbf{H}\left({x}^{c}\right)s& =& -\nabla e\left({x}^{c}\right)\\ {x}^{+}& =& {x}^{c}+s\end{array}$

where $\mathbf{H}\left({x}^{c}\right)$ and $\nabla e\left({x}^{c}\right)$ correspond to the Hessian and gradient of $e\left(x\right)$ as defined previously, evaluated at the current point ${x}^{c}$ . Since the $p$ -norm is convex for $1 , problem [link] is convex. Therefore Newton's method will converge to the global minimizer ${x}^{☆}$ as long as $\mathbf{H}\left({x}^{c}\right)$ is not ill-conditioned.

memory of development brain of the human psychologist
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lord
hlo
Ananya
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Shilan
hi
zge
hello
Emm
usef
hi
Gil
Haw are you?
Shilan
Aha ok
Shilan
hi
Daniella
waasup
Isaiah
hello
Sara
hello
androi
Haw are you
Shilan
im ok you
Daniella
Me to
Shilan
Thanks
Shilan
What's going?
Shilan
working
Daniella
hey everyone this Mahmoud
Mahmoud
is
Mahmoud
Fine
Shilan
Hello
Jobe
how is it going people?
Tomasz
kind
Alter
thanks
Alter
hi..!
Alter
yes we'll
Alter
yes we'll
Alter
how did psychology begin?
of psychologys commencement, the traces can be seen in the work of Aristotle, where he talk about soul and body, likewise work in durrant, de anima, all these were somewhere supporting dualism, in which soul could exist separately from body
amaan
but if you talk about the moder psychology, Gustav fechner, is credited with performing scientific experiments, basis of his experiments in psychology with his studies perception.
amaan
hi, can you help me to find research related to child truma
Heba
yes same here I have cptsd and am looking for more info since my doctor doesn't know what to do I want to know what I can
Mark
does psychology deal with love?
Maybe, i think
edem
I definitely would say yes
Clara
how so
Isaiah
*triarchic
Meredith
there are so many different reasons why you can fall in love with someone, many of them develope subconsciously -> psychology
Clara
love messes with the brain, a lot, ergo I believe that Psychology does indeed deal with love
I would like an example as to what and how you think it deals with.
Tyler
how can I discover that this individual has a long-term memory and shot- term memory?
Namuaha
what is synapse
In the central nervous system, a synapse is a small gap at the end of a neuron that allows a signal to pass from one neuron to the next. synapse are found where nerve cells connect with other nerve cells
Najeem
a synapse the connection is where a neuron cell connects to another neuron cell.
Shaun
good
Jobe
what is psychology
Jobe
can you do auto book auto
WHT u mean?
usef
yes
MD
heyy, may i join the conversation please?
yes of course
Najeem
how to avoid theory of bias confirmation in real life
Scar
yes
Jobe
hello
Bhavin
hello to all
Genevieve
hey, may I join the conversation?
samra
salaam
Ibrahim
so i cn even do this after ba hons in psychology?
Avneet
the only eligibility criteria is that you should have 50% of aggregate in your psychology papers. (bachelors)
syeda
okay thank you so much❤ have a lovely day🙂
Avneet
you're welcome. glad it helped ^_^
syeda
To pursue a career as a psychotherapist you'll have to do your bachelors in psychology. (bsc honors is preferable). since there are many fields and you've chosen as a therapist. a masters degree in clinical psychology or therapy and family counseling is preferable.
syeda
so i cn even do this after ba hons in psychology?
Avneet
yes you can.
syeda
you're welcome. glad it helped ^_^
syeda
hello
Vhikkie
yh
Parker
hello
Avneet
is there any psychotherapist here? i need to know the qualification one hve to pursue.
Avneet
I'm clinic psychologist...
Shilan
Hey. I'm pursuing BA in Clinical Psychology
Aakarshan
the only eligibility criteria is that you should have 50% of aggregate in your psychology papers. (bachelors)
syeda
hello
Vhikkie
who is the father of psychology
aristatil
and please, how would you guys, describe the study of psychology at college ?
edem
psychologist student?
Aspen
i mean not yet but am about to start college so wanna know how is it(college in general and psychology course) please
edem
Psychology is the study of mind and behaviour. So if you will take psychology as a subject so you will get to know how your everything (physical, mental, social, spiritual aspects) effects your behaviour
sakina
With this brief knowledge you can help people to cope up with their problems and only you can guide them correctly
sakina
And if you go for further specialisations you can study hypnosis, face reading, body language etc
sakina
Thanks a lot🙏🏾 And ik some of the stuffs u said but i am also going to write thesis, right ?
edem
ok no prob, thanks a lot🙏🏾✨
edem
cerebellum
Khan
hae everyone, hope you are well this evning my question is what is the difference between drive and motivation
Michael
good question
Rainee
drive is more like an impulse or urge and i think they both go together (drive and motivation) even if there is a slight difference
edem
@ Michael Drive is delivered to be innate without the use of an external stimuli, motivation normally evolves an outside stimuli which may include praise, appreciate, or reward.
Reginald
*believed...sorry for typo
Reginald
@Reginald, can't the motivation come from the inner self?
edem
Good question, please give an example.
Reginald
can we say desire of success for example
edem
Wilhelm Wundt is the father of psychology
ipau
Wilhem Wundt thank you for the road that you opened.
Qwanta
You mean who is the father of having a great educated argumentative guess? nothing is more wrong than this question. The question is you should ask yourselfs is, how sure are you abour their scientific studying? one's percieved assimilated approach to judging another person and saying they are
Roger
the biggest problem with scientific research and data is that ya you could get the same result 1000 times then it could go the other way 1000 times, but we would never know that and we did, we would still say ya but the proof is there. The only thing science proves is that humanity has
Roger
no facts about human behavior in the scientific context, but more in the trial and error.. sorry to tell you, but so far no one has proven Father of anything, thats up to you and i, judgement is bias, science is good enough lazy
Roger
cognitive development is the growing and development of the brain.
Ecofascism is a theoretical political model in which an authoritarian government would require individuals to sacrifice their own interests to the "organic whole of nature". The term is also used as a rhetorical pejorative to undermine the environmental movement.
ipau
what's the big difference between prejudice and discrimination?
A prejudiced person may not act on their attitude.  Therefore, someone can be prejudiced towards a certain group but not discriminate against them.  Also, prejudice includes all three components of an attitude (affective, behavioral and cognitive), whereas discrimination just involves behavior
Nancy Lee
hi
basher
hello
Rahul
what is all about cognitive development?
Kamohelo
cognitive development is the growing and development of the brain
Jessy
how do you control a variable when using spss whilst running a pearsons correlation analysis?
it dependa on your study. according to what you want to say and explain your result
Pouran
Hello
Jobe
why does it say her and she
stages of cognitive development
sensory preoperatinal concrete formal
Rajendra
What's mental memory?
Namuaha
Memory is our ability to encode, store, retain and subsequently recall information and past experiences in the human brain. It can be thought of in general terms as the use of past experience to affect or influence current behaviour.
Shilan
What I mental memory?
Namuaha
What's mental memory?
Namuaha
my thankful
Namuaha
Jobe
what is psychology
the study of insecurities and the effect on the host .
Sera
Psychology is the scientific study of behavior & mental processes
Angela
psychology is science about learning human behaviour
Zhamshid
behaviorosm
Khan
is the study of human behaviour and mental processes
Jobe
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