A mathematical refresher





Exponents


Definition of exponentiation


Multiplying the number 3 by itself four times gives


<SPAN role=presentation tabIndex=0 id=MathJax-Element-1-Frame class=MathJax style="POSITION: relative" data-mathml='3×3×3×3= 81′>3×3×3×3= 813×3×3×3= 81
3×3×3×3= 81


which can be written more simply with the shorthand notation


<SPAN role=presentation tabIndex=0 id=MathJax-Element-2-Frame class=MathJax style="POSITION: relative" data-mathml='34= 81′>34= 8134= 81
34= 81


The notation 3 4 is read as “3 raised to the fourth power” or “3 to the fourth power.” In this example, the number 3 is referred to as the base, and the number 4 is called the exponent. In the natural sciences the most frequently used base is an irrational number with the symbol e , whose value to three decimal places is 2.718. An irrational number is a number that cannot be written as a fraction. e.g., the square root of 2 ( ) and the number pi (π).


The most general representation of exponentiation is


<SPAN role=presentation tabIndex=0 id=MathJax-Element-3-Frame class=MathJax style="POSITION: relative" data-mathml='am=c’>??=?am=c
am=c


where a and m can be arbitrary numbers. A few numerical examples are


<SPAN role=presentation tabIndex=0 id=MathJax-Element-4-Frame class=MathJax style="POSITION: relative" data-mathml='34= 81;2.73= 19.683;101.3= 19.952…;3.11.7= 6.844…’>34= 81;2.73= 19.683;101.3= 19.952;3.11.7= 6.84434= 81;2.73= 19.683;101.3= 19.952…;3.11.7= 6.844…
34= 81;2.73= 19.683;101.3= 19.952…;3.11.7= 6.844…


The first two examples can be verified by hand calculation; the last two are easy to verify on a calculator.


Multiplication of exponentials


The rule for multiplying exponentials follows from the definition of exponentials:



<SPAN role=presentation tabIndex=0 id=MathJax-Element-5-Frame class=MathJax style="POSITION: relative" data-mathml='am×an=a(m+n)’>??×??=?(?+?)am×an=a(m+n)
am×an=a(m+n)


This rule is easily verified by checking an example:


<SPAN role=presentation tabIndex=0 id=MathJax-Element-6-Frame class=MathJax style="POSITION: relative" data-mathml='32×34= (3 ×3) × (3 × 3 ×3 × 3) =36=3(2 + 4)’>32×34= (3 ×3) × (3 × 3 ×3 × 3) =36=3(2 + 4)32×34= (3 ×3) × (3 × 3 ×3 × 3) =36=3(2 + 4)
32×34= (3 ×3) × (3 × 3 ×3 × 3) =36=3(2 + 4)


All the other properties of exponentials follow directly from the rule for combining exponents.


Meaning of the number 0 as exponent


We use the previous rule to deduce what a to the 0th power ( a 0 ) means. To make the point concrete, take a = 3. Equation AB.1 allows us to write


<SPAN role=presentation tabIndex=0 id=MathJax-Element-7-Frame class=MathJax style="POSITION: relative" data-mathml='32×30=3(2 + 0)=32′>32×30=3(2 + 0)=3232×30=3(2 + 0)=32
32×30=3(2 + 0)=32


This expression is true if, and only if, 3 0 = 1. In general, any non-0 number raised to the 0th power is equal to 1:



<SPAN role=presentation tabIndex=0 id=MathJax-Element-8-Frame class=MathJax style="POSITION: relative" data-mathml='a0= 1(as long asa≠0)’>?0= 1(as long as?0)a0= 1(as long asa≠0)
a0= 1(as long asa≠0)


The reason for excluding 0 from the definition will be clear shortly.


