# The Essay Examples

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## Problem Solving

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The points of inflections are (1, 1/√e) and (-1, 1/√e) b).f(x)=x+2sinxfirst derivative =1+2cos x Second derivative =2sinxGetting the local maxima and minima we equate the first derivative to zero 1+2cos x=0cos x=-12x=2π3, 4π3 evaluating the second derivative at x=2π3 and x=4π3-2sin2π3 =-2sin32=-√3. Therefore x=2π3 is a maximum value -2sin4π3 =-2sin-32=√3And x=4π3 minimum C. f(x)=x13(x+4)The first derivative =x13ddx(x+4)+x+4ddxx13=x13+(x+4)(13x23)=x13+(x3x23+43x23)f'(x)=43x23+4x133Using the product rule to get the second derivative =43ddxx-23+x13+x-23+x13ddx43=43(-23x-53+13x-23)f''x=-89x53+49x23d. The intervals on which the function increases or decreases and local...

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## HOME WORK

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The length of a 12-foot by 8-foot rectangle is increasing at a rate of 3 feet per second and the width is decreasing at 2 feet per second (see figure below). How fast is the perimeter changing? Solution Given dl/dt=3 And dw/dt=-2 Then dp when l=12 ft and w=8ft Perimeter= 2l+2w Dp/dt=d/dt{2l+2w} =2dl/dt+2dw/dt=2(3)+2(-2) =2 ft/sec How fast is the area changing? Solution Given dl/dt=3, dw/dt=-2 Da/dt when l=12 ft , w=8ft Area=lxwDa/dt=d/dt{lxw} =Dl/dt*w+dw/dt*l =3(8)+ 12(-2) =0 ft2/sec (c) Find all critical points and local extremes of the following function on the given intervals. f (x) = 2-x3 on Solution To find the critical points, take...

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## 983176648 test 3 revised

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the function is given below Fill in the following table with approximate values (if they exist) 2 4 6 8 10 12 14 1 0 1 (b) For the function find and an equation of the tangent line to the graph of the above function at In the questions c to h , find (C ) y = x3 – x2 – x5/3 = 3x2 – 2x – 5/3x (d) y=x3 ln(1+x2) You may use (e) y'= 2+cosx+ ddxsinx-sinx. ddx(2+cosx)(2+cosx)2y'= 2+cosx.cosx-sinx. (-sin⁡(x))(2+cosx)2y'= 2+cosx+cos2 x+sin2(x)(2+cosx)2cos2 x+sin2x=1sin2x=1- cos2 xy'= 2+cosx+cos2 x+1- cos2 x(2+cosx)2y'= 1+ 2cosx(2+cosx)2(f) y=sin-1(ex)siny=(ex)cosyy'=(ex)y'=excosyy=ex1-e2x(g) Find if i.e. = ddxsin3(2x)=3sin22x.cos2x.2(2) (a) Find an equation...

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## quantitative finance

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The expected value of the unhedged firm is: 0.5*200/(1+0.05)+0.5*(500-(500-300)*0.4)/(1+0.05)=95.24+200=295.24 Problem (b) Let’s buy put options for gold pricing at 500 with premium P. The expected value of the hedged firm is: (500-(500-300)*0.4)/(1+0.05)-P=420-P Problem (c) The safe debt is the one that pays off not more than 200 in a year. The value of the firm is:0.5*(200-200)/(1+0.05)+0.5*(500-200-(500-200-300)*0.4)/(1+0.05)=142.86 Problem (d) The safe debt is the one that pays off not more than 200 in a year. Let’s buy put options for gold pricing at 500 with premium P. The value of the firm is: (500-200-(500-200-300)*0.4)/(1+0.05)-P=285.71-P Problem (e) The firm’s value...

• Words: 275
• Pages: 1

## Mathematical Finance

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the function V (t, z) lim V( t, z ) / (t, z) = ∞ is said to is said to be decreasing in convex and in z if and only if V < 1. Q.3 V(t,x,y) = dXt = rXt dt + t(µ-r) + tdWt – Ct dt + Yt dtdXt = rXt dt + t(µ-r) + tdWt – Ct dt + Yt dt = stochastic differential equation let f(t, x, y) = logeXft = 0, fx = 1/x, fxx = -1/x2 df = 0 dt + ½ dx - 1/2x * 1/2 (dx)2 = 1/x (dXt = rXt dt + t(µ-r) + tdWt – Ct dt + Yt dt) – 1/2x * ½ (2x2 dt) HJB = V( x,t,y) / t = ½ ydW2t + rXt –2 (V(x, t, y)/ x) + 2/2 * 2V(x, t, y) /...

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## Mathematical Finance

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the time limit Such that Rearranginging Deviding by h Taking limits as h0 s.t The economic interpretation of g(t)y is labor income at the time, t. Part 2 2. 1. Optimal Trading Strategy Wealth process dXtπ=πtTμtdt+σtdWt+Xtπ-1nTπtX0=x0y=EβTI(βTST, where I is the optimal trading. where , and is uniquely determined by 2.2. Optimal Consumption in terms of X(t)+g(t)y= Such that,left952500 = Reference Bensoussan, A., & Zhang, Q. (2009). Mathematical modeling and numerical methods in finance. Amsterdam:...

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## Problem 2

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the MMs’ strategy is to not acquire information (pn = 0 for n = 1, 2). Argue that in this case, bn = 1/2 in equilibrium. Lack of value information implies that both players are bound to make a bid that is defined by the expectation of the market. In essence, the prospect of attaining bn-V would remain equally distributed among the two market markers. Arguably, the attribute of the trader stands dismissed along the implication of noise. Considering the input of the two market markers, n1 and n2, their ability to achieve bn when the value is unknown would be split evenly MMn1 = bn – V MMn2 = bn – V But V = 0 (since v is undisclosed) Thus, gains from bid would equate to bn. The probability...

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## problems

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the entries made into the books of accounts relating to manufacturing accounts. Various classifications of costs in the manufacturing sector have been used to record each cost item. P15-3A comprises of two cases in which manufacturing costs and selling expenses are listed. The paper seeks to show how the cost of goods manufactured is arrived at, the accompanying income statement and a list of selected current assets. Problem 14-5A looks into the computation of costs of goods manufactured presented in a schedule as well as the income statement for the financial year. Problem 15-4A also looks into the manufacture cost schedule and list of current assets for the accounting period ending 31st October...

• Words: 825
• Pages: 3