Unit FP2: Further Pure Mathematics A2 2

An algebraic inequality is solved by finding every point where its truth value can change, then determining which intervals satisfy the original relation. Denominator zeros and modulus boundaries must be recorded before simplifying.

For example, $$\frac1{x-1}>\frac{x}{x+2}\quad\Longleftrightarrow\quad\frac{-x^2+2x+2}{(x-1)(x+2)}>0.$$ The numerator is zero at $x=1\pm\sqrt3$; the denominator is zero at $x=-2,1$. A sign chart across these four ordered critical points gives $$-2<x<1-\sqrt3\quad\text{or}\quad1<x<1+\sqrt3.$$ The denominator points are excluded even though they separate sign intervals.

For the official model inequality $|x^2-1|>2(x+1)$, the right side is negative when $x<-1$, so that whole interval works automatically. For $x\ge-1$, solve $$x^2-1>2(x+1)\quad\text{or}\quad x^2-1<-2(x+1).$$ These become $(x-3)(x+1)>0$ or $(x+1)^2<0$, giving $x>3$. Hence the full solution is $$x<-1\quad\text{or}\quad x>3.$$

The method of differences rewrites each term as a difference of shifted expressions. When consecutive terms are displayed, the interior contributions cancel and only a small number of boundary terms remain.

For the official model, partial fractions give $$\frac1{r(r+1)}=\frac1r-\frac1{r+1}.$$ Therefore $$\sum_{r=1}^{n}\frac1{r(r+1)}=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1n-\frac1{n+1}\right).$$ Every interior fraction cancels, leaving $$1-\frac1{n+1}=\frac{n}{n+1}.$$

If a closed form $S(n)=\sum_{r=p}^{n}f(r)$ is already known, a later finite block is found by subtraction: $$\sum_{r=a}^{b}f(r)=S(b)-S(a-1).$$ This preserves the correct first included term $r=a$.

Euler's relation represents a rotation through a real angle $\theta$ as a complex exponential: $e^{i\theta}=\cos\theta+i\sin\theta$. Its real and imaginary parts are the coordinates of the corresponding point on the unit circle.

The second relation is obtained by replacing $\theta$ with $-\theta$ and using that cosine is even while sine is odd. Adding the two equations eliminates the imaginary terms; subtracting them eliminates the cosine terms.

For example, $$e^{3i\theta}+e^{-3i\theta}=2\cos3\theta,$$ while $$e^{3i\theta}-e^{-3i\theta}=2i\sin3\theta.$$ These conversions let a trigonometric expression be handled with laws of indices and then converted back.

De Moivre's theorem says that multiplying a complex number in polar form $n$ times multiplies its argument by $n$ and raises its modulus to the power $n$. For every integer $n$,$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta).$$More generally, $[r(\cos\theta+i\sin\theta)]^n=r^n[\cos(n\theta)+i\sin(n\theta)]$.

For positive integers, the result follows by induction. The case $n=1$ is immediate. If it holds for $n=k$, multiplication by $\cos\theta+i\sin\theta$ and the angle-addition formulae give $\cos((k+1)\theta)+i\sin((k+1)\theta)$. For $n=0$, both sides equal $1$. For negative $n$, take the reciprocal of the positive-power result; the reciprocal of $\cos\phi+i\sin\phi$ is $\cos\phi-i\sin\phi$. This proves the theorem for every integer $n$.

For instance, equating imaginary parts when $n=3$ gives $$\sin3\theta=3\sin\theta\cos^2\theta-\sin^3\theta=3\sin\theta-4\sin^3\theta.$$ In the reverse direction, Euler's forms give $$\cos^2\theta=\frac{1+\cos2\theta}{2}.$$ Thus the same theorem connects multiple angles with powers in both directions.

The $m$ values have equal modulus $R^{1/m}$ and arguments separated by $2\pi/m$, so they form a regular $m$-gon centred at the origin. Substitute one root into the original equation to check the modulus and multiplied argument.

On an Argand diagram, $|z-a|$ is the distance from the point $z$ to the fixed point $a$, while $\arg(z-a)$ is the directed angle of the displacement from $a$ to $z$. Translate each complex condition into distance or angle geometry before sketching.

For an exact equation, set $z=x+iy$ and write each modulus as a squared distance. For example, $$|z|=2|z-3|$$ becomes $$x^2+y^2=4[(x-3)^2+y^2],$$ so $$(x-4)^2+y^2=4.$$ The locus is therefore the circle with centre $(4,0)$ and radius $2$.

