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# Why can future forward interest rates be assumed to be lognormally distributed in the standard market model?

This seems to be the underlying assumption that allows us to use the standard market model/Black's framework in order to value interest rate derivatives, but I haven't found any understandable explanation that explains why this is an assumption that can be made. Interest rates themselves don't follow a Geometric Brownian motion, which I think is implied by a log-normal distribution? So why would forward rates?

SHORT STORY: forward Libor rates need not be assumed to be log-normally distributed. For example, they can be assumed to be normally distributed (and indeed, on Bloomberg, Swaption implied vols are quoted both, in terms of normal as well as log-normal models).

The only condition required is that the forward Libor rate process needs to be a martingale under the T-forward measure (I show this below). As long as the chosen modelling process satisfies the martingality condition, the distribution of this process can (in theory) be whatever we want it to be.

LONG STORY:

Part 1: Notation:

Denote a forward Libor rate at some time $$t$$, that sets at time $$T_i$$ and matures at time $$T_{i+1}$$, as $$L(t, T_i, T_{i+1})$$ (note that it only makes sense to discuss this random variable for $$t\leq T_i < T_{i+1}$$). The annual fraction over which this Libor compounds is $$\tau$$.

To make the notation clear, some examples:

$$(i) L(t_0,t_0,T_1)$$ would be the spot Libor rate maturing at some future time $$T_1$$ (we can suppose that $$T_1=t_0+\tau$$)

$$(ii) L(t_0,T_1,T_2)$$ would be today's value of forward Libor rate setting at time $$T_1$$ and maturing at time $$T_2$$ (i.e. this would be today's FRA on that libor)

$$(iii) L(t,T_1,T_2)$$ would be a future value at time "$$t$$" of the same forward Libor as in (ii): i.e. this would be a random variable, value of which is unknown today.

Part 2: Mechanics of Libor rates:

Suppose that you can lend and borrow at these Libor rates freely: that is to say that you can agree today at time "$$t_0$$" to borrow or lend any amount "$$x$$" at time $$T_i$$ and you'll then have to repay (or you'll receive) an amount $$x*(1+\tau L(t_0, T_i, T_{i+1}))$$ at time $$T_{i+1}$$.

Suppose you can do this for amount $$x=\frac{1}{1+\tau L(t_0, T_i, T_{i+1})}$$. Then, at time $$T_{i+1}$$, you'd have to repay (or you'd receive) exactly 1 unit of currency: in other words, you can effectively trade (forward) zero coupon bonds that pay 1 unit of currency at some specific maturity. Suppose you can do this at any time, not just today at "$$t_0$$", but at any time "$$t$$".

Suppose that at time "$$t$$" you want to trade some (spot) zero coupon bonds of various maturities. Denote a zero coupon bond that matures at time $$T_i$$ as $$P(t,T_i):=\frac{1}{1+\tau L(t, t, T_i)}$$. Denote another zero coupon bond that matures at time $$T_{i+1}$$ as $$P(t,T_{i+1}):=\frac{1}{1+\tau L(t, t, T_{i+1})}$$. Note that:

$$\frac{P(t,T_i)}{P(t,T_{i+1})}=1+\tau L(t, T_i, T_{i+1})$$

I.e. the above says that we can express a forward Libor rate as a ratio of two spot zero coupon bonds (these bonds, of course, we have constructed from spot Libor rates at time $$t$$: so basically, we are saying that we can construct a forward Libor rate from spot Libor rates: no big deal, really).

Part 3: Martingale condition

Re-arrange the above equation as:

$$\frac{P(t,T_i)}{P(t,T_{i+1})}-1=\tau L(t, T_i, T_{i+1})$$

$$\frac{P(t,T_i)-P(t,T_{i+1})}{P(t,T_{i+1})}=\tau L(t, T_i, T_{i+1})$$

$$\left(P(t,T_i)-P(t,T_{i+1})\right)\frac{1}{\tau}=P(t,T_{i+1}) L(t, T_i, T_{i+1})$$

Now, the LHS of the above is a linear combination of traded and liquid securities (as per our assumptions): therefore, the LHS has to be a martingale under a suitable Numeraire as per the Fundamental Theorem of Asset Pricing. Choosing $$P(t,T_{i+1})$$ as Numeraire, we get:

$$\mathbb{E}\left[\frac{\left(P(t,T_i)-P(t,T_{i+1})\right)\frac{1}{\tau}}{P(t,T_{i+1})} | \mathcal{F}_{t_0}\right]=\frac{\left(P(t_0,T_i)-P(t_0,T_{i+1})\right)\frac{1}{\tau}}{P(t_0,T_{i+1})}$$

But $$\left(P(t,T_i)-P(t,T_{i+1})\right)\frac{1}{\tau}=P(t,T_{i+1}) L(t, T_i, T_{i+1})$$

So we get:

$$\mathbb{E}\left[\frac{P(t,T_{i+1}) L(t, T_i, T_{i+1})}{P(t,T_{i+1})} | \mathcal{F}_{t_0}\right]=\mathbb{E}\left[L(t, T_i, T_{i+1})| \mathcal{F}_{t_0}\right]=L(t_0, T_i, T_{i+1})$$

Therefore, under the $$T_{i+1}$$ forward measure, associated with the bond $$P(t,T_{i+1})$$ as numeraire, the forward Libor $$L(t, T_i, T_{i+1})$$ must be a martingale.

Part 4: Lognormality or Normality???

The martingale condition above does not tell us anything about the distribution of the forward Libor, except for the fact that whatever process we chose, it must be a martingale under the forward measure. Indeed, with many rates being in the negative territory, Normal models have become acceptable as well as log-normal models. On Bloomberg, Swaptions implied vols are quoted in terms of normal as well as log-normal models.

A lognormal distribution has three valuable properties (I) It ensures that the rate is only allowed to be positive; (II) the changes in the interest rate are proportional to the interest rate; and (III) the option price is analytically solvable.

BTW, just to be precise, note that in Black's model, it is an assumption that the distribution of the interest rate in the forward measure is lognormal.

In the past, (I) was considered to be essential as negative interest rates were seen as anomalous and so unlikely. How times have changed ! It was also believed that changes in higher levels of interest rates were roughly proportional to their own level in accordance with (II), although this was no longer so true at low rates. And (III) is very useful as an analytically tractable option price is faster to calculate and more numerically smooth than one computed by a tree or Monte Carlo. So for all these reasons Black's model became the industry standard for European-style interest rate options (caps/floors and European swaptions). However this does NOT mean that it is used by banks for their internal risk and pricing models.

Currently, other models such as SABR, Bachelier (normal), LMM and shifted variations of all of these models are used by banks for their internal pricing and hedging.

Black's model is mainly used for volatility quotations. In this way, Black's model is simply a translator between a volatility quote and the cap/floor price. The actual model used by the bank to hedge it must be calibrated to reprice these option prices. In doing this, we are moving beyond using Black's model as a model of interest rates and to it being used as a quotation tool which could encompass any model.

For example, we know that interest rates tend to mean-revert. Black's model does not take this into account explicitly. However Black volatilities are quoted for different expiry dates (for caps and floors) and this market term structure of volatilities embeds within it expectations of mean reversion. Also, black vols are quoted for different cap/floor strikes and this also adjusts for the shape of the true distribution and allows it to be non-lognormal.

Lastly, by definition, Black's model will exactly refit the market prices of options at the quoted strikes and maturities. It may also be used by less sophisticated users to get a quick and dirty approximate value of option greeks - as the quotation means that the price will be right, it is hoped that the delta will not be too wrong.


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# Physics textbooks that distinguish between laws and definitions?

Often when I am learning physics I start to think about whether the laws I'm learning are mere definitions or experimentally determined, and usually the textbook does not make this clear. As Thomas Kuhn writes in The Structure of Scientific Revolutions,

These generalizations look like laws of nature, but their function for group members is not often that alone. Sometimes it is: for example the Joule-Lenz Law, $H = RI^2$. When that law was discovered, community members already knew what $H$, $R$, and $I$ stood for, and these generalizations simply told them something about the behavior of heat, current, and resistance that they had not known before. But more often, as discussion earlier in the book indicates, symbolic generalizations simultaneously serve a second function, one that is ordinarily sharply separated in analyses by philosophers of science. Like $f = ma$ or $I = V/R$, they function in part as laws but also in part as definitions of some of the symbols they deploy. Furthermore, the balance between their inseparable legislative and definitional force shifts over time. In another context these points would repay detailed analysis, for the nature of the commitment to a law is very different from that of commitment to a definition. Laws are often corrigible piecemeal, but definitions, being tautologies, are not. For example, part of what the acceptance of Ohm's Law demanded was a redefinition of both 'current' and 'resistance'; if those terms had continued to mean what they had meant before, Ohm's Law could not have been right; that is why it was so strenuously opposed as, say, the Joule-Lenz Law was not.

Another example is conservation of momentum; is momentum conserved because it is defined as a quantity that is conserved or because it has been observed to be conserved?

Are there any physics textbooks (maybe for classical mechanics) that take a more axiomatic approach, clearly distinguishing definitions from laws determined by experiment?

Before answering, please see our policy on resource recommendation questions. Please write substantial answers that detail the style, content, and prerequisites of the book, paper or other resource. Explain the nature of the resource so that readers can decide which one is best suited for them rather than relying on the opinions of others. Answers containing only a reference to a book or paper will be removed!