Negative numbers as exponents


To deduce the meaning of a negative exponent, again take a = 3 as the base and use Equation AB.1 to write


<SPAN role=presentation tabIndex=0 id=MathJax-Element-9-Frame class=MathJax style="POSITION: relative" data-mathml='32×3- 2=3[2 +( – 2)]=30= 1′>32×3– 2=3[2 +( – 2)]=30= 132×3- 2=3[2 +( – 2)]=30= 1
32×3- 2=3[2 +( – 2)]=30= 1


This expression is true if, and only if, 3 −2 = 1/3 2 . In general,



<SPAN role=presentation tabIndex=0 id=MathJax-Element-10-Frame class=MathJax style="POSITION: relative" data-mathml='a-m=1am(as long asa≠0)’>??=1??(as long as?0)a-m=1am(as long asa≠0)
a-m=1am(as long asa≠0)


Division of exponentials


The definition in Equation AB.3 extends the rule for combining exponents ( Equation AB.1 ) to include division of exponentials:



<SPAN role=presentation tabIndex=0 id=MathJax-Element-11-Frame class=MathJax style="POSITION: relative" data-mathml='aman=am×a-n=a(m-n)’>????=??×??=?(??)aman=am×a-n=a(m-n)
aman=am×a-n=a(m-n)


We now see why in the definition of the 0th power ( Equation AB.2 ), the base a cannot be 0: a = 0 in Equation AB.4 would force a division of 0 by 0—an operation that has no meaning.


Exponentials of exponentials


Combining exponents shows what happens when an exponential is raised to another power, for example:


<SPAN role=presentation tabIndex=0 id=MathJax-Element-12-Frame class=MathJax style="POSITION: relative" data-mathml='(72)3=72×72×72=7(2+2 +2)=76′>(72)3=72×72×72=7(2+2 +2)=76(72)3=72×72×72=7(2+2 +2)=76
(72)3=72×72×72=7(2+2 +2)=76


In general,



<SPAN role=presentation tabIndex=0 id=MathJax-Element-13-Frame class=MathJax style="POSITION: relative" data-mathml='(am)n=a(m×n)’>(??)?=?(?×?)(am)n=a(m×n)
(am)n=a(m×n)


Fractions as exponents


We now investigate the case in which the exponent is a fraction (i.e., a m / n ). Taking a concrete example with a = 7, we ask what is meant by 7 1/2 . Applying Equation AB.1 gives


<SPAN role=presentation tabIndex=0 id=MathJax-Element-14-Frame class=MathJax style="POSITION: relative" data-mathml='71/2×71/2=7(1/2 + 1/2)=71= 7′>71/2×71/2=7(1/2 + 1/2)=71= 771/2×71/2=7(1/2 + 1/2)=71= 7
71/2×71/2=7(1/2 + 1/2)=71= 7


which immediately shows that 7 1/2 = . In general:



<SPAN role=presentation tabIndex=0 id=MathJax-Element-15-Frame class=MathJax style="POSITION: relative" data-mathml='a1n=an’>?1?=??a1n=an
a1n=an


A direct consequence of combining Equations AB.6 and AB.4 is that



<SPAN role=presentation tabIndex=0 id=MathJax-Element-16-Frame class=MathJax style="POSITION: relative" data-mathml='amn=a(m×1n)=(am)1n=amn=(a1n)m=(an)m’>???=?(?×1?)=(??)1?=???=(?1?)?=(??)?amn=a(m×1n)=(am)1n=amn=(a1n)m=(an)m
amn=a(m×1n)=(am)1n=amn=(a1n)m=(an)m


Because any rational decimal number can always be written as a fraction (e.g., 1.5 = 3/2, 0.47 = 47/100), decimal numbers in the exponent can be dealt with as fractions.


Equations AB.1 to AB.7 constitute the properties of exponentials that are important for computation. These properties are summarized in Box AppB.1 .