For an inequality, first draw its equality boundary, test a point not on that boundary to choose the correct side, and use a solid boundary for $\le$ or $\ge$. For an argument condition, also respect the stated argument range so that the correct ray or arc is selected.

A transformation assigns each permitted point $z$ in the source plane a point $w$ in the image plane. To find the image of a locus, connect $z=x+iy$ and $w=u+iv$ through the transformation, substitute into the source condition, and simplify until the result is a condition on $u$ and $v$.

For example, let $$w=\frac{z-1}{z+1},\qquad z\ne-1.$$ If $z$ lies on the imaginary axis, then $z=iy$ for real $y$, so $$|w|=\frac{|iy-1|}{|iy+1|}=1.$$ The image lies on the unit circle. Solving $z=(1+w)/(1-w)$ shows that $w=1$ has no finite preimage, so the image is the unit circle with that point excluded.

For a non-degenerate fractional linear transformation, $ad-bc\ne0$. If $c\ne0$, the source value $z=-d/c$ is a pole and is excluded; the inverse also reveals any missing image value. Lines and circles commonly map to lines or circles, but the algebra and exclusions decide which one in a particular problem.

After obtaining the image equation, check it with one easy source point: transform that point directly and verify that its $w$-coordinates satisfy the new equation. This catches sign errors introduced while rationalising or completing the square.

A first-order equation is separable when it can be written as $$\frac{dy}{dx}=g(x)h(y),$$ so every factor involving $y$ can be placed with $dy$ and every factor involving $x$ with $dx$. The constant of integration produces a family of solution curves; an initial condition selects one member.

For $$\frac{dy}{dx}=x(1+y),\qquad y(0)=0,$$ first record the equilibrium solution $y=-1$. For the other solutions,$$\int\frac{1}{1+y}\,dy=\int x\,dx,$$so $$\ln|1+y|=\frac{x^2}{2}+C\quad\Longrightarrow\quad y=Ae^{x^2/2}-1.$$The condition $y(0)=0$ gives $A=1$, hence $y=e^{x^2/2}-1$.

For the full family $y=Ae^{x^2/2}-1$, every curve is symmetric about the $y$-axis and has $dy/dx=0$ at $x=0$. Curves with $A>0$ lie above the equilibrium and have a minimum at $x=0$; curves with $A<0$ lie below it and have a maximum there. This qualitative information should agree with any plotted member.

A first-order linear equation must first be written in the standard form $$\frac{dy}{dx}+P(x)y=Q(x).$$ Its integrating factor is $$\mu(x)=e^{\int P(x)\,dx}.$$ Because $\mu'=P\mu$, multiplying by $\mu$ turns the entire left side into the product derivative $\dfrac{d}{dx}(\mu y)$.

For $$\frac{dy}{dx}-2y=x,$$ we have $P=-2$ and $\mu=e^{-2x}$. Thus $$(e^{-2x}y)'=xe^{-2x}.$$Integration by parts gives $$e^{-2x}y=-e^{-2x}\left(\frac{x}{2}+\frac14\right)+C,$$ so $$y=-\frac{x}{2}-\frac14+Ce^{2x}.$$ If $y(0)=\tfrac34$, then $C=1$.

The term containing $C$ is the homogeneous part of the family, while the remaining terms form one particular response to $Q(x)$. Their sum is still a solution because the equation is linear.

A given substitution is useful when it converts a non-standard first-order equation into either a separable equation or a linear equation. The essential move is to differentiate the substitution with the chain rule, replace every occurrence of the old dependent variable consistently, and solve the resulting equation in the new variable.

After differentiating, substitute into the original equation and simplify until only $x$, $v$ and $dv/dx$ remain. Identify the transformed equation as separable or linear, solve it by the corresponding earlier method, restore the integration constant, and finally replace $v$ to express the result in the requested variable and form.