• Are there any physics textbooks (maybe for classical mechanics) that take a more axiomatic approach, clearly distinguishing definitions from laws determined by experiment? "Physics textbooks" is pretty broad. Are we talking about freshman texts? Graduate-level texts like Jackson? Best-selling freshman texts like Halliday are written for people who don't care about these issues. An axiomatic approach is not synonymous with rigor, and axiomatizations are non-unique. Something that's a definition in one axiomatization can be a law in another axiomatization. For freshman mechanics, try Kleppner. – Ben Crowell Sep 11 '13 at 5:24
• I am asking as an undergraduate, so a textbook at that level would be preferable. I understand that axiomatizations need not be unique; I wonder though, how can you be rigorous without indicating which laws are axioms and which are empirical? – Hypercube Sep 11 '13 at 5:30
• @Hypercube Well in classical mechanics one is best advised to use the Lagrangian or Hamiltonian formulation. This provides a basic mathematical template for analyzing any physical system in the universe. The constraints and structure of the Lagrangian thus constitute the experimentally determined axioms. I don't know about books on the freshman level, but I believe Goldstein's book on Mechanics covers the Lagrangian approach really well. It also delves it classical field theory. – dj_mummy Sep 11 '13 at 7:08
• I am asking as an undergraduate, so a textbook at that level would be preferable. Upper division? Lower division? how can you be rigorous without indicating which laws are axioms and which are empirical? I don't really follow you here. Axioms are typically empirical. For instance, the reason Euclid chose the parallel postulate as an axiom was that it seemed to be valid empirically. With hindsight, this was because he was an intelligent being living in a region of relatively low spacetime curvature. – Ben Crowell Sep 11 '13 at 15:53
• Lower division. In any case, I have started Goldstein and it's excellent so far. Thank you! – Hypercube Sep 13 '13 at 6:28

An axiomatic approach is not as valuable as you think.

This is true even in math...

Some math textbooks will define the algebra of real 2x2 matrices as "four real numbers in a 2x2 grid with the following addition and multiplication rules", and then later they will prove that matrices are equivalent to the linear operators on a 2D real vector space with a specific basis.

Other math textbooks will define the algebra of real 2x2 matrices as "the set of linear operators on a 2D real vector space with a specific basis", and then later they will prove that matrices are equivalent to four real numbers in a 2x2 grid with the following addition and multiplication rules.

Neither presentation is incorrect. It's just that one or the other presentation might work better pedagogically and in context.

After the presentation of this definition and proof, the textbook would hopefully tell you that the other presentation is also possible. These are two alternate ways to define the same thing.

In physics you have the same situation. See my answer to What's the best definition of energy.

I think what you're really looking for is not a "textbook that distinguishes laws from definitions", but a "textbook that is very clearly written and that emphasizes the historical or modern experiments that validate and motivate these concepts."

By the way you won't and can't get a whole consistent and correct picture in a lower-division physics course. The truth will emerge, clearer and clearer, as you take more courses, such as Lagrangian mechanics and quantum mechanics and quantum field theory etc. etc. They will shed new light on all these concepts, letting you see them from different angles.

The situation (definitions vs. axiomatic interpretation of laws, theories, etc.) is even less satisfying as you suggest in your question.

When Newton gave us his second law, F=ma, these quantities are defined in terms of length, time, mass, and energy, but I would point out that it does so without ever defining what a length or a time actually is. Einstein's later formulation of Special Relativity is proof that evidently, length and time are not exactly what Newton thought they were (as in, relative to the observer in different inertial reference frames).

This answer is not meant to discourage you from making either axiomatic structures in math or in physics, nor to try and define something new that you may observe in either discipline. Just be aware, there are limitations of both that are associated with the fact that you are doing so with your manifestly finite mind, and this will require you to use a certain amount of circular reasoning, no matter how airtight your math, definitions, or axioms might be.

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algebraic function sols - Maple Help

algcurves

 algfun_series_sol
 determine Puiseux expansions of an algebraic function with "nice" coefficients

 Calling Sequence algfun_series_sol(F, x, y, opts) algfun_series_sol(F, x=a, y, opts)

Parameters

 F - irreducible polynomial in two variables x - name of the independent variable y - name of the dependent variable opts - sequence of optional arguments of the form 'option'='value', where option is one of coeffs, msparse_order, mpoints_search, or RNF x=a - point of expansion

Description

 • An irreducible polynomial F of degree $N$ in the variable y with coefficients in a field $L\left(x\right)$ gives an algebraic extension L(x)[y]/(F) of $L\left(x\right)$, that is, it gives an algebraic function $y\left(x\right)$ which is a multivalued function. The algebraic function has $N$ Puiseux expansions at x=a. Each Puiseux expansion is of the form ${\sum }_{n=-K}^{\mathrm{\infty }}v\left(n\right){\left(x-a\right)}^{\frac{n}{r}}$ for some integer $r$ ($r$ is called the ramification index of the Puiseux expansion), and some integer $K$ and elements $v\left(n\right)$ in the algebraic closure of $L$.
 • The algcurves[algfun_series_sol] command determines those Puiseux expansions whose coefficients $v\left(n\right)$ have special properties ("nice" expansions). It can determine polynomial, rational, hypergeometric, d'Alembertian, m-sparse, or m-sparse m-hypergeometric sequences $v\left(n\right)$ (where m is an integer greater than $1$ and can be given as an optional argument or, otherwise, be chosen as the maximal value for each expansion by the command).
 • If the point of expansion is given and is of the form x=a, the command returns a set of "nice" expansions at a. Otherwise, the command finds all the "nice" points. These are the points where the "nice" expansions exist. In this case, the command returns a set of lists ['Point'=the point $A$, a set of "nice" expansions at $x=A$]. The point $A$ is an algebraic number or the symbol _a  which denotes any regular point of the given algebraic function.
 • The command determines the field $L$ from the input. The ground field $L$ of the computation is the smallest field such that F and the given or constructed points of expansion are in L(x)[y]. The field $L$ must be a rational number field or its algebraic extension by an algebraic number or several parameters.
 • The command builds a linear ordinary differential equation which the given algebraic function satisfies. Then it uses the package Slode for finding "nice" points and for building "nice" expansions for the constructed linear ordinary differential equation. The "nice" Puiseux expansion at the point x=a is represented in one of the following forms:
 1 For a polynomial, rational, hypergeometric, or d'Alembertian sequence $v\left(n\right)$:

$v\left(-K\right){\left(x-a\right)}^{-\frac{K}{r}}+...+v\left(\mathrm{\mu }-1\right){\left(x-a\right)}^{\frac{\mathrm{\mu }-1}{r}}+\sum _{n=\mathrm{\mu }}^{\infty }\gamma \left(n\right){\left(x-a\right)}^{\frac{n}{r}}$

 where $-K\le \mathrm{\mu }$ and $\mathrm{\gamma }\left(n\right)$ is a polynomial, rational, hypergeometric, or d'Alembertian term in $n$.
 2 For an m-sparse m-hypergeometric sequence $v\left(n\right)$:

$v\left(-K\right){\left(x-a\right)}^{-\frac{K}{r}}+...+v\left(m\mathrm{\mu }+M-1\right){\left(x-a\right)}^{\frac{m\mathrm{\mu }+M-1}{r}}+\sum _{n=\mathrm{\mu }}^{\infty }\gamma \left(n\right){\left(x-a\right)}^{\frac{mn+M}{r}}$

 where $0<=M and $\mathrm{\gamma }\left(n\right)$ is a hypergeometric term in $n$.
 3 For an m-sparse sequence $v\left(n\right)$:

$\mathrm{FPSstruct}\left(v\left(-K\right){\left(x-a\right)}^{-\frac{K}{r}}+...+v\left(m\mathrm{\mu }+M-1\right){\left(x-a\right)}^{\frac{m\mathrm{\mu }+M-1}{r}}+\sum _{n=\mathrm{\mu }}^{\infty }v\left(mn+M\right){\left(x-a\right)}^{\frac{mn+M}{r}},\mathrm{rec}\right)$

 where $\mathrm{rec}$ is a homogeneous linear recurrence with polynomial coefficients for the sequence $v\left(mn+M\right)$.

Options

 • Each optional argument is of the type 'option' = 'value'. The following options are supported.
 coeffs
 Specifies the type of the coefficients; the default value is hypergeom.
 – dAlembertian, that is, $v\left(n\right)$ is annihilated by a linear recurrence operator that can be written as a composition of first-order operators (see LinearOperators).
 – hypergeom, that is, $\frac{v\left(n+1\right)}{v\left(n\right)}$ is a rational function in $n$ for all large enough $n$.
 – rational, that is, $v\left(n\right)$ is a rational function in $n$ for all large enough $n$.
 – polynomial, that is, $v\left(n\right)$ is a polynomial in $n$ for all large enough $n$.
 – sparse, that is, for all large enough $n$ only each m-th coefficient $v\left(mn+M\right)$ is nonzero, where $2\le m$ is an integer (it is called the sparse order) and $M$ is an integer such that $0\le M.
 – mhypergeom, that is, $v\left(n\right)$ satisfies the property of the option sparse and $\frac{v\left(\left(n+1\right)m+M\right)}{v\left(mn+M\right)}$ is a rational function in $n$ for all large enough $n$.
 msparse_order
 Specifies the sparse order. This option is given as 'msparse_order'=n, where n is an integer greater than 1. This option is useful only if one of the options 'coeffs'='mhypergeom' or 'coeffs'='sparse' is given.
 mpoints_search
 Specifies the algorithm for finding the "nice" points. This option is useful only in the case when the point of expansion is not specified and one of the options 'coeffs'='mhypergeom' or 'coeffs'='sparse' is given. The default value is full.
 – full, that is, the algorithm, which gives all "nice" points
 – quick, that is, the algorithm, which may miss some "nice" points but takes considerably less time than the full algorithm
 RNF
 Specifies the representation of $v\left(n\right)$. This option makes sense only if one of the options 'coeffs'='hypergeom' or 'coeffs'='mhypergeom' is given. The default value is false.
 – false, that is, $v\left(n\right)$ is represented as in the package LREtools.
 – true, that is, $v\left(n\right)$ is the first minimal representation of a hypergeometric term (see RationalNormalForms).