BOX AppB.1

Properties of Exponentials



<SPAN role=presentation tabIndex=0 id=MathJax-Element-17-Frame class=MathJax style="POSITION: relative" data-mathml='am×an=a(m+n)aman=am×a-n=a(m-n)a0=1(as long asa≠0)’>??×??=?(?+?)????=??×??=?(??)?0=1(as long as?0)am×an=a(m+n)aman=am×a-n=a(m-n)a0=1(as long asa≠0)
am×an=a(m+n)aman=am×a-n=a(m-n)a0=1(as long asa≠0)

<SPAN role=presentation tabIndex=0 id=MathJax-Element-18-Frame class=MathJax style="POSITION: relative" data-mathml='a−m=1am(as long asa≠0)(am)n=a(m×n)’>??=1??(as long as?0)(??)?=?(?×?)a−m=1am(as long asa≠0)(am)n=a(m×n)
a−m=1am(as long asa≠0)(am)n=a(m×n)

<SPAN role=presentation tabIndex=0 id=MathJax-Element-19-Frame class=MathJax style="POSITION: relative" data-mathml='a1n=an’>?1?=??a1n=an
a1n=an

<SPAN role=presentation tabIndex=0 id=MathJax-Element-20-Frame class=MathJax style="POSITION: relative" data-mathml='amn=a(m×1n)=(am)1n=amn=(a1n)m=(an)m’>???=?(?×1?)=(??)1?=???=(?1?)?=(??)?amn=a(m×1n)=(am)1n=amn=(a1n)m=(an)m
amn=a(m×1n)=(am)1n=amn=(a1n)m=(an)m





Logarithms


Definition of the logarithm


The logarithm can be seen as the inverse of exponentiation. The general way of representing a logarithm is


<SPAN role=presentation tabIndex=0 id=MathJax-Element-21-Frame class=MathJax style="POSITION: relative" data-mathml='logac=m’>log??=?logac=m
logac=m


This expression is read as “the logarithm ‘to the base a ’ of the number c is equal to m. ” The meaning of this expression is “ m is the power to which a must be raised to get the number c. ” This is really just another way of representing exponentiation, a m = c. The definition of the logarithm may be stated as


<SPAN role=presentation tabIndex=0 id=MathJax-Element-22-Frame class=MathJax style="POSITION: relative" data-mathml='alogac=c’>?log??=?alogac=c
alogac=c


Examples corresponding to those in the exponentiation section are


<SPAN role=presentation tabIndex=0 id=MathJax-Element-23-Frame class=MathJax style="POSITION: relative" data-mathml='log381= 4;log2.719.683 = 3;log1019.925…= 1.3,log3.16.844…= 1.7′>log381= 4;log2.719.683 = 3;log1019.925= 1.3,log3.16.844= 1.7log381= 4;log2.719.683 = 3;log1019.925…= 1.3,log3.16.844…= 1.7
log381= 4;log2.719.683 = 3;log1019.925…= 1.3,log3.16.844…= 1.7


Historically, the number 10 is the most commonly used base for logarithms. For this reason, base-10 logarithms are referred to as common logarithms. In the natural sciences the irrational number e is usually used as the base for logarithms. Just as the exponential function using e as the base is called the natural exponential function, a logarithm using the base e is called the natural logarithm and is given the symbol ln. Thus the natural logarithm of x is symbolized as ln x.


The definitions given previously, along with the properties of exponentials, imply three rules governing logarithms.


Logarithm of a product



<SPAN role=presentation tabIndex=0 id=MathJax-Element-24-Frame class=MathJax style="POSITION: relative" data-mathml='loga(b×c)=logab+logac’>log?(?×?)=log??+log??loga(b×c)=logab+logac
loga(b×c)=logab+logac


Examining the definition of a logarithm immediately shows where this rule comes from:


<SPAN role=presentation tabIndex=0 id=MathJax-Element-25-Frame class=MathJax style="POSITION: relative" data-mathml='b=alogabandc=alogac’>?=?log??and?=?log??b=alogabandc=alogac
b=alogabandc=alogac


so


<SPAN role=presentation tabIndex=0 id=MathJax-Element-26-Frame class=MathJax style="POSITION: relative" data-mathml='b×c=alogab×alogac=a(logab+logac)’>?×?=?log??×?log??=?(log??+log??)b×c=alogab×alogac=a(logab+logac)
b×c=alogab×alogac=a(logab+logac)


(the last step is a consequence of the rule for combining exponents— Equation AB.1 ). Taking log a of both sides gives Equation AB.8 .