Consider $$\frac{dy}{dx}+2y=xy^3$$ with the given substitution $v=y^{-2}$. Since $$\frac{dv}{dx}=-2y^{-3}\frac{dy}{dx},$$ multiplying the original equation by $-2y^{-3}$ gives $$\frac{dv}{dx}-4v=-2x.$$This is linear. With integrating factor $e^{-4x}$,$$ve^{-4x}=\int-2xe^{-4x}\,dx=e^{-4x}\left(\frac{x}{2}+\frac18\right)+C,$$so $$v=\frac{x}{2}+\frac18+Ce^{4x},\qquad y^2=\frac{1}{\frac{x}{2}+\frac18+Ce^{4x}}.$$

For a constant-coefficient equation $$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=f(x),\qquad a\ne0,$$ the general solution is the sum of the complementary function (CF), which solves the homogeneous equation, and one particular integral (PI), which accounts for $f(x)$: $y=\mathrm{CF}+\mathrm{PI}$.

The auxiliary equation comes from trying $y=e^{mx}$ in the homogeneous equation. Each independent root solution contributes an arbitrary constant, so a second-order equation has two constants before conditions are imposed.

Differentiate the trial, substitute it into the full equation and compare coefficients to determine its unknowns. If any part of the trial is already in the CF, multiply the whole trial by $x$; use another factor of $x$ if the overlap is repeated.

For the official resonance model $$y''+4y=\sin2x,$$ the auxiliary roots are $\pm2i$, so $$\mathrm{CF}=A\cos2x+B\sin2x.$$ A trial $C\cos2x+D\sin2x$ duplicates the CF and contributes zero. Use an $x$ factor instead: $y_p=Kx\cos2x$. Since $(x\cos2x)''+4x\cos2x=-4\sin2x$, $K=-\tfrac14$. Hence $$y=A\cos2x+B\sin2x-\frac14x\cos2x.$$

Apply values of $y$ and $y'$ only after CF and PI have been combined, producing two simultaneous equations for $A$ and $B$. A final substitution into the differential equation checks both the PI coefficients and any resonance factor.

A given substitution can turn a variable-coefficient second-order equation into the constant-coefficient type just solved. The decisive step is to transform both derivatives with the chain rule before simplifying; every derivative must be taken with respect to the variable displayed in its denominator.

Differentiate the supplied relation, replace $y'$, $y''$ and every explicit occurrence of the old variable, then cancel common non-zero factors using the stated domain. Check that the resulting equation is exactly the claimed constant-coefficient form. Solve it as CF + PI in the new variable, and finally back-substitute both the variable and any dependent-variable relation.

For $$x^2y''+3xy'-3y=2\ln x,\qquad x>0,$$ take the given substitution $t=\ln x$. Then $$y'=\frac1x y_t,\qquad y''=\frac1{x^2}(y_{tt}-y_t),$$ so the equation becomes $$y_{tt}+2y_t-3y=2t.$$ Its auxiliary roots are $1$ and $-3$. A PI $at+b$ gives $a=-\tfrac23$ and $b=-\tfrac49$, hence $$y=Ae^t+Be^{-3t}-\frac23t-\frac49.$$ Returning to $x$ gives $$y=Ax+Bx^{-3}-\frac23\ln x-\frac49,\qquad x>0.$$

The transformed equation and the final answer can be checked independently: substitute the derivative relations into the original equation for the first check, then replace the final $y$ and its $x$-derivatives in the original equation for the second.

The $n$th derivative of $y=f(x)$ is written $f^{(n)}(x)$ or $\dfrac{d^ny}{dx^n}$. Higher derivatives are found by differentiating successively, but patterns, recurrences and a derivative table prevent repeated algebra from becoming unmanageable.

For $y=\ln(2+x)$,$$y'=\frac1{2+x},\qquad y''=-\frac1{(2+x)^2},\qquad y'''=\frac2{(2+x)^3}.$$The powers, alternating signs and factorial factors give $$y^{(n)}=\frac{(-1)^{n-1}(n-1)!}{(2+x)^n},\qquad n\ge1.$$ Substituting $n=1,2,3$ checks the general expression.

When a function satisfies a differential relation, differentiating that relation can be shorter than starting again from the original function. Keep every product-rule term: differentiating $x,y''$ gives $y''+x,y'''$. Values of lower derivatives at a point can then determine a required higher derivative there.

A Maclaurin series is a Taylor series centred at $0$. Its coefficients are chosen so that the polynomial and the function have the same successive derivatives at $x=0$:$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots+\frac{f^{(n)}(0)}{n!}x^n+\cdots.$$

If $f(x)=a_0+a_1x+a_2x^2+\cdots$, then setting $x=0$ gives $a_0=f(0)$. Differentiate once and set $x=0$ to get $a_1=f'(0)$; differentiating $r$ times gives $f^{(r)}(0)=r!a_r$. Hence $a_r=f^{(r)}(0)/r!$.

