- Level: AS and A Level
- Subject: Maths
- Word count: 3122
Repeated Differentiation
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Introduction
Repeated Differentiation I am going to investigate what effect differentiation has on several different functions of x and try to find general rules for the nth differential for each function. I'm interested in investigating this particular topic as I enjoy differentiation and integration in the A-Level class work I do and would like to take it further to see I can find any patterns and hopefully find general formulae for nth differentials. Repeated differentiation to A-Level standard is really only used up to the second differential of a function, as this gives the turning points on the graph of the function. In this piece of coursework I will look into what happens when functions are differentiated n times and the results achieved when different functions are multiplied together. I am going to first study repeated differentiation on a simple function of x, for example a.xb , and then look at increasingly more complicated functions including sine and cosine. I could use the " Differentiation of products " technique to incorporate two different functions and I could may be find a way of using the " R, ??formula " and looking to see if there is a rule for the nth differential of a function of x in that form. I could prove any general rules I might find using the method known as "Proof by induction". Notation in use throughout coursework: yI = dy/dx yII = d2y/dx2 yIII = d3y/dx3 ... etc. yn = dny/dxn nCr = n! / [ r! (n-r)! ] In general, unless otherwise stated, for an equation to be true both n and r must be a positive non-zero integers where n is the differential number and r is the term number. The simplest function of x to differentiate is xb . This is an example of what happens to such a function upon differentiation... start: y = x11 yI = 11x10 = 11x10 yII = 11.10x9 = 110x9 yIII = 11.10.9x8 = 990x8 yIV = 11.10.9.8x7 = 7920x7 yV = 11.10.9.8.7x6 = 55440x6 ... ...read more.
Middle
= cos(x + ?/2 ) equation correct for n=1 (see previous work) [2] Assume true for n=k yk = cos(x + k.?/2?? [3] Differentiate both sides w.r.t. x yk+1 = -sin(x + k.?/2 ) = cos(x + (k+1)??/2 ) Therefore if equation true for n=k also true for n=(k+1). Since true for n=1, then by Mathematical Induction also true for all positive integer values of n. Differentiation of products centres on a very simple equation: dy/dx = uv' + vu' where y = v.u . One of the simplest examples of product differentiation is y = ex .sin(x), (because ex isn't changed by differentiation). For example... start: y = ex sin(x) yI = ex (sin(x) + cos(x) ) = ex (sin(x) + sin(x + ?/2 ) ) yII = ex (sin(x) + sin(x + ?/2 ) ) + ex (cos(x) + cos(x +?/2 ) ) = ex (sin(x) + 2.sin(x + ?/2 ) + sin(x + 2?/2 ) ) yIII = ex (sin(x) + 2.sin(x + ?/2 ) + sin(x + 2?/2 ) ) + ex (cos(x) + 2.cos(x + ?/2 ) + cos(x + 2?/2 ) ) = ex (sin(x) + 3.sin(x + ?/2 ) + 3.sin(x + 2?/2 ) + sin(x + 3?/2 ) ) As you can see I have simplified each term in cos into a term in sin and grouped all of the same terms together. The multiplying factors of each term belong to a pattern know as Pascal's Triangle: 1 1,2,1 1,3,3,1 1,4,6,4,1 1,5,10,5,1 ...etc. The numbers in this pattern can be written algebraically as nCr . For example in the 5th row the 3rd number is 5C3 = 5! / (3! 2!) = 10 . The rest of the general solution is easily derived using the fact that ex differentiates to ex (ie. it isn't changed by differentiation), as well as the rule obtained above from a simplified version of the formula for the nth differential of sin(bx) ...read more.
Conclusion
dn ( axb ) = a.b! x(b-n) dxn (b-n)! dn ( sin(bx) ) = bn sin(bx + n?/2 ) dxn dn ( ex sin(x) ) = ex ( nC0 sin(x) + nC1 sin(x + ?/2 ) + ... + nCr-1 sin (x + (r-1).?/2?????????nCn sin (x + n.?/2?? dxn dn ( eax sin(bx) ) = eax ( an sin(bx) + nC1 a(n-1) b.sin(bx + ?/2 ) + ... + nC(r-1) an-(r-1) br-1 sin(bx + (r-1).?/2 ) + ... dxn + bn sin(bx + n.?/2?) ) dn ( eax (sin(bx) + cos(bx) ) = ?2.eax ( an sin(bx + ????) + nC1 a(n-1) b.sin(bx + 3?/4 ) + ... dxn + nC(r-1) an-(r-1) br-1 sin(bx + (r-1).?/2 + ?/4 ) + ... + bn sin(bx + n.?/2 + ?/4?) ) Further Investigation Using the rules above I could investigate repeated differentiation further. I could may be try to find a rule for the nth differential of functions like eax (b.sin(x) ??d.cos(x) ) possibly by using the " R, ??formula ". I could investigate more simple functions of the form ( f(x) )-1 and see what happens to them. Or I could even bring in other trigonometric functions like sec(x) , cosec(x) or cot(x) and try and find some sort of general rule for those too. A general formula for this last example could be a combination of a rule for the nth differential of ( f(x)-1 ) and the rule for the nth differential of cosine or sine, so this is what I'd try first after finding a rule for inverse functions. I could also look into the differentiation of functions like sin-1(x) - which, despite it immediately differentiating away from a function that includes a trigonometric function like sin-1(x) or even sin(x) , there might possibly be some sort of pattern which could be interesting to investigate. Other examples of this are ln(f(x)) or tan-1(f(x)) . ?? ?? ?? ?? Repeated Differentiation - Winter 2001 Page 1 of 13 Ben Leavett 13E ...read more.
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