Examples

 > $f≔{x}^{11}-2{x}^{8}+{x}^{5}-{y}^{2}{x}^{6}+2{y}^{2}{x}^{3}-{y}^{2}+2y{x}^{3}-2y-1$
 ${f}{≔}{{x}}^{{11}}{-}{2}{}{{x}}^{{8}}{-}{{y}}^{{2}}{}{{x}}^{{6}}{+}{{x}}^{{5}}{+}{2}{}{{y}}^{{2}}{}{{x}}^{{3}}{+}{2}{}{y}{}{{x}}^{{3}}{-}{{y}}^{{2}}{-}{2}{}{y}{-}{1}$ (1)
 > $\mathrm{algcurves}\left[\mathrm{algfun_series_sol}\right]\left(f,x=0,y,'\mathrm{coeffs}'='\mathrm{mhypergeom}'\right)$
 $\left\{{-}{1}{-}{{x}}^{{5}}{{2}}}{-}{{x}}^{{3}}{-}{{x}}^{{6}}{-}{{x}}^{{9}}{-}\left({\sum }_{{\mathrm{_n}}{=}{4}}^{{\mathrm{\infty }}}{}{\left(\sqrt{{x}}\right)}^{{6}{}{\mathrm{_n}}}\right)\right\}$ (2)
 > $f≔{x}^{5}{y}^{6}-1$
 ${f}{≔}{{x}}^{{5}}{}{{y}}^{{6}}{-}{1}$ (3)
 > $\mathrm{algcurves}\left[\mathrm{algfun_series_sol}\right]\left(f,x,y\right)$
 $\left\{\left[{\mathrm{Point}}{=}{0}{,}\left\{\frac{{1}}{{{x}}^{{5}}{{6}}}}\right\}\right]{,}\left[{\mathrm{Point}}{=}{\mathrm{_a}}{,}\left\{\frac{{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{6}}{-}{\mathrm{_a}}\right){}\left({\sum }_{{\mathrm{_n}}{=}{0}}^{{\mathrm{\infty }}}{}\frac{{\left({-}\frac{{1}}{{\mathrm{_a}}}\right)}^{{\mathrm{_n}}}{}{\mathrm{\Gamma }}{}\left(\frac{{5}}{{6}}{+}{\mathrm{_n}}\right){}{\left({x}{-}{\mathrm{_a}}\right)}^{{\mathrm{_n}}}}{{\mathrm{\Gamma }}{}\left({\mathrm{_n}}{+}{1}\right)}\right)}{{\mathrm{_a}}{}{\mathrm{\Gamma }}{}\left(\frac{{5}}{{6}}\right)}\right\}\right]\right\}$ (4)
 > $f≔x{y}^{2}-1-2y-{y}^{2}$
 ${f}{≔}{{y}}^{{2}}{}{x}{-}{{y}}^{{2}}{-}{2}{}{y}{-}{1}$ (5)
 > $\mathrm{algcurves}\left[\mathrm{algfun_series_sol}\right]\left(f,x,y,'\mathrm{coeffs}'='\mathrm{polynomial}'\right)$
 $\left\{\left[{\mathrm{Point}}{=}{0}{,}\left\{{-}\left({\sum }_{{\mathrm{_n}}{=}{0}}^{{\mathrm{\infty }}}{}{\left(\sqrt{{x}}\right)}^{{\mathrm{_n}}}\right)\right\}\right]\right\}$ (6)
 > $\mathrm{algcurves}\left[\mathrm{algfun_series_sol}\right]\left(f,x=4,y,'\mathrm{coeffs}'='\mathrm{dAlembertian}'\right)$
 $\left\{{-}\frac{{\sum }_{{\mathrm{_n}}{=}{0}}^{{\mathrm{\infty }}}{}{\left({-1}\right)}^{{\mathrm{_n}}}{}{{3}}^{{-}{\mathrm{_n}}}{}\left({-}\left({\sum }_{{\mathrm{_n1}}{=}{0}}^{{\mathrm{_n}}{-}{1}}{}\frac{{{3}}^{{\mathrm{_n1}}{+}{1}}{}{{4}}^{{-}{\mathrm{_n1}}}{}{\mathrm{\Gamma }}{}\left({\mathrm{_n1}}{+}\frac{{1}}{{2}}\right)}{{\mathrm{\Gamma }}{}\left({\mathrm{_n1}}{+}{2}\right)}\right){+}{4}{}\sqrt{{\mathrm{\pi }}}\right){}{\left({x}{-}{4}\right)}^{{\mathrm{_n}}}}{{12}{}\sqrt{{\mathrm{\pi }}}}{,}\frac{{\sum }_{{\mathrm{_n}}{=}{0}}^{{\mathrm{\infty }}}{}{\left({-1}\right)}^{{\mathrm{_n}}}{}{{3}}^{{-}{\mathrm{_n}}}{}\left({-}\left({\sum }_{{\mathrm{_n1}}{=}{0}}^{{\mathrm{_n}}{-}{1}}{}\frac{{{3}}^{{\mathrm{_n1}}{+}{1}}{}{{4}}^{{-}{\mathrm{_n1}}}{}{\mathrm{\Gamma }}{}\left({\mathrm{_n1}}{+}\frac{{1}}{{2}}\right)}{{\mathrm{\Gamma }}{}\left({\mathrm{_n1}}{+}{2}\right)}\right){+}{12}{}\sqrt{{\mathrm{\pi }}}\right){}{\left({x}{-}{4}\right)}^{{\mathrm{_n}}}}{{12}{}\sqrt{{\mathrm{\pi }}}}\right\}$ (7)


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Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What

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Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What  [#permalink]

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15 Mar 2015, 23:06
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Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What is the value of Q?

A. 3
B. 3 1/7
C. 3 2/7
D. 3 3/7
E. 3 1/2

Kudos for a correct solution.

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Re: Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What  [#permalink]

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22 Mar 2015, 09:44
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Bunuel wrote:
Attachment:
ppocg_img3.png
Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What is the value of Q?

A. 3
B. 3 1/7
C. 3 2/7
D. 3 3/7
E. 3 1/2

Kudos for a correct solution.

Oops , i am missing a chance to earn a easy kudo .
here we go ,
Slope of line when calculated with (-5,-1) (2,2) = $$3/7$$
Slope of line when calculated with (2,2) (5,q) = $$\frac{q-2}{5-2}$$
$$\frac{q-2}{3} = \frac{3}{7}$$
$$q= 3 \frac{2}{7}$$
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Re: Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What  [#permalink]

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16 Mar 2015, 22:09
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Bunuel wrote:
Attachment:
ppocg_img3.png
Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What is the value of Q?

A. 3
B. 3 1/7
C. 3 2/7
D. 3 3/7
E. 3 1/2

Kudos for a correct solution.

I am surprised no one has solved this question yet!

Using the slope concept, if x co-ordinate increases by 7 (from (– 5, – 1) to (2, 2)), y co-ordinate increases by 3 (from -1 to 2). So when x co-ordinate increases by further 3 (from 2 to 5), the y co-ordinate will increase by (3/7)*3 = 9/7 = 1 (2/7)
So y co-ordinate will become 2 + 1(2/7) = 3 (2/7)
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Re: Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What  [#permalink]

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17 Mar 2015, 09:50
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Here is my approach:

m= 2+1 / 2+5 = 3/7

y = (3/7)x + b
2 = (3/7)2 + b
b = 8/7

y = (3/7)5 + (8/7)
y = 23/7 = 3 2/7

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Re: Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What  [#permalink]

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22 Mar 2015, 12:04
C

slopes are same for all the points

so we can equate the slope3/7 to the one with Q co-ordianare

getting 23/7
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Re: Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What  [#permalink]

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23 Mar 2015, 05:33
Bunuel wrote:

Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What is the value of Q?

A. 3
B. 3 1/7
C. 3 2/7
D. 3 3/7
E. 3 1/2

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

Think about this visually. From (–5, –1) to (2, 2), the line moves right 7 and up 3, so that’s a slope of 3/7.

From the point (2, 2) to the point (5, Q), there’s a horizontal distance of 3, an unknown vertical distance — call it h, so that h = Q – 2, or Q = h + 2. The ratio of this vertical and horizontal distance must equal the slope.

h/3 = 3/7 --> h = 9/7.

That’s h. Now, add 2 to get Q.

Q = h + 2 = 3 2/7.

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Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What  [#permalink]

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13 May 2015, 10:10
Let the 3 points be A(-5, -1) B(2, 2) and C(5, Q).

Equation of line passing through A and B is :
(y-y1) = M(x-x1)
y + 1 = 3/7( x + 5) .. here slope(M) is calculated as : y2 - y1/x2 -x1( gives 3/7)

Since point C(5, Q) passes through the line, so it should satisfy the above equation.
Q + 1 = 3/7 (5 + 5)
Q = 30/7 - 1
Q = 23/7 or 3 2/7

Option C
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Re: Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What  [#permalink]

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25 Feb 2017, 06:54
As slope for the line remains constant, slope between (-5,-1) and (2,2) would be the same.
thus, {2- (-1)}/{2-(-5)} = (Q-2)/(5-2)
=> 3/7=(Q-2)/3
=> Q-2=9/7 => Q=23/7 = 3 2/7
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Re: Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What  [#permalink]

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10 Apr 2017, 00:38
VeritasPrepKarishma wrote:
Bunuel wrote:
Attachment:
ppocg_img3.png
Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What is the value of Q?

A. 3
B. 3 1/7
C. 3 2/7
D. 3 3/7
E. 3 1/2

Kudos for a correct solution.

I am surprised no one has solved this question yet!