The logarithmic expansion follows because for $r\ge1$, $$\left.\frac{d^r}{dx^r}\ln(1+x)\right|_{x=0}=(-1)^{r-1}(r-1)!.$$ Standard series may then be adapted and combined. For example,$$\ln\frac{1+x}{1-x}=\ln(1+x)-\ln(1-x)=2x+\frac{2x^3}{3}+O(x^5).$$ Even powers cancel, so only the needed terms should be retained.

A Taylor series expands $f(x)$ about a chosen centre $x=a$ in powers of the displacement $h=x-a$:$$f(x)=f(a)+f'(a)h+\frac{f''(a)}{2!}h^2+\frac{f'''(a)}{3!}h^3+\cdots.$$The coefficients therefore describe the function's value and successive local changes at $a$, not at $0$.

For the official example $f(x)=\sin x$ about $a=\pi$, let $h=x-\pi$. The derivative values are$$f(\pi)=0,\quad f'(\pi)=-1,\quad f''(\pi)=0,\quad f'''(\pi)=1.$$Therefore, up to and including $h^3$,$$\sin x=-(x-\pi)+\frac{(x-\pi)^3}{6}+O\!\left((x-\pi)^4\right).$$ Directly using $\sin(\pi+h)=-\sin h$ confirms the signs.

The Taylor-series method solves a differential equation locally by using the equation and its derivatives to generate $y(a),y'(a),y''(a),\ldots$ at the expansion centre. A term through $(x-a)^n$ requires derivative values through $y^{(n)}(a)$.

For the official model$$y''+xy'+y=0,\qquad y(0)=1,\quad y'(0)=0,$$the original equation at $0$ gives $y''(0)=-1$. Differentiating once gives$$y'''+2y'+xy''=0,$$so $y'''(0)=0$. Differentiating again gives$$y''''+3y''+xy'''=0,$$so $y''''(0)=3$. Hence, through $x^4$,y=1-rac{x^2}{2!}+\frac{3x^4}{4!}+O(x^5)=1-\frac{x^2}{2}+\frac{x^4}{8}+O(x^5).

Equivalently, substitute $y=a_0+a_1x+a_2x^2+\cdots$ and its differentiated series into the equation, then equate coefficients. The derivative-at-the-centre method is usually shorter when initial values are supplied explicitly.

A polar point $(r,\theta)$ lies a distance $r\ge0$ from the pole in the direction measured by $\theta$ from the initial line. Cartesian and polar descriptions are connected by$$x=r\cos\theta,\qquad y=r\sin\theta,\qquad r^2=x^2+y^2.$$At the pole $r=0$, the angle is not unique.

To sketch reliably, first state the permitted $\theta$ interval. Test symmetry by replacing $\theta$ with $-\theta$ or $\pi-\theta$, solve $r=0$ for passages through the pole, find maximum and minimum $r$, and calculate a small table at exact angles. Plot in increasing $\theta$ so the order of loops or arcs is clear.

Cartesian conversion can confirm the shape. For example,$$r=2a\cos\theta\quad\Longrightarrow\quad r^2=2ar\cos\theta,$$so $$x^2+y^2=2ax\quad\Longrightarrow\quad (x-a)^2+y^2=a^2.$$This proves that the polar curve is the circle described in the table.

Curves drawn over the same ray intersect where their admissible radii are equal. Solve $r_1(\theta)=r_2(\theta)$, retain angles in the stated range, and substitute back for the exact radius. The pole may be a shared point even when it arises from different angle values.

A narrow polar sector of angle $d\theta$ has area approximately $\tfrac12r^2d\theta$. Therefore the area swept once as $\theta$ runs from $\alpha$ to $\beta$ is$$A=\frac12\int_{\alpha}^{\beta}r^2\,d\theta.$$The limits must describe the actual boundary rays or curve intersections of the required region.

For the upper half of the cardioid $r=a(1+\cos\theta)$, $0\le\theta\le\pi$,$$A=\frac{a^2}{2}\int_0^\pi(1+2\cos\theta+\cos^2\theta)\,d\theta.$$Using $\cos^2\theta=\tfrac12(1+\cos2\theta)$ gives$$A=\frac{a^2}{2}\left(\pi+\frac\pi2\right)=\frac{3\pi a^2}{4}.$$