Exercise 2.10: Fourier series of a triangle waveConsider the Fourier sine series approximation for the triangle wave depicted in Fig-ure 2.32.MfM (x) = an sin(nnx)хе [0, 11n=1(a) Find the coefficients an, n = 1,2,... To save time you may find the following 1integral formulas useful%3D(mx + b) sin(nnx)dx =тx + bsin(nnx)cos(nnx) +(nT)2sin(nnx) sin(Mnx)dx =Snm.2n, m = 1,2, ...2.10 Exercises2230 ..+2а1Figure 2.32: Triangle wave on [0, 1].a Plot the function fM (x) for M 5,10, 50 with parameter a 0.1 to demonstrateconvergence to f(x). How many terms are required to obtain good accuracy?

Hi! You have posted multiple subparts. We will be solving only the first part. If you need an answer to the other then kindly repost the question by specifying it. For (a), First, we will find the equation of the given triangular figure. After including the points and their names the figure looks like the following. The function easily can be found by finding the line joining Q & R and R & S.The line passing through Q and R is y-0x-b=1-0b+a-b=>y=1ax-b. Similarly, the line passing through R and S is y-0x-(b+2a)=1-0b+a-(b+2a)=>y=-1ax-(b+2a). Thus, the function looks like, f(x)=0,0≤x≤b1ax-b,b≤x≤b+a-1ax-(b+2a),b+a≤x≤b+2a0b+2a≤x≤1.As per the question, fM(x)=∑n=1Man sin(nπx) for x∈[0, 1]. Now, we need to find the coefficients an, n=1, 2, 3, ... . So, basically, we are doing the odd extension of the function and finding out the coefficients. Here, an=21∫01f(x) sin(nπx)dx. Thus, an=21∫01f(x) sin(nπx)dx =2∫0b0 sin(nπx)dx+∫bb+a1a(x-b) sin(nπx)dx+∫b+ab+2a-1a(x-(b+2a)) sin(nπx)dx+∫b+2a10 sin(nπx)dx =2∫bb+a1a(x-b) sin(nπx)dx+∫b+ab+2a-1a(x-(b+2a)) sin(nπx)dx Now, ∫bb+a1ax-bsinnπxdx=1a·∫bb+ax-bsinnπxdx=1a·∫bb+axsinπnx-bsinπnxdx=1a∫bb+axsinπnxdx-∫bb+absinπnxdx=1a-b1πncosπnb+a-a1πncosπnb+a+1π2n2sinπnb+a+b1πncosπbn-1π2n2sinπbn-b1πn-cosπnb+a+cosπbn=-1πncosπna+b-1π2an2sinπbn+1π2an2sinπna+bAnd ∫b+ab+2a-1ax-b+2asinnπxdx=-1a·∫b+ab+2ax-b+2asinnπxdx=-1a·∫b+ab+2axsinπnx-bsinπnx-2asinπnxdx=-1a∫b+ab+2axsinπnxdx-∫b+ab+2absinπnxdx-∫b+ab+2a2asinπnxdx=-1a-b1πncosπnb+2a-a2πncosπnb+2a+1π2n2sinπnb+2a+b1πncosπnb+a+a1πncosπnb+a-1π2n2sinπnb+a-b1πn-cosπnb+2a+cosπnb+a-a2πn-cosπnb+2a+cosπnb+a=-1πncosπna+b+1π2an2sinπna+b+2πncosπna+b-1π2an2sinπn2a+b Thus, an=2∫bb+a1a(x-b) sin(nπx)dx+∫b+ab+2a-1a(x-(b+2a)) sin(nπx)dx =2-1πncosπna+b-1π2an2sinπbn+1π2an2sinπna+b+-1πncosπna+b+1π2an2sinπna+b+2πncosπna+b-1π2an2sinπn2a+b =2-sinπnb+2sinπnb+a-sinπnb+2aπ2an2 Answer: So, an=2-sinπnb+2sinπnb+a-sinπnb+2aπ2an2.