Using the slope concept, if x co-ordinate increases by 7 (from (– 5, – 1) to (2, 2)), y co-ordinate increases by 3 (from -1 to 2). So when x co-ordinate increases by further 3 (from 2 to 5), the y co-ordinate will increase by (3/7)*3 = 9/7 = 1 (2/7)
So y co-ordinate will become 2 + 1(2/7) = 3 (2/7)

Responding to a pm:

Quote:

I really did not understand how did u get 2 ahead of 9/7.
Also, the concept of slopes that you used, can u explain it to me. I do understand that slopes is rise/ run. or I have read it in school as Tan angle that line makes with x -axis. Now, the approach you used I drew inference that Slope is with an increase in an x coordinate the y coordinate increases with according to slope .

If you write $$\frac{9}{7}$$ in mixed fractions, you get $$1\frac{2}{7}$$.

As for the slope concept, I have explained it in detail in this blog post: https://www.veritasprep.com/blog/2016/0 ... line-gmat/
Take a look and get back to me if any doubts remain.
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Re: Line J passes through the points (– 5, – 1), (2, 2), and (5, Q). What &nbs [#permalink] 10 Apr 2017, 00:38
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text:
# Margin of Error Calculator

## How does Margin of Error calculator work?

The working principle of the margin of error calculator is very easy;
• Enter the sample size
• Enter the population size
• Enter the sample proportion (max = 1)
• Select the confidence level
• Hit the "calculate" button to get the step-by-step solution.
• You can erase all the values by clicking on the "reset" button.

## Margin of error calculator

The margin of error calculates the MOE using sample size (n), population size (P), sample proportion (p), and confidence level. It gives a step-by-step solution to the problem.

## What is the margin of error?

The margin of error (MOE) for a survey tells you ways close you'll be able to expect the survey results to be to the proper population price. E.g., In a random survey, we came to know that 55% of the population favored Brand X over Brand Y with 2% MOE.

According to this survey, the actual population percentage that voted for Brand X falls within the range of 55% ± 2%. MOE refers to the degree of error in the results that we get from surveys of random sampling.

## The formula for the margin of error:

There are two different cases to calculate the margin of error of a raw sample.

### Margin of error with FPC:

The formula to calculate the MOE with FPC is as follows:

MOE with finite population correction :

Now let’s have a look at the FPC, in the sample survey statistics, this concept plays a vital role. It is a factor that is used to balance the estimated variance for a hypothetical total or mean.

In the above formula,

• “P hat” is the sample proportion
• “P” is the population size
• “z” is the z-score
• “n” is the sample size

### Simple margin of error:

The simple MOE can be calculated using the following formula:

• “P hat” is the sample proportion
• “z” is the z-score
• “n” is the sample size

## Procedure to calculate the margin of error:

In this section, with the assistance of examples, the procedure is explained briefly.

Example 1:

Calculate the MOE with FPC if the sample proportion is 0.2, the population size is 1029, the sample size is 112, and the confidence interval is 95%.

Solution:

Step 1: Extract the data

Sample proportion =  = 0.2

Population size = p = 1029

The confidence interval of z at 95% = 1.960

Sample size = n = 112

Step 2: enter the values in the following formula

MOE with FPC =

MOE with FPC = 1.960 {{0.2 (1 - 0.2)}/ {112(1029 – 1)/ (1029 – 112)}

MOE with FPC = 1.960 {0.2 (0.8) / {(112) * (1028) / (917)}}}

MOE with FPC = 1.960 { 0.2(0.8)/ {112(1.121)}

MOE with FPC = 1.960 {0.16/ {125.55}}

MOE with FPC = 1.960 * 0.03569761

MOE with FPC = 0.06996

MOE with FPC = 6.997%

## References:

Margin of error | Statistics How To.

Your guide to margin of error | Qualtrics


text:
Question

# In the given figure, $$\rho_{1} ,\rho _{2}$$ density of liquid in container (1) & (2), and $$\rho_{3}$$ density of liquid (2) of manometer The level of fluid in both the container is same. Which is the correct condition?

A
ρ2>ρ1
B
ρ2<ρ1
C
ρ2=ρ1
D
cant say

Solution

## The correct option is C $$\rho_2 < \rho_1$$From the height of the liquid in the manometer we see that the pressure exerted by the liquid in the container 1 is greater than that exerted by the liquid in container 2. Thus the density of the liquid in 1 is greater than the density in container 2.Physics

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# Exponential relationship issue

I read this relation and I am not sure why this is true, is it I can't see why it would be?

$$(e^{ -i\pi/2})^{ -ix}\approx ie^{-\pi x/2}$$ I get that $e^{i\pi/2}=-i$, but I can't see why this relation would be true.

-
There must be an error. The quantity on the left is equal to $e^{-\pi x/2}$ exactly; the extra $i$ is not correct. –  Michael C. Grant Apr 26 at 12:17
I don't think the relation is true. The expression on the left is $(-i)^{-ix}$. There is no universal convention for raising $-i$ to a non-integer power, which is what you are trying to do. –  Stefan Smith Apr 26 at 23:31

You have to be careful working with complex exponentials because of problems with multivalued logarithms. The exponential $a^b$ for $a,b \in \mathbb{C}$ is usually defined in terms of the multivalued logarithm as $e^{b \log a}$.

As a multivalued expression, we have \begin{align*} (e^{-i \pi/2 })^{-ix} &= e^{-ix \log (e^{-i \pi/2 })} \\ &= e^{-ix (-i\pi/2 + 2\pi i k) } && (k \text{ an integer; } k = 0 \text{ for principle log})\\ &= e^{-x\pi/2 + 2\pi x k} \\ &= e^{2\pi k x}e^{-x\pi/2} \end{align*}

Therefore, for your statement to be true (in the sense that for one branch it works) we would have to have \begin{align*} e^{2\pi k x}e^{-x\pi/2} &= i e^{-x\pi/2} \\ \iff e^{2\pi k x} &= i \\ \iff e^{2\pi k x} &= e^{i \pi / 2} \\ \iff 2 \pi k x &\equiv i \pi / 2 \pmod {2\pi i} \end{align*}

For this to be true, $x$ must be purely imaginary. Let $x = i(m + r)$ where $m$ is an integer and $0 \le r < 1$ real. \begin{align*} 2 \pi k (i(m + r)) &\equiv i \pi / 2 \pmod {2\pi i} \\ \iff 2 \pi i m k + 2 \pi i k r &\equiv i \pi / 2 \pmod {2\pi i} \\ \iff mk + kr &\equiv 1 / 4 \pmod {1} \\ \iff kr &\equiv 1 / 4 \pmod {1} \end{align*}

In summary: the only time your statement could actually be sort of true is when $x$ is purely imaginary, and the fractional part of the imaginary part of $x$ (i.e. $r$ above) multiplied by some integer $k$ has a fractional part of $1/4$.

-

Given $z$, $\omega \in \mathbb{C}$, $z\neq 0$

$$z^{\omega}=e^{\omega Log(z)}$$

where $$Log(z)=log(|z|)+i(arg(z))$$

Let $z = e^{-i\frac{\pi}{2}},-$ $\omega=-ix$

Then

$$(e^{-i\frac{\pi}{2}})^{-ix}=e^{-ix(Log(e^{-i\frac{\pi}{2}}))}$$

$$Log(e^{-i\frac{\pi}{2}})=log|1|+i(-\frac{\pi}{2})=-i\frac{\pi}{2}$$

so

$$(e^{-i\frac{\pi}{2}})^{-ix}=e^{-ix(-i\frac{\pi}{2})}=e^{-\frac{x\pi}{2}}$$

I'm sorry I don't get the factor $i$

-
yeah that is what i get but the factor i is important, mm il have another think –  user73685 Apr 26 at 12:07
@user73685 can you give some context? It may be very important –  Nivalth Apr 26 at 12:09
$$(e^a)^b = e^{ab}$$ What do you get when you do this?
This isn't true in general for complex $a$ and $b$ -- not even as a multivalued expression. –  Goos Apr 26 at 13:30


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# Thread: i need help in integration question?

1. ## i need help in integration question?

solve this question.......

2. Originally Posted by khhratmath123
solve this question.......
Show us your work so far and then we can guide you through where you're stuck.

As a pointer, the way I would approach this is to factorise the denominator, and then use partial fractions to get two fractions which are easy to integrate.

3. Originally Posted by khhratmath123
solve this question.......
Also, if you are having problems factorising the fraction, remember you can take a factor of -1 out, so your problem becomes:

$(-1)\int^{-2}_{-3}\frac{1}{x^2 + 6x+5}dx$

4. i have solved this question but i have doubt....plz check..is it right?

5. this was easy:

∫((dx)/(x²+6x+5))
=∫((dx)/((x+1)(x+5)))

now use partial fractions ,get the constants and put the limits.

your answer is definitely wrong as it should come in terms of logs.

6. Originally Posted by khhratmath123
i have solved this question but i have doubt....plz check..is it right?
Correct.

7. Originally Posted by khhratmath123
i have solved this question but i have doubt....plz check..is it right?
No. I see what you have done, but your answer should be:

$[\frac{1}{2}tanh^{-1}(\frac{x+3}{2})]^{-2}_{-3}$

Or the way that me and Pulock suggested, this way you would get the actual logs instead of the inverse hyperbolic.

8. Originally Posted by craig
No.
$\int\frac{dx}{(2)^2-(x+3)^2}$

Let $u=x+3\Rightarrow{du}=dx$

$\Rightarrow\int\frac{dx}{(2)^2-(x+3)^2}=\int\frac{du}{(2)^2-(u)^2}$

9. Originally Posted by VonNemo19
$\int\frac{dx}{(2)^2-(x+3)^2}$

Let $u=x+3\Rightarrow{du}=dx$

$\Rightarrow\int\frac{dx}{(2)^2-(x+3)^2}=\int\frac{du}{(2)^2-(u)^2}$
I see how you derived that. However, if it was:

$\int\frac{du}{\sqrt{(2)^2-(u)^2}}$ that would make his answer right.