Logarithm of an exponential


The logarithm of an exponential is given by



<SPAN role=presentation tabIndex=0 id=MathJax-Element-27-Frame class=MathJax style="POSITION: relative" data-mathml='logacm=m×logac’>log???=?×log??logacm=m×logac
logacm=m×logac


This rule is a consequence of the rule governing logarithm of a product ( Equation AB.8 ) combined with the definition of an exponential:


<SPAN role=presentation tabIndex=0 id=MathJax-Element-28-Frame class=MathJax style="POSITION: relative" data-mathml='logacm=loga(c×c×…×c)mtimes=(logac+logac+…+logac)mtimes=m×logac’>log???=log?(?×?××?)?times=(log??+log??++log??)??????=?×log??logacm=loga(c×c×…×c)mtimes=(logac+logac+…+logac)mtimes=m×logac
logacm=loga(c×c×…×c)mtimes=(logac+logac+…+logac)mtimes=m×logac


A special case of the logarithm of an exponential occurs frequently and is sometimes given as a separate rule:



<SPAN role=presentation tabIndex=0 id=MathJax-Element-29-Frame class=MathJax style="POSITION: relative" data-mathml='loga1c=logac-1=-1×logac=-logac’>log?1?=log??1=-1×log??=-log??loga1c=logac-1=-1×logac=-logac
loga1c=logac-1=-1×logac=-logac


Incidentally, the product rule together with the rule for taking the logarithm of an exponential ( Equation AB.9 ) gives the logarithm of a quotient:


<SPAN role=presentation tabIndex=0 id=MathJax-Element-30-Frame class=MathJax style="POSITION: relative" data-mathml='logabc=loga[b×c-1]=logab+loga(c-1)=logab+(-1)logac=logab-logac’>log???=log?[?×?1]=log??+log?(?1)=log??+(1)log??=log??log??logabc=loga[b×c-1]=logab+loga(c-1)=logab+(-1)logac=logab-logac
logabc=loga[b×c-1]=logab+loga(c-1)=logab+(-1)logac=logab-logac


Changing the base of a logarithm


For any number c , its logarithms using two different bases, a and b , are related by the expression



<SPAN role=presentation tabIndex=0 id=MathJax-Element-31-Frame class=MathJax style="POSITION: relative" data-mathml='logac=(logab) ×logbc’>log??=(loga?) ×log??logac=(logab) ×logbc
logac=(logab) ×logbc


This rule is the result of combining the definition of a logarithm, c = b log b c , with the rule for taking the logarithm of an exponential ( Equation AB.9 ).


The rules of logarithms are summarized in Box AppB.2 .



BOX AppB.2

Properties of Logarithms



<SPAN role=presentation tabIndex=0 id=MathJax-Element-32-Frame class=MathJax style="POSITION: relative" data-mathml='loga(b×c)=logab+logaclogabc=logab-logaclogacm=m×logac’>log?(?×?)=log??+log??log???=log??log??log???=?×log??loga(b×c)=logab+logaclogabc=logab-logaclogacm=m×logac
loga(b×c)=logab+logaclogabc=logab-logaclogacm=m×logac

<SPAN role=presentation tabIndex=0 id=MathJax-Element-33-Frame class=MathJax style="POSITION: relative" data-mathml='loga1c=-logaclogac=(logab)×logbc’>log?1?=-log??log??=(log??)×log??loga1c=-logaclogac=(logab)×logbc
loga1c=-logaclogac=(logab)×logbc





Solving quadratic equations


Any equation that can be put into the form



<SPAN role=presentation tabIndex=0 id=MathJax-Element-34-Frame class=MathJax style="POSITION: relative" data-mathml='ax2+bx+c= 0′>??2+??+?= 0ax2+bx+c= 0
ax2+bx+c= 0


can be solved for the value of x through the quadratic formula :



<SPAN role=presentation tabIndex=0 id=MathJax-Element-35-Frame class=MathJax style="POSITION: relative" data-mathml='x=-b±b2- 4ac2a’>?=?±?2– 4??2?x=-b±b2- 4ac2a
x=-b±b2- 4ac2a