But as far as I remember, $\int\frac{du}{1-u^2} = tanh^{-1} u$ ?

10. Originally Posted by craig
I see how you derived that. However, if it was:

$\int\frac{du}{\sqrt{(2)^2-(u)^2}}$ that would make his answer right.

But as far as I remember, $\int\frac{du}{1-u^2} = tanh^{-1} u$ ?
My mind saw a radical where there was none.

11. Originally Posted by VonNemo19
My mind saw a radical where there was none.
Easy to do

12. thanks for all
i have solved by partial fraction method


text:
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text:
Following the tendencies for improved web experience we at Telerik are pushing more and more operations to be performed on the client. Now with the Q2 2013 Beta release we have turned our Grid for ASP.NET AJAX into client-side editing machine bringing improved performance and better user experience with the new Batch Editing feature. The fun comes when you see the functionality in action and try to customize   by integrating grid editing and data visualization but before that let’s go through the main use cases.

## What is batch editing?

Batch editing is a new type of inline editing which brings the editing experience of RadGrid closer to this of Excel. Upon a click/double-click (whatever the developer specifies), the user will be able to open the cell for editing and edit as many cells as they like. By hitting the Save button just once, all their changes will be preserved.

## Getting started

I should admit – the best part is turning the feature on. The only thing you need to do is set the GridTableView.EditMode to Batch. It supports automatic operations so you could easily switch it with your current editing mode. If you need more customization over the behavior you could always use the GridTableView.BatchEditingSettings.

## Using templates

We have done our best to make everything automatic without the need for additional JavaScript code. If GridTemplateColumn is created and the EditItemTemplate is specified and one edit control is placed in the template, the grid will automatically find the control and get/set values from\to it. In rare cases when there are multiple edit controls in the template and control over them is required, the RadGrid exposes four client-side events for manual control over the editors.

## Using manual CRUD operations

Performing manual operations using batch editing is similar to the current code that you would use.  Subscribe to RadGrid UpdateComamnd, DeleteCommand, InsertCommand commands or use the new GridBatchEditingCommand where you could get all commands at once that have been performed on the client. If you choose to not use the new event you will need to cast the CommandArgument property to GridBatchEditingEventArgument and use the NewValues, OldValues hash tables properties to perform CRUD operations.

## How to integrate batch editing and data visualization

Batch editing enables advanced editing functionalities to be developed purely on the client without performing a postback. The implementation that we will go through shows how real time data visualization using Telerik’s  ASP.NET Chart based on HTML5 integrates with RadGrid’s editing capabilities. The idea is to dynamically refresh the chart during changes in the product price.

The implementation is straightforward and easy to implement. The RadGrid and RadHtmlChart are bound with the same data source and the batch editing functionality is turned on. The main client-side code involved is in updating the chart by detecting changes in the numeric text box as shown below.  In the KeyDown event the numeric value is taken and provided to the updateChart function which loops through the chart data source and sets the ProductPrice value. Additionally, for better look and feel when selecting a row the associated chart item will pop out. You could take a look at the main JavaScript functions below and if you are interested in the whole implementation you could download the source code from here.

function OnKeyDown(numericElement, e)
{
    var dataItem = $find($telerik.$(numericElement).closest("tr")[0].id);  var id = parseInt(dataItem.getDataKeyValue("Id"));  setTimeout(function ()  {  var productPrice =$find(numericElement.id).get_textBoxValue();
        updateChart(id, productPrice);
    });
}

function RowSelected(sender, args)
{
    var dataItem = args.get_gridDataItem();
    var id = parseInt(dataItem.getDataKeyValue("Id"));
    updateChart(id);
}

function updateChart(id, productPrice)
{
    productPrice = parseInt(productPrice);
    var htmlChart = $find("<%= RadHtmlChart1.ClientID %>");  var dataSource =$telerik.\$.parseJSON(htmlChart.get_dataSourceJSON());
    for (var i = 0; i < dataSource.length; i++)
    {
        dataSource[i].IsExploded = false;
        if (dataSource[i].Id === id)
        {
            dataSource[i].IsExploded = true;
            if (productPrice && productPrice !== 0)
            {
                dataSource[i].ProductPrice = productPrice;
                break;
            }
        }
    }
    htmlChart.set_dataSource(dataSource);
    htmlChart.set_transitions(false);
    htmlChart.repaint();
}

## How do you like it?

This feature is still in Beta and we are open to your feedback to make the feature better for our official release in mid-June. Comment below or share your insights in the Beta forums. You can download the Beta to play with the new feature or check it out on the Beta demos


text:
# Mathematics Extension 2 – Polynomials

Polynomials

This topic extends the basis knowledge built in the Extension 1 course to encompass complex numbers. We shall break this topic into sections for study.

## Polynomials with Integer Roots

Here, the most important theorem to remember is the integer root theorem. The theorem is stated as

If a polynomial with integer coefficients has an integer root, then the root is a factor of the constant term.

We shall not prove this theorem, but shall consider examples to illustrate use of this theorem.

### Example 1

Find all the zeros of the polynomial $Latex formula$.

### Solution 1

Here, we firstly must find any roots. It is customary to check $Latex formula$ initially. It happens that:

$Latex formula$

and hence we have that $Latex formula$ is a factor of the polynomial (by the factor theorem).

That is, after performing long division (or simply by inspection) we obtain,

$Latex formula$

Now, upon factoring the quadratic we have,

$Latex formula$

Hence we have that the zeros of the polynomial are given by $Latex formula$.

### Example 2

Factorise the polynomial $Latex formula$ over the real numbers.

### Solution 2

To factorise successfully there are two methods we may use here; either we factorise using the method of grouping, or we test factors of the constant term. We shall try the latter to illustrate the techniques involved.

So, we test $Latex formula$. We find that:

$Latex formula$

Hence we have that $Latex formula$ is a factor by the factor theorem.

Performing long division, and factorising the polynomial gives:

$Latex formula$

That is,

$Latex formula$

over the $Latex formula$ numbers.

Note: At times, it is recommended that a similar theorem regarding rational roots be memorised as well. We do not recommend this, as this theorem is not stated in the syllabus. Only cases involving integer roots may be examined.

## Fundamental Theorem of Algebra

This theorem simply states that:

A polynomial of degree $Latex formula$ will have exactly zeros over the Complex field allowing for multiplicities.

This theorem is of immense importance in this topic, and should b committed to memory (verbatim).

Now, the importance of this theorem does not lie in its ability to solve equations. In fact, in that regard it is absolutely useless. The importance lies in its ability to predict the number of roots of the polynomial. This proof of existence turns out to be quite important even though the theorem gives no indication of how to find the roots.

### Example 3

Find all the roots of the polynomial $Latex formula$.

### Solution 3

We know by the fundamental theorem of algebra, that there must exist 4 roots for this Quartic polynomial, allowing for multiplicities. We hence have that:

$Latex formula$

Now, we treat the polynomial as a quadratic in terms of $Latex formula$. Factorising gives:

<=>$Latex formula$

Now, solving the equations $Latex formula$ and $Latex formula$ gives us,

$Latex formula$

Hence we have that the roots of the Quartic are $Latex formula$ only (We know there exist no other roots by the fundamental theorem of algebra).

## Complex Conjugate Root Theorem

This theorem simply states that;

If $Latex formula$ is a complex (non-real) root of the polynomial $Latex formula$ then the complex conjugate of $Latex formula$, is also a root of $Latex formula$.

We shall investigate a proof of this theorem as it is required knowledge by the syllabus.

 Theorem:If $Latex formula$ is a complex (non-real) root of the polynomial $Latex formula$ where $Latex formula$ has real coefficients, then the complex conjugate of $Latex formula$, is also a root of $Latex formula$ .Proof:Let $Latex formula$ be a non- real root of a real polynomial $Latex formula$. Then we have $Latex formula$ Taking complex conjugates of both sides gives; $Latex formula$ But, $Latex formula$ and $Latex formula$ since $Latex formula$ is real. Thus, using the fact that $Latex formula$ we now have $Latex formula$ But $Latex formula$. Thus; $Latex formula$ But the expression on the left is simply $Latex formula$. Thus $Latex formula$ That is, $Latex formula$ is also a root of $Latex formula$.

Students should memorise the theorem and proof shown here. We shall now consider an example which illustrates the use of this theorem.

### Example 4

Given that $Latex formula$ is a root of the equation latex]P(z)=2z^{3}-az^{2}+bz-2[/latex] where $Latex formula$ and $Latex formula$ are real:

a) Find the values of $Latex formula$ and $Latex formula$.

b) Hence, factorise $Latex formula$ over the real numbers.

### Solution 4

a) So we simply make the substitution into the equation, upon which we equate real parts and imaginary parts. i.e.

$Latex formula$

Upon expanding we obtain,

$Latex formula$

That is,

$Latex formula$

Equating real and imaginary parts gives:

$Latex formula$

That is,

$Latex formula$

b) To factorise $Latex formula$ over the real numbers, we simply note that since $Latex formula$ is a root of the polynomial with real coefficients then it follows that the complex conjugate of $Latex formula$, $Latex formula$ is also a root of the polynomial. Hence we have that the two of the roots of the polynomial are given by $Latex formula$ and $Latex formula$. Hence upon consideration of the sum of the roots one at a time, we have that the third root of the cubic, $Latex formula$ is given by,

$Latex formula$

That is, we have that:

$Latex formula$

However this is over the complex field.

To obtain the factorisation over the reals, we consider the product of the two complex factors.

$Latex formula$

=> $Latex formula$

$Latex formula$

Which is the factorisation over the real field.

Note: It should be noted that the product of $Latex formula$ is always real. This is because, $Latex formula$. Obviously $Latex formula$ which is real, and $Latex formula$ which is real. Hence we have that the product of two conjugate factors gives a real quadratic factor. Consequently, we can deduce that all real polynomials may be expressed as the product of real linear and quadratic factors, since all complex roots occur in pairs for complex roots of a real polynomial, and since the product of the two conjugate factors equals to a real quadratic polynomial.

## Multiple Root Theorem

The multiple root theorem simply states that;

If $Latex formula$ has $Latex formula$ where $Latex formula$ as a root of $Latex formula$multiplicity, then $Latex formula$ has $Latex formula$ as a root of multiplicity $Latex formula$.

This theorem is easily proved, and both the theorem and proof should be memorised. Below is a proof.

### Theorem

If $Latex formula$\textit{ has }$Latex formula$ where $Latex formula$ as a root of multiplicity $Latex formula$ , then $Latex formula$ has $Latex formula$ as a root of multiplicity $Latex formula$.

### Proof

Let $Latex formula$ be a polynomial given by

$Latex formula$

Where $Latex formula$ is a root of multiplicity $Latex formula$ and $Latex formula$ is some polynomial. Then, by the product rule we have

$Latex formula$

$Latex formula$

$Latex formula$

Where $Latex formula$ is some polynomial. Thus, $Latex formula$ has $Latex formula$ as a root of multiplicity $Latex formula$.

Here are some commonly asked questions regarding his theorem.

### Example 5

$Latex formula$ has a root of multiplicity $Latex formula$, find all the roots of $Latex formula$.

### Solution 5

So, if $Latex formula$ has a root of multiplicity $Latex formula$, then $Latex formula$ has a root of multiplicity $Latex formula$. Differentiating $Latex formula$ and solving the subsequent equation formed upon setting $Latex formula$ gives;

$Latex formula$

=> $Latex formula$

Now, substituting these both back into the original equation shows that $Latex formula$. Hence we have that $Latex formula$ is a double root by the multiple roots theorem. Now, using the sum of the roots one at a time gives:

$Latex formula$

Hence we have that the roots of $Latex formula$ are given by $Latex formula$.

### Example 6

If the given polynomial $Latex formula$ has a multiple root at $Latex formula$, show that

$Latex formula$

### Solution 6

It is given that the root $Latex formula$ is a multiple root but the multiplicity is not given. Thus we can only assume that it is the root of the first derivative as well.

Let $Latex formula$.

Thus, $Latex formula$ for the multiple root.

$Latex formula$

i.e.

$Latex formula$

Now, $Latex formula$ since $Latex formula$ is a root of $Latex formula$.

Thus,

$Latex formula$

But,

$Latex formula$

$Latex formula$

## Roots and Coefficients

We shall firstly introduce a table giving the relationship between the coefficients of each polynomial, and the roots.

 Order of Polynomial $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$

This notion was first introduced in the Extension 1 Mathematics topic Polynomials. The pattern is very easy to see, and students should commit this to memory.

Now, questions involving roots and coefficients are generally difficult, although they may be occasionally easier. We shall study an example to illustrate use of these techniques.

### Example 7

It is known that two of the four roots of the equation $Latex formula$ are equal but opposite in sign (i.e. $Latex formula$ ). Show that the other two roots are of the form $Latex formula$ and $Latex formula$.

### Solution 7

Let the roots be $Latex formula$. We have that:

$Latex formula$

We also have that:

$Latex formula$

Hence we have,

$Latex formula$

Now, we also have that:

$Latex formula$

Hence we have that

$Latex formula$

Hence we have two simultaneous equations to solve, namely;

$Latex formula$

$Latex formula$

Now, rearranging $Latex formula$ and substituting into $Latex formula$ gives:

$Latex formula$

=> $Latex formula$

$Latex formula$

Hence we have that

$Latex formula$

Without the loss of generality.

Hence we have that the other two roots, $Latex formula$ and $Latex formula$ are of the form $Latex formula$ where

$Latex formula$

## Finding a Polynomial given its Roots

To find the equation of a polynomial given its roots is relatively simple. If given the roots directly, then simply create a product of the factors and expand. If given the roots in terms of the roots of another polynomial, then we simply apply the algebraic trick of equating the roots of the polynomial to be found to some variable $Latex formula$ say, and then simply find the value of the original root in terms of the variable $Latex formula$, then simply substitute and rearrange to obtain in polynomial form. This trick is illustrated in the next examples. Bear in mind however that you specifically may be asked to find the equation of a polynomial using the sum and product of the roots. In this case, the algebraic trick that we apply here will not obtain you any marks.

### Example 8

Given that the equation $Latex formula$ has roots $Latex formula$, find the polynomial equation with roots $Latex formula$.

### Solution 8

In this method we shall use the algebraic substitution.

Let $Latex formula$

$Latex formula$

Now, substituting the value of $Latex formula$ into the original equation gives:

$Latex formula$

$Latex formula$

$Latex formula$ has roots $Latex formula$.

(We must express the equation in polynomial form).

### Example 9

Find the equation with roots $Latex formula$ given that the equation $Latex formula$ has roots $Latex formula$,in polynomial form.

### Solution 9

Let $Latex formula$

Inserting this value of $Latex formula$ into the given equation gives:

$Latex formula$

$Latex formula$

The question requires the equation to be in polynomial form.

Thus,

$Latex formula$

Squaring both sides gives:

$Latex formula$

$Latex formula$

$Latex formula$ Expressing the answer in polynomial form.

### Example 10

Given that $Latex formula$ has roots $Latex formula$, find the polynomial equation with roots, $Latex formula$.

### Solution 10

Let $Latex formula$.

$Latex formula$

Inserting $Latex formula$ into the given equation gives:

$Latex formula$

$Latex formula$

$Latex formula$

Expressing the equation in polynomial form.

## Finding Expressions involving the Roots of a Polynomial

In this section, we are essentially required to find the value of certain expressions given the roots of a polynomial. To illustrate the notions involved in this section, we shall answer an example question.

### Example 11

The cubic equation $Latex formula$ has 3 real non zero roots $Latex formula$. Find in terms of the constants $Latex formula$ and $Latex formula$, the values of:

a) $Latex formula$

b) $Latex formula$

c) $Latex formula$

### Solution 11

a) Now, to find the value of this we consider the following;

$Latex formula$

Using the relationship between the rots and coefficients we have that:

$Latex formula$

b) To find the next value, we must consider the following “trick”. Note that this “trick” is quite standard and must be known by every student.

Since $Latex formula$ is a root of the polynomial equation, then we have that:

$Latex formula$

The same condition holds for $Latex formula$ and $Latex formula$:

$Latex formula$

$Latex formula$

Now, performing $Latex formula$ gives:

$Latex formula$

=> $Latex formula$

c) We now perform the operation $Latex formula$ which gives:

$Latex formula$

=> $Latex formula$

In general, questions involving this subtopic are very similarly answered. The only formula to be memorised is the expansion for the sum of the squares of the roots. That is,

$Latex formula$]\alpha ^{2}+\beta ^{2}+\gamma ^{2}=(\alpha +\beta +\gamma )^{2}-2(\alpha \beta +\alpha \gamma +\beta \gamma )[/latex]

We shall now look at a further example to further reinforce technique.

### Example 12

Given that the equation $Latex formula$ has roots $Latex formula$, find:

a) $Latex formula$

b) $Latex formula$

c) $Latex formula$

d) $Latex formula$

e) $Latex formula$

### Solution 12

a) $Latex formula$

b) $Latex formula$

c) $Latex formula$

$Latex formula$

d) $Latex formula$is best found by the following method,

$Latex formula$

Now, $Latex formula$, are solutions of the above equation, thus,

$Latex formula$ (1)

$Latex formula$(2)

$Latex formula$ (3)

Now, (1) $Latex formula$ (2) $Latex formula$ (3) gives:
$Latex formula$

Thus, $Latex formula$

$Latex formula$

e) For this we shall use the previous expression.

$Latex formula$

$Latex formula$ (Multiplying both sides by $Latex formula$)

Thus, inserting $Latex formula$, into this expression gives:

$Latex formula$ (1)

$Latex formula$ (2)

$Latex formula$ (3)

Now, (1) $Latex formula$ (2) $Latex formula$ (3) gives:

$Latex formula$

$Latex formula$

$Latex formula$

## Solving Polynomials

This is the most important aspect of the polynomials topic, due to the high number of marks given in such questions. The best method to illustrate the techniques involved in solving polynomials, is to solve examples. To ensure understanding of this topic, consider the below examples.

Note that the questions in this section involve parts of the complex number topic, and it is for this reason that the mark values are high.

### Example 13

Factorise the polynomial $Latex formula$ over $Latex formula$.

### Solution 13

With polynomials that have symmetry about their coefficients, there is a set way of solving these equations. This method applies to equations of the form $Latex formula$.

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

Now, $Latex formula$ $Latex formula$,

$Latex formula$

$Latex formula$

Thus, $Latex formula$

But the question asks for the factorisation to be over $Latex formula$.

$Latex formula$ over $Latex formula$.

### Example 14

a) By considering 2 different expansions for $Latex formula$ find an expression for $Latex formula$ in terms of $Latex formula$.

b) Hence solve the equation $Latex formula$.

c) By solving the equation above using another method, find surdic expressions for $Latex formula$ and $Latex formula$.

### Solution 14

a) By De Moivre’s theorem, $Latex formula$.

By the Binomial theorem, $Latex formula$

Now, equating real and imaginary parts gives:

$Latex formula$ and $Latex formula$

Now:

$Latex formula$

Thus:

$Latex formula$

Dividing top and bottom by $Latex formula$ gives:

$Latex formula$

b) Now, we must rearrange the equation $Latex formula$ to obtain it in some form similar to the
above expression for $Latex formula$.