The ± sign means that x may take on two values, one corresponding to using the + sign in the numerator, and one corresponding to using the – sign. Solving the quadratic equation 4 x 2 – 9 = 0 illustrates the technique. Comparing this equation with the general form of the quadratic equation given previously shows the coefficients to be a = 4, b = 0, and c = –9. The two solutions that satisfy the equation are:


<SPAN role=presentation tabIndex=0 id=MathJax-Element-36-Frame class=MathJax style="POSITION: relative" data-mathml='x=-(0)±02- 4(4)( – 9)2(4)=±1448=±128=±32′>?=(0)±02– 4(4)( – 9)2(4)=±1448=±128=±32x=-(0)±02- 4(4)( – 9)2(4)=±1448=±128=±32
x=-(0)±02- 4(4)( – 9)2(4)=±1448=±128=±32


The two answers are x = 3/2 and x = –3/2.




Differentiation and derivatives


The slope of a graph and the derivative


The derivative of a function at a particular point is most easily viewed as the slope of the function at that point. A slope is just a rate of change of one variable relative to another. For a straight line, the slope, being the rate of change of y relative to xyx ), is everywhere the same and is easy to compute ( Fig. AppB.1 ).




Fig. AppB.1


Graph of a line represented by the equation y = mx + b . The slope of the line is m , and b is the y -intercept; that is, the line crosses the y -axis at the point (0, b ). If two points lying on the line, ( x 1 , y 1 ) and ( x 2 , y 2 ), are known, the slope can be calculated: m = Δ y x = ( y 2 y 1 )/( x 2 x 1 ).


For functions with nonlinear graphs, determining the slope at a particular point on the curve is slightly more involved. Assume, for the moment, that the function is the parabola, y = x 2 , with the graph shown in Fig. AppB.2 . The problem of finding the slope of the parabola at any particular point ( x , y ) is to find the slope of the tangent line that just touches the parabola at that point.




Fig. AppB.2


Graph of the parabola y = x 2 . Two lines are also shown: (1) a tangent line ( solid ) that passes through one point, ( x , y ) , on the parabola; and (2) a test line ( dashed ) that passes through ( x , y ) , as well as a second point, ( x x , y+Δ y ). As Δ x (and thus Δ y ) grows ever smaller, the test line approaches the tangent line ever more closely. In the limit of infinitesimally small Δ x (i.e., as Δ x → 0), the test line becomes the tangent line. At the point ( x , y ) the derivative of the function has a value equal to the slope of the tangent line.


The general features of the problem are shown in Fig. AppB.2 . To determine the slope of the parabola at the point ( x , y ), we can first pass a test line through a second, arbitrary point ( xx , yy ). As Δ x becomes smaller, so will Δ y , and eventually the test line should become the tangent line as Δ x approaches 0. Because the points ( x , y ) and ( xx , yy ) both lie on the parabola y = x 2 , we can write the two points as ( x , x 2 ) and ( xx , ( xx ) 2 ), respectively. The slope of the line connecting these points is


<SPAN role=presentation tabIndex=0 id=MathJax-Element-37-Frame class=MathJax style="POSITION: relative" data-mathml='ΔyΔx=y2-y1x2-x1=(x+Δx)2-x2(x+Δx) -x=(x2+2xΔx+Δx2) -x2Δx=2x+Δx’>Δ?Δ?=?2?1?2?1=(??)2?2(??) –?=(?2+2?Δ??2) –?2Δ?=2??ΔyΔx=y2-y1x2-x1=(x+Δx)2-x2(x+Δx) -x=(x2+2xΔx+Δx2) -x2Δx=2x+Δx
ΔyΔx=y2-y1x2-x1=(x+Δx)2-x2(x+Δx) -x=(x2+2xΔx+Δx2) -x2Δx=2x+Δx

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Apr 18, 2020 | Posted by in NEUROLOGY | Comments Off on A mathematical refresher

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