$Latex formula$

$Latex formula$

Now, let $Latex formula$

$Latex formula$

Thus, $Latex formula$

$Latex formula$

Thus, $Latex formula$

c) Now, the equation $Latex formula$ can be solved by the method of grouping.

$Latex formula$

$Latex formula$

$Latex formula$

Now, since $Latex formula$, then $Latex formula$ and $Latex formula$.

### Example 15

$Latex formula$, where $Latex formula$ is an integer. $Latex formula$ has two rational roots which are opposites of each other, and two non-real roots.

a) If $Latex formula$ is a non-real root of $Latex formula$, show that $Latex formula$ and $Latex formula$.

b) If the rational roots are $Latex formula$, deduce that $Latex formula$.

c) Find the rational roots and the value of $Latex formula$.

### Solution 15

a) $Latex formula$ is a Quartic polynomial and hence has four roots only by the Fundamental Theorem of Algebra. Now, let the two rational roots be $Latex formula$ and $Latex formula$, and the complex root be $Latex formula$. Since the coefficients of the polynomial are real, and since $Latex formula$ is a non-real root, then it follows that the complex conjugate of $Latex formula$, is also a root of the polynomial by the complex conjugate complex conjugate theorem.

That is, we have that the roots of the polynomial are of the form $Latex formula$. Now, consider the sum of the roots one at a time. That is, consider:

$Latex formula$

That is, we have that:

$Latex formula$

Now, $Latex formula$. Hence we have that:

$Latex formula$

=> $Latex formula$

Now, since $Latex formula$, then it follows that since:

$Latex formula$

=> $Latex formula$

Now, since we have that $Latex formula$ is non-real, then it follows that $Latex formula$ and consequently that:

$Latex formula$

b) Consider the product of the roots:

$Latex formula$

$Latex formula$

$Latex formula$

Now:

$Latex formula$

from the result in (a).

So we have:

$Latex formula$

c) So we must consider the product of the roots and the sum of the roots two at a time,

$Latex formula$

Hence we have that:

$Latex formula$

Also, from (b) we have:

$Latex formula$

Now, we let $Latex formula$ and $Latex formula$. We then have that:

$Latex formula$

and

$Latex formula$

Solving simultaneously gives:

$Latex formula$

=> $Latex formula$

=> $Latex formula$

Now, $Latex formula$ and hence we have that

$Latex formula$

We then have that:

$Latex formula$

That is, the rational roots are $Latex formula$ and $Latex formula$.

Substituting $Latex formula$ into $Latex formula$and then solving for $Latex formula$ gives:

$Latex formula$

=> $Latex formula$

=> $Latex formula$

## Finding the Number of Real Roots of a Polynomial

At times we do not wish to solve for the roots of a polynomial, but instead require the number of real roots the polynomial possesses. There is a method of doing so and its use involves consideration of the product of the $Latex formula$-values of the stationary points. This method is mainly applied to cubic polynomials, however the results may be extended to include special cases of other polynomials. Consider the below diagrams:

Students should note that no memorisation is required. When required, simply sketch the required situation, after which the other results follow. Consider the below example which illustrates use of this powerful technique.

### Example 16

Consider the function $Latex formula$, where $Latex formula$ and $Latex formula$ are positive. Show that the function $Latex formula$ has 3 distinct real roots if and only if $Latex formula$.

### Solution 16

For the function to have three distinct real roots, we must have that the product of the $Latex formula$-values of the stationary points must be less than $Latex formula$.

So, firstly we differentiate the function and find the stationary points.

$Latex formula$

For stationary points.

=> $Latex formula$

$Latex formula$ and $Latex formula$. Hence we have that the stationary points are $Latex formula$ and $Latex formula$. Now, the product of the $Latex formula$-values of the stationary points must be less than zero for polynomial to have three distinct real roots. That is,

$Latex formula$

$Latex formula$

Now, $Latex formula$ is positive as is $Latex formula$. Hence we have that:

$Latex formula$

$Latex formula$

Hence there exists three distinct real roots for $Latex formula$ if and only if $Latex formula$.

## Partial Fractions

The method of partial fractions is a pivotal part of your mathematical education. This technique essentially involves expressing a rational function (polynomial divided by polynomial) as the sum of two or more other rational functions. The use of such a technique becomes apparent in the topic Integration in which rational functions are integrated using this method.

The method of partial fractions essentially involves the equating of the rational function to the required sum. Now, in order to do so, we must set the correct numerator on each expression. In general, when placing a numerator on a linear expression, place a constant $Latex formula$etc. $Latex formula$, and when placing a numerator on a quadratic place a linear expression $Latex formula$. Be sure that the same constant is not used twice. Now, these are the only two cases required. Upon doing so, we simply consider the sum and rearrange to find the value of the constants by making substitutions and equating coefficients. We shall illustrate these in the below examples.

Note: We do not need to consider any other case (other than quadratic and linear denominators) since all real polynomials may be expressed as the product of real linear and quadratic expressions.

Now before we move on to examples, be careful to note that the degree of the numerator must be less than the degree of the denominator before undergoing the partial fraction decomposition. To solve this problem, we simply perform the long division, and express the polynomial as the sum of a quotient and a remainder over the divisor polynomial. The below examples illustrate this method.

### Example 17

Express the rational function:

$Latex formula$

In the form:

$Latex formula$

where $Latex formula$.

### Solution 17

Firstly, we notice that the denominator is $Latex formula$. Using this fact, we may express $Latex formula$ in a more useful form

$Latex formula$

Which is the required form.

Note: In general questions do not involve prior preparation of a rational function for the use of the method of partial fractions.

We shall now consider example questions which illustrate the method involved in obtaining the partial fraction expansion.

### Example 18

Express the following rational functions as the sum of partial fractions;

a) $Latex formula$

b) $Latex formula$

c) $Latex formula$

### Solution 18

a) As stated above we simply equate the expression to the required expression, whilst placing constants for the values in the numerator (since both denominators are linear).

$Latex formula$

Now, multiplying through by $Latex formula$ gives:

$Latex formula$

Now, letting $Latex formula$ gives:

$Latex formula$

Letting $Latex formula$ gives:

$Latex formula$

Hence we have that:

$Latex formula$

b) Here we equate the rational function to the required final partial fraction expression, taking care to place a linear factor for the numerator on the quadratic expression.

$Latex formula$

Multiplying through by $Latex formula$ gives:

$Latex formula$

Now, let $Latex formula$. We thus have that:

$Latex formula$

Now, we equate the coefficient of $Latex formula$ on the LHS and the RHS. Doing so gives:

$Latex formula$

Now, $Latex formula$. Hence we have that:

$Latex formula$

Now, equating the coefficients of the constant term on LHS and RHS gives:

$Latex formula$

Hence we have that:

$Latex formula$

Hence we have that:

$Latex formula$

Note: The use of the method of equating coefficients and substitution is usually faster than simply using a single method.

c) Here we perform the same sequence of steps as before:

$Latex formula$

Now, multiplying through by $Latex formula$ gives:

$Latex formula$

Now, the method of substitution proves to be difficult here. Instead, we shall equate coefficients.

Coefficient of $Latex formula$:

$Latex formula$

Coefficient of $Latex formula$:

$Latex formula$

Coefficient of $Latex formula$:

$Latex formula$

Coefficient of constant:

$Latex formula$

Now solving $Latex formula$ and $Latex formula$ simultaneously gives:

$Latex formula$

Solving $Latex formula$ and $Latex formula$ simultaneously gives:

$Latex formula$

Hence we have that the partial fraction expansion is given by:

$Latex formula$

Almost every question falls into a similar category to one of the three questions solved above. Be sure to understand the methods involved. Partial fraction expansions involving squares in the linear terms are not included in the syllabus, however they are asked in HSC questions. Students are not expected to be able to predict the final partial fractions expansion, and are instead given the expression required and students are only required to find the values of the constants.

Note: In the above examples, the use of the method of substitution may seem contradictory. We are substituting a point into the expression, although the expression is derived from another expression within which we are unable to substitute those points (due to a division by zero). We are in actual fact not making a direct substitution, but instead taking a limit as $Latex formula$ approaches the required value. However, this produces the same value as if we had simply substituted the point in, and hence we skip the rigor of the method, and simply make the substitution rather than take the limit every time we use the technique.


text:
# When (if at all) would an arm cannon be practical?

Having a weapon attached to your arm that restricts the use of your hand is a common design trope in sci-fi-themed video games, such as Mega Man or Metroid. They even appear in the Marvel movies sometimes. But could such a cannon ever be more practical than just carrying a gun? In other words, what assumptions would you have to make about the world, the wielder, or the use case in order for an arm cannon to be the best choice of weapon?

For the purpose of this question, assume that:

• By "arm cannon", I mean a weapon that encloses or replaces a humanoid forearm and fires deadly projectiles.
• The arm cannon should be reliable enough to graduate from an R&D lab. It doesn't need to be cheap or mass-produced, but it needs to be useful.
• The user should still be fairly agile. If mobility was not a concern, they could just mount a turret.
• It can do other things (like serve as a replacement hand) if you want it to, but it's not a requirement.
• The extent that an arm cannon inhibits or requires normal use of the hand is unspecified. Perhaps the arm cannon relies on deft hand motions to operate, or it might even replace the arm entirely.
• The means by which an arm cannon is equipped or unequipped (if at all) is unspecified. Maybe it slips on like a glove at will, or maybe it's part of a larger exoskeleton that takes an hour to put on. Don't forget about mobility.
• The nature of the user is unspecified. Maybe an arm cannon wouldn't be practical for humans, but it'd be fine for humanoid robots that normally use hands, or some humanoid alien with more convenient biology.
• More practical version with little or no modification: arm cannonball, whose arm is unspecified. – user6760 Feb 7 at 6:02
• "By "arm cannon", I mean a weapon that encloses or replaces a humanoid forearm and fires deadly projectiles." Here's your problem pal, attaching cannon to your joint, be it elbow or shoulder joint, is suboptimal, and requires an really odd pose to deal with recoil. You'd have far better chance with recoilless weapon arm, such as: Flamethrower, Plasma cannon, Laser cannon, etc. Still human anotomy would limit your aim, compared to just having normal handheld version of the weapon(which wouldn't affect flamethrower much, but would have effect on longer ranged use of other). – Failus Maximus Feb 7 at 6:17
• Commenting, because this is redundant with other answers, but... the most obvious way such a weapon is likely to be better than a stand-alone firearm is in the same way that your phone is better than an expensive DSLR or mirrorless camera. Namely, not at all... except for the (not so) minor fact that you have the one on you, and left the other at home / on the kitchen counter / somewhere you just can't get to it fast enough. – Matthew Feb 7 at 16:59
• Oh, you didn't mean a cannon that shoots human arms? – user253751 Feb 7 at 17:25
• I can't remember his name so it's making it difficult to google. I vaguely recall a knight, French I think, who lost a hand early in his career and actually did have a gun installed on the stump. If anyone knows more, it may make a good answer. – Ryan_L Feb 7 at 17:51

Sounds like you mostly covered the bases in the bullet points. But I'll give you my take on it. The first two are unlikely uses, and the third is what I think is it's most practical usage.

Prosthetic Arm Weapon (Least Practical)

Now if you were in a culture where making a fully functional prosthetic hand is not possible, and you are going into battle, you might as well weaponize it, as was done occasionally throughout history. For inspiration, the patta is a historical gauntlet-sword that turns the blade into an extension of the arm. BUT, in a high tech sci-fi setting this seems unnecessary. The prosthetic technology would likely be advanced enough to allow for a hand which is more multipurpose.

Non Human Users (Maybe Practical)

Now the main problem is twofold. One, an arm cannon is hard to aim, because we can't naturally look down our forearms very easily. Two, a cannon covering the arm or hand is a bit of an issue because hands are very useful, and we only have two.

Perhaps you have an alien with double jointed limbs that has no trouble aiming down the sights of such a weapon. Or perhaps they have four limbs and have arms to spare. One less thing to drop, I guess.

Arm Mounted Backup Weapon (Most Practical)

Here's where I think it could shine. Now there is a caveat, that I think it should be arm or wrist mounted rather than covering the hand, because the hand is just too important and versatile to monopolize like that.

But should you need a quick, instinctual point and shoot weapon - going about your tasks, or you lost your main weapon, what could be quicker than having one mounted on your arm? You could make it melee capable as well, perhaps with a laser sight if you wanted. That, I think, would be the best use. Not as a main weapon or armament, but as a quick and intuitive backup weapon that's always handy.

• hard to aim can be easily overcome with training, many special forces train away from using sights, due to the problems of holding a gun out in front of you in tight quarters. With HUDs or AI it is a non-issue. – John Feb 7 at 11:14
• The prosthetic weapon might be practical for people who have lost a limb due to injury. While not as versatile as a hand, an arm cannon could be preferable to a stump (at least if you're going to continue to work in a profession that requires carrying a weapon). – Darrel Hoffman Feb 7 at 14:23
• en.wikipedia.org/wiki/Pata_(sword) is interesting. Perhaps you could link to it into your answer. – CJ Dennis Feb 7 at 21:48

the biggest advantage of a weapon firmly attached to it's owner over one held in hand is that a weapon typically does not care who owns or holds it.
That means, if i manage to take away your pistol, i can aim it at you.
So, a weapon that cannot be taken away has great advantages.

And then there is, of course, the fact that something that is firmly attached to my body is always with me. Maybe you've seen pulp fiction, and remember the scene where John Travolta leaves his weapon on the kitchen counter to go to the toilet? That mistake won't happen if your weapon is attached to you.

There is real life Sedgley OSS .38 pistol used during WW2. Actually, its a glove with pistol attached to it. It fits into weapon that encloses or replaces a humanoid forearm and fires deadly projectiles. It was meant as a covert operations and assassination weapon in the Pacific Theater. It had low accuracy and was realy hard to aim, but during trench combat it could work as "one punch - one shot - one kill" weapon. So, arm cannon can be practical as concealed weapon, during close quarters combat to improve melee strikes.

• Amazing! Truly worthy of the Gun Jesus Ian on Forgotten Weapons. – Gustavo Feb 7 at 20:05

Aiming the gun

With a conventional firearm, the shooter aligns the eye, the sights on the weapon, and the target. The weapon is held in one or both hands. Counting the shoulder, there are three joints to get into a comfortable and secure hold. With a forearm-mounted gun, there are only two joints, shoulder and ellbow. This could lead to a less comfortable shooting stance.

Take a pencil and a bit of tape and tape the pencil first on your index finger, then on your forearm. Try to peer along the full length of the pencil with one eye. Which is more comfortable?

This could be overcome with electronic sights and a head-up display.

Fixing the gun

For changing magazines, clearing jams, and the like, the weapon is not held in the shooting stance. Both hands can be used on the conventional gun. With a forearm-mounted weapon, there is only one hand, the other is permanently fixed where it cannot reach the gun.

This could be overcome if there are no user-serviceable parts inside, e.g. a futuristic laser weapon or railgun.

It could work, and many answers have stated some solid pros and cons.

A solid Pro would be that you are armed at all times. Another is the element of surprise, "Aha, you thought I had nothing but now I have a gun at your head". But there are some major cons to what you may have seen in these guns from Comics and Video Games.

If you must have a forearm type prosthetic slug thrower, you have to consider some things.

Ammunition loading. Remember that rounds don't magically appear in the chamber. You have to load the weapon, and you have to do it with one hand, so you have to account for that.

Ammunition size effects lots of things. The bigger the round and powder charge, the more robust the mechanism and barrel must be. That means size. If you don't want a Popeye like forearm, this has to be taken into account.

Recoil. This is a big one. When you fire a normal rifle or pistol, the recoil normally impacts the shoulder or the wrist, both of which can distribute the force in lots of little ways. A forearm gun is going to go right into the bones of the forearm. Most automatics are therefore going to pound the heck out of the stump, possible causing stress fractures and other trauma. Recoil charts are available, but just consider Newtons third law and ke=1/2 mv^2 if you are looking for something futuristic.

Heat. Where will the heat go? You tube has videos of people cooking bacon on the barrel of an ak-47 or other rifles. I know my own bolt action rifle (.308 Winchester) gets too hot to touch after shooting 20 rounds, and that is with stopping to reload every 5 rounds, one round at a time with a pretty long pause between shots. So you have to account for heat.

Here is the solution though. A Cannon is probably out for the above reasons, but you could theoretically put a good .25 Magnum round weapon in a forearm prosthetic. The rounds are small enough where you could load several without looking like Popeye. The mechanism is fairly small, so could be concealable. Kel Tech weapons makes a neat little .25 which is the same size as an ordinary pistol, but can hold 30 rounds in a small magazine. So you get a very small form factor that can be hidden, and a surprise weapon that is effective.

That will give you the best bang for your buck. (so sorry for the pun. I couldn't help it. I have a problem)

I don't understand the problem. There seem to be so many solutions.

• In a sci-fi setting, you wouldn't let a human control a spaceship. This would be done by an AI that is far more intelligent and technically skilled. The same would be with shooting and pretty much anything else where stakes are high. Humans suck at it, don't let them do it. There would probably be a special interface for the human. He could look at people, see their name, status, power level, etc, see if the person carried a weapon and either confirm the kill or mark him/her as unimportant.
• Have the canon arm defend or surveil its user no matter if he wants it or not, as protection or punishment. The human would have to fight against his arm at some point. He'd purposefully crush it with a car or something like that. This may already be a cliché, but I really like it.
• Maybe you also don't want people to know that you have a gun. That's why it's built into your prosthetic.
• Make some shit up and explain it with 'people being crazy'. It's not that hard.

I suspect it's only practical for

• concealment in a prosthetic
• training individuals or troops to understand it as being part of their role in society

One who has had their arm replaced with a gun (especially at a young age) without option to choose or have it removed will be sure to understand that they are and have always been meant to harm.

The thing is: it's not practical for you.

Nobody in his right mind wants to lose a hand. Yet, in some extreme contexts, it could make sense. You're a convicted criminal condemned to mine the moon? Enjoy your new personal mining tool.

You're a cyborg grown in a vat to only ever know war and killing fields? I'm sure that the hand cannon is as close to you as a samurai's sword to it's weaboo.

You're exceptionally determined to be the best quantum mechanic in the fleet? Those freakish hands which let you manipulate force fields will come in handy.

Well, there's a thing you mentioned: the cannon could serve as a replacement hand. That's important, because now we have two new crowds:

1. Amputees: most people without a hand will accept to wear a hand cannon if they regain a hand as a side-effect. I know I would.

2. Elite troops/spies/mercenaries: ever played Shadowrun or the like? Watch some cyberpunk and you'll see heavily augmented people. If the hand cannon doesn't make you lose the hand, people who want to gear up may just think that it's a great deal.

If you mean something to keep always attached to your arm it will never be practical. Even if the weapon will be very light the weight would be felt over a long time, the arm is a long lever and the oscillations would be on a relatively long arc. What would happen is that walking for a long time, paying a lot of attention to the external dangers rather than the own sensations, it could eventually cause some nausea because of the altered balance.

Another problem is that the arm may get tired, the soldier cannot choose when he in the mood for fighting and when he isn't. So he may find himself with the movements and the reaction times slowed down at the wrong time.

If instead you mean something that can be easily attached/detached it would make sense if the soldier has a lot of tools like remote controls for companion drones, special sensors, other weapons and